NCERT Exemplar Class 12th Maths Solutions Chapter 11 Three Dimensional Geometry

# NCERT Exemplar Class 12th Maths Solutions Chapter 11 Three Dimensional Geometry

Edited By Ravindra Pindel | Updated on Sep 15, 2022 05:16 PM IST | #CBSE Class 12th

NCERT exemplar Class 12 Maths solutions chapter 11 Three Dimensional Geometry - If you observe the world around you, you see everything in three dimensions! Even a tiny strand of hair has dimensions of length, width, and depth. 3D geometry refers to the mathematics of perception, direction, and shape. NCERT exemplar Class 12 Maths chapter 11 solutions initiates that there is always a requirement of three parameters to work with the comprehensive concepts of three-dimensional geometry. NCERT exemplar solutions For Class 12 Maths chapter 11 would help you out in any three-dimensional Mathematical problem that you're stuck on and get you back in your study loop. NCERT exemplar Class 12 Maths solutions chapter 11 PDF download is helpful for students to learn offline when there is a slow internet connection. Also, read - NCERT Class 12 Maths Solutions

Question:1

Find the position vector of a point A in space, so that $\overline{OA}$ is inclined at 60? to $\overline{OX}$ and 45? to $\overline{OY}$ and $\left |\overline{OA} \right |$= 10 units.

Given, $\overline{OA}$ is inclined at 600 to and at $\overline{OX}$ 450 to $\overline{OY}$
$\overline{OA}$ = 10 units.
We want to find the position vector of point A in space, which is nothing but $\overline{OA}$
We know, there are three axes in space: X, Y, and Z.
Let OA be inclined with OZ at an angle α.
We know, directions cosines are associated by the relation:
l² + m² + n² = 1 ….(i)
In this question, direction cosines are the cosines of the angles inclined by on $\overline{OA}$, $\overline{OX}$, $\overline{OY}$ and $\overline{OZ}$
So,$l=\cos 60^{\circ},m=\cos 45^{\circ},n=\cos \alpha$
Substituting the values of l, m, and n in equation (i),
$\left (\cos 60^{\circ} \right )^{2}+\left (m=\cos 45^{\circ} \right )^{2}+\left (n=\cos \alpha \right )^{2}=1$
We know the values of $\cos 60^{\circ}$and $\cos 45^{\circ}$, i.e. 1/2 and 1/√2 respectively.
Therefore, we get
$\left ( \frac{1}{2} \right )^{2}+\left ( \frac{1}{\sqrt{2}} \right )^{2}+\cos^{2}\alpha =1$
$\Rightarrow \frac{1}{4}+\frac{1}{2}+\cos^{2}\alpha =1$
$\Rightarrow \cos^{2}\alpha =1-\frac{1}{4}-\frac{1}{2}$
$\Rightarrow \cos^{2}\alpha =\frac{4-1-2}{4}$
$\Rightarrow \cos^{2}\alpha =\frac{1}{4}$
$\Rightarrow \cos \alpha =\pm \sqrt{\frac{1}{4}}$
$\Rightarrow \cos \alpha =\pm \frac{1}{2}$
So $\overrightarrow{OA}$ is given as
$\overrightarrow{OA}=\overrightarrow{OA}\left ( l\hat{i}+m\hat{j}+n\hat{k} \right )$..........(ii)
We have,
$l = \cos 60^{\circ} = \frac{1}{2}\\ m =\ cos 45^{\circ} = \frac{1}{\sqrt{2}}\\ n = \cos \alpha = \pm \frac{1}{2}\\$
Inserting these values of l, m and n in equation (ii),
$\overrightarrow{OA}=\left |\overrightarrow{OA} \right |\left ( \frac{1}{2} \hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{2}\hat{k} \right )$
Also ,Put $\Rightarrow \overrightarrow{OA}=10\left ( \frac{1}{2} \hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{2}\hat{k} \right )$
$\Rightarrow \overrightarrow{OA}=10\times \frac{1}{2} \hat{i}+10\times\frac{1}{\sqrt{2}}\hat{j}+10\times\frac{1}{2}\hat{k}$
$\Rightarrow \overrightarrow{OA}=5i+10\times\frac{\sqrt{2}}{\sqrt{2}}\times+10\times\frac{1}{\sqrt{2}}\hat{j}\pm 5\hat{k}$
$\Rightarrow \overrightarrow{OA}=5i+\frac{10\times\sqrt{2}}{2}\hat{j}\pm 5\hat{k}$
$\Rightarrow \overrightarrow{OA}=5i+5\sqrt{2}\hat{j}\pm 5\hat{k}$
Thus, position vector of A in space $=5i+5\sqrt{2}\hat{j}\pm 5\hat{k}$

Question:2

Find the vector equation of the line parallel to the vector $3\hat{i}-2 \hat{j}+3\hat{k}$ and which passes through the point (1, -2, 3).

Given, vector = $3\hat{i}-2\hat{j}+6\hat{k}$

Point = (1, -2, 3)

We can write this point in vector form as $\hat{i}-2\hat{j}+3\hat{k}$

Let ,

$\overrightarrow{a}=\hat{i}-2\hat{j}+3\hat{k}$

$\overrightarrow{b}=3\hat{i}-2\hat{j}+6\hat{k}$

We must find the vector equation of the line parallel to the vector $\overrightarrow{b}$ and passing through the point

We know, equation of $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$ a line passing through a point and parallel to a given vector is denoted as

Where, $\lambda \epsilon \mathbb{R}$

In other words, we need to find $\overrightarrow{r}$

This can be achieved by substituting the values of the vectors in the above equation. We get

$\overrightarrow{r}=\left ( \hat{i}-2\hat{j}+3\hat{k} \right )+\lambda \left ( 3\hat{i}-2\hat{j}+6\hat{k} \right )$

$\Rightarrow \overrightarrow{r}=\left ( \hat{i}-2\hat{j}+3\hat{k} \right )+\lambda \left ( 3\hat{i}-2\hat{j}+6\hat{k} \right )$

This can be further rearranged, upon which we get:

$\Rightarrow \vec{r}=\hat{i}-2\hat{j}+3\hat{k}+3\lambda \hat{j}+3\hat{k}+6\lambda\hat{k}$

$\Rightarrow \vec{r}=\hat{i}+3\lambda \hat{i}-2\hat{j}-2\lambda\hat{j}+3\hat{k}+6\lambda\hat{k}$

$\Rightarrow \vec{r}=\left ( 1-3\lambda \right )\hat{i}+\left ( -2-2\lambda \right )\hat{j}+\left ( 3+6\lambda \right )\hat{k}$

Thus, the require vector equation of line is $\vec{r}=\left ( \hat{i}-2\hat{j}+3\hat{k} \right )+\lambda\left ( 3\hat{i}-2\hat{j}+6\hat{k} \right )$

which can also be written as $\left ( 1-3\lambda \right )i+\left ( -2-2 \lambda \right )\hat{j}+\left ( 3+6\lambda \right )\hat{k}$

Question:3

Show that the given lines, $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-4}{5}=\frac{y-1}{2}=z$ intersect.
Also, find the point of intersection of the lines.

We have the lines,
$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$
$\frac{x-4}{5}=\frac{y-1}{2}=z$
Let us denote these lines as L1and L2, such that
$L_{1}:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda$
$L_{2}=\frac{x-4}{5}=\frac{y-1}{2}=z=\mu$
where $\lambda ,\mu \epsilon \mathbb{R}$
We must show that the lines L1and L2 intersect.
To show this, let us first find any point on line L1 and line L2
For L1:
$L_{1}:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda$
$\Rightarrow \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda$
$\Rightarrow \frac{x-1}{2}=\lambda,\frac{y-2}{3}=\lambda,\frac{z-3}{4}=\lambda$
We must find the values of x, y, and z. Therefore, let us take $\Rightarrow \frac{x-1}{2}=\lambda$
$\Rightarrow x-1=2\lambda$
$\Rightarrow x=2\lambda+1$
Take $\frac{y-2}{3}=\lambda$
$\Rightarrow y-2=3\lambda$
$\Rightarrow y=3 \lambda+2$
Take $\frac{z-3}{4}=\lambda$
$\Rightarrow z-3=4\lambda$
$\Rightarrow z=4\lambda+3...(i)$
Therefore, any point on L1 can be represented as $(2\lambda + 1, 3\lambda + 2, 4\lambda + 3)$.
Now,
For L2:
$L_{2}=\frac{x-4}{5}=\frac{y-1}{2}=z=\mu$
$\Rightarrow \frac{x-4}{5}=\frac{y-1}{2}=z=\mu$
$\Rightarrow \frac{x-4}{5}=\mu,\frac{y-1}{2}=\mu,z=\mu$
We must find the values of x, y, and z. Therefore,
Take $\frac{x-4}{5}=\mu$
$\Rightarrow x-4=5\mu$
$\Rightarrow x=5\mu+4$
Take $\frac{y-1}{2}=\mu$
$\Rightarrow y-1=2\mu$
$\Rightarrow y=2\mu+1$
Take $z=\mu$
$\Rightarrow z=\mu........(ii)$
Hence, any point on line L? can be represented as (5μ + 4, 2μ + 1, μ).
If lines L1 and L2 intersect, then there exist λ and μ such that
$\left ( 2\lambda+1,3\lambda+3,4\lambda+3 \right )\equiv \left ( 5\mu+4,2\mu+1,\mu \right )$
$\Rightarrow 2\lambda+1= 5\mu+4......(iii)$
$3\lambda+2=2\mu+1.....(iv)$
$4\lambda+3=\mu.....(iv)$
Substituting the value of μ from equation (v) into equation (iv),
$3\lambda+2=2\left ( 4\lambda+3 \right )+1$
$\Rightarrow 3\lambda+2=8\lambda+6+1$
$\Rightarrow 3\lambda+2=8\lambda+7$
$\Rightarrow 8\lambda-3\lambda=2-7$
$\Rightarrow 5\lambda=-5$
$\Rightarrow \lambda=-\frac{5}{5}$
$\Rightarrow \lambda=-1$
Putting this value of $\lambda$ in eq (v),
$4\left ( -1 \right )+3=\mu$
$\Rightarrow \mu=-4+3$
$\Rightarrow \mu=-1$
To check, we can substitute the values of $\lambda$ and $\mu$ in equation (iii), giving us:
$2(-1) + 1 = 5(-1) + 4$
$\Rightarrow -2 + 1 = -5 + 4$
$\Rightarrow -1 = -1$
Therefore $\lambda$ and $\mu$ also satisfy equation (iii).
So, the z-coordinate from equation (i),
$z=4\lambda +3$
$\Rightarrow z=4\left ( -1 \right )+3 \left [ \because \lambda=-1 \right ]$
$\Rightarrow z=-4+3$
$\Rightarrow z=-1$
And the z-coordinate from equation (ii),
$z=\mu$
$z=-1\left [ \because \mu=-1 \right ]$
So, the lines intersect at the point
$(5\mu + 4, 2\mu + 1, \mu) = (5(-1) + 4, 2(-1) + 1, -1).\\ Or, (5\mu + 4, 2\mu + 1, \mu) = (-5 + 4, -2 + 1, -1)\\ Or (5\mu + 4, 2\mu + 1, \mu) = (-1, -1, -1)$
Therefore the lines intersect at the point (-1, -1, -1).

Question:4

Given, lines:
$\vec{r}=3\hat{i}-2\hat{j}+6\hat{k}+\lambda\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )$
$\vec{r}=\left (2\hat{i}-5\hat{k} \right )+\mu\left ( 6\hat{i}+3\hat{j}+2\hat{k} \right )$
We are instructed to find the angle between the lines.
The line $\vec{r}=3\hat{i}-2\hat{j}+6\hat{k}+\lambda\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )$ is parallel to the vector
$2\hat{i}+\hat{j}+2\hat{k}$
Let
$\vec{b_{1}}=2\hat{i}+\hat{j}+2\hat{k}$

Then, we can say the line $\vec{r}=3\hat{i}-2\hat{j}+6\hat{k}+\lambda\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )$ is parallel to vector $\vec{b_{1}}=2\hat{i}+\hat{j}+2\hat{k}$
Similarly, let $\vec{b_{2}}=6\hat{i}+3\hat{j}+2\hat{k}$
Then, we can say is $\vec{r}=2\hat{j}-5\hat{k}+\mu \left ( 6\hat{i}+3\hat{j}+2\hat{k} \right )$ parallel to the vector $\vec{b_{2}}=6\hat{i}+3\hat{j}+2\hat{k}$
If we take θ as the angle between the lines, then cosine θ is:
$\cos \theta = \frac{\vec{b_{1}}\vec{b_{2}}}{\left |\vec{b_{1}} \right |\left |\vec{b_{2}} \right |}$
Substituting the values of $\vec{b_{1}}=2\hat{i}+\hat{j}+2\hat{k}$ and $\vec{b_{2}}=6\hat{i}+3\hat{j}+2\hat{k}$ in the above equation,
We get
$\cos \theta=\frac{\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )\left ( 6\hat{i}+3\hat{j}+2\hat{k} \right )}{\left | 2\hat{i}+\hat{j}+2\hat{k} \right |\left | 6\hat{i}+3\hat{j}+2\hat{k} \right |}$
Here,
$\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )\left ( 6\hat{i}+3\hat{j}+2\hat{k} \right )=\left ( 2 \times 6 \right )+\left ( 1 \times 3 \right )+\left ( 2 \times 2 \right )$
$\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )\left ( 6\hat{i}+3\hat{j}+2\hat{k} \right )=12+3+4$
$\left ( 2\hat{i}+\hat{j}+2\hat{k} \right )\left ( 6\hat{i}+3\hat{j}+2\hat{k} \right )=19...........(i)$
Also,
$\left |2\hat{i}+\hat{j}+2\hat{k} \right |\left | 6\hat{i}+3\hat{j}+2\hat{k} \right |=\sqrt{2^{2}+1^{2}+2^{2}}\sqrt{6^{2}+3^{2}+2^{2}}$
$\left |2\hat{i}+\hat{j}+2\hat{k} \right |\left | 6\hat{i}+3\hat{j}+2\hat{k} \right |=\sqrt{4+1+4}\sqrt{36+9+4}$
$\left |2\hat{i}+\hat{j}+2\hat{k} \right |\left | 6\hat{i}+3\hat{j}+2\hat{k} \right |=\sqrt{9}\sqrt{49}$
$\left |2\hat{i}+\hat{j}+2\hat{k} \right |\left | 6\hat{i}+3\hat{j}+2\hat{k} \right |=3 \times 7$
$\left |2\hat{i}+\hat{j}+2\hat{k} \right |\left | 6\hat{i}+3\hat{j}+2\hat{k} \right |=21............(ii)$
Substituting the values of $\cos \theta$ in equation (i) and (ii), we get
$\cos \theta=\frac{19}{21}$
$\Rightarrow \theta=\cos^{-1}\left (\frac{19}{21} \right )$
Therefore, the angle between the lines is $\cos^{-1}\left (\frac{19}{21} \right )$

Question:5

Prove that the line through points A (0, -1, -1) and B (4, 5, 1) intersects the line through C (3, 9, 4 ) and D (-4, 4, 4).

Given: A (0, -1, -1), B (4, 5, 1), C (3, 9, 4), D (-4, 4, 4).
To prove: The line passing through A and B intersects the line passing through C and D.
Proof: We know, equation of a line passing through two points (x1 , y1 , z1) and (x2 , y2 , z2) is:
$\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$
Hence, the equation of the line passing through A (0, -1, -1) and B (4, 5,1) is:
$\frac{x-0}{4-0}=\frac{y-(-1)}{5-(-1)}=\frac{z-(-1)}{1-(-1)}$
, where x1 = 0, y1 = -1, z1 = -1; and x2 = 4, y2 = 5, z2 = 1
$\Rightarrow \frac{x-0}{4}=\frac{y+1}{6}=\frac{z+1}{2}\\ \Rightarrow \frac{x}{4}=\frac{y+1}{6}=\frac{z+1}{2}$
Let
$L_{1}: \frac{x}{4}=\frac{y+1}{6}=\frac{z+1}{2}=\lambda\\ \\ \frac{x}{4}=\lambda, \frac{y+1}{6}=\lambda,\frac{z+1}{2}=\lambda\\$
We must find the values of x, y, and z. Therefore,
$Take \frac{x}{4}=\lambda\\ \Rightarrow x=4\lambda\\ \\ Take \frac{y+1}{6}=\lambda\\ \Rightarrow y+1=6\lambda\\ \Rightarrow y=6\lambda-1\\ \\ Take \frac{z+1}{2}=\lambda\\ \Rightarrow z+1=2\lambda\\ \Rightarrow z=2\lambda-1\\ \\$
This implies that any point on the line L1 is (4λ, 6λ – 1, 2λ – 1).
The equation of the line passing through points C (3, 9, 4) and D (-4, 4, 4) is:
$\frac{x-3}{-4-3}=\frac{y-9}{4-9}=\frac{z-4}{4-4}$
, where x1 = 3, y1 = 9, z1 = 4; and x2 = -4, y2 = 4, z2 = 4
$\frac{x-3}{-7}=\frac{y-9}{-5}=\frac{z-4}{0}$
Let
$L_{2}:\frac{x-3}{-7}=\frac{\left (y-9 \right )}{-5}=\frac{z-4}{0}=\mu$
$\Rightarrow \frac{x-3}{-7}=\mu,\frac{\left (y-9 \right )}{-5}=\mu,\frac{z-4}{0}=\mu$
We must find the values of x, y, and z. Therefore,
$Take \, \frac{x-3}{-7}=\mu\\ \Rightarrow x-3=-7\mu\\ \Rightarrow x=-7\mu+3\\ \\ Take \,\frac{\left (y-9 \right )}{-5}=\mu \\ \Rightarrow y-9=-5\mu\\ \Rightarrow y=-5\mu+9\\ \\ Take\frac{z-4}{0}=\mu \\ \Rightarrow z-4=0\\ \Rightarrow z=4$
This implies that any point on line L2 is (-7μ +3, -5μ + 9, 4).
If the lines intersect, then there must exist a value of λ and for μ, for which
$\left (4\lambda, 6\lambda - 1, 2\lambda -1\right) \equiv \left(-7\mu + 3, -5\mu + 9, 4\right ) \\ \Rightarrow 4\lambda = -7\mu + 3...(i)\\ 6\lambda - 1 = -5\mu + 9..(ii)\\ 2\lambda - 1 = 4 %u2026(iii)\\$
From equation (iii), we get
$2\lambda - 1 = 4 \\ \Rightarrow 2\lambda=4+1\\ \Rightarrow 2\lambda=5\\ \Rightarrow \lambda=\frac{5}{2}$
Substituting the value of λ in equation (i),
$4\left ( \frac{5}{2} \right )=-7\mu+3\\ \Rightarrow 2 \times 5=-7\mu+3\\ \Rightarrow 10=-7\mu+3\\ \Rightarrow 7\mu=3-10\\ \Rightarrow 7\mu=-7 \\ \Rightarrow -\frac{7}{7}\\ \Rightarrow \mu=-1$
Substituting these values of λ and μ in equation (ii),
$6\left ( \frac{5}{2} \right )-1=-5\left ( -1 \right )+9\\ \Rightarrow 3 \times 5 - 1 = 5 + 9\\ \Rightarrow 15 - 1 = 14\\ \Rightarrow 14 = 14$
Since the values of λ and μ satisfy eq (ii), the lines intersect.
Hence, proved that the line through A and B intersects the line through C and D.

Question:6

Prove that lines x = py + q, z = ry + s, and x = p’y + q’, z = r’y + s’ are perpendicular if pp’ + rr’ + 1 = 0.

Given: x = py + q, z = ry + s, and x = p’y + q’, z = r’y + s’ are perpendicular.
To Prove: pp’ + rr’ + 1 = 0.
Proof:
Let us take x = py + q and z = ry + s.
From x = py + q;
py = x - q
$\Rightarrow y=\frac{x-q}{p}$
From z = ry + s;
ry = z - s
$\Rightarrow y=\frac{z-s}{r}$
So, $\frac{x-q}{p}=y=\frac{z-s}{r}$
$\frac{x-q}{p}=\frac{y}{1}=\frac{z-s}{r}$ Or, … (i)
Now, if we take x = p’y + q’ and z = r’y + s’
From x = p’y + q’;
p’y = x - q’
$\Rightarrow y=\frac{x-{q}'}{{p}'}$
From z = r’y + s’;
r’y = z - s’
$\Rightarrow y=\frac{z-{s}'}{{r}'}$
So,
$\frac{x-{q}'}{{p}'}=y=\frac{z-{s}'}{{r}'}$
Or,
$L_{2}:\frac{x-{q}'}{{p}'}=\frac{y}{1}=\frac{z-{s}'}{{r}'}.......(ii)$
From (i),
Line L1 is parallel to $p\hat{i}+\hat{j}+r\hat{k}$ (from the denominators of the equation (i))
From (ii),
Line L2 is parallel to ${p}'\hat{i}+\hat{j}+{r}'\hat{k}$ (from the denominators of the equation (ii))
According to the question, L1 and L2 are perpendicular.
Therefore, the dot product of the vectors should equate to 0.
Or,
$\left (p\hat{i}+\hat{j}+r\hat{k} \right ).\left ({p}'\hat{i}+\hat{j}+{r}'\hat{k} \right )\\ \Rightarrow p{p}'+1+r{r}'=0$
(since, in vector dot product, $\left (x\hat{i}+y\hat{j}+z\hat{k} \right )\left ({x}'\hat{i}+{y}'\hat{j}+{z}'\hat{k} \right )= x{x}'+y{y}'+z{z}'=0$
Or,
$p{p}'+r{r}'+1=0$
Therefore, the lines are perpendicular if pp’ + rr’ + 1 = 0.

Question:7

Find the equation of a plane which bisects perpendicularly the line joining A (2, 3, 4) and B (4, 5, 8) at right angles.

Given, there exists a plane which perpendicularly bisects the line joining A (2, 3, 4) and B (4, 5, 8) at right angles. We must find the equation of this plane.
First, let us find the midpoint of AB.
Since the midpoint of any line is halfway between the two end points,
$Midpoint \: of \: AB=\left ( \frac{2+4}{2},\frac{3+5}{2},\frac{4+8}{2} \right )$
$Midpoint \: of \: AB=\left ( \frac{6}{2},\frac{8}{2},\frac{12}{2} \right )$
= (3, 4, 6).
We can represent this as a position vector, $\vec{a}=3\hat{i}+4\hat{j}+6\hat{k}$
Next, we must find the normal of the plane, $\vec{n}$
$\vec{n}=\left ( 4-2 \right )\hat{i}+\left ( 5-3 \right )\hat{j}+\left ( 8-4 \right )\hat{k}\\ \Rightarrow \vec{n}=2\hat{i}+2\hat{j}+4\hat{k}$
We know, the equation of the plane which perpendicularly bisects the line joining two given points is
$\left ( \vec{r}-\vec{a} \right )\vec{n}=0$
Where,
$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$
Substituting the values in the above equation,
$\left (\left (x\hat{i}+y\hat{j}+z\hat{k} \right ) -\left (3\hat{i}+4\hat{j}+6\hat{k} \right ) \right ).\left (2\hat{i}+2\hat{j}+4\hat{k} \right )=0\\ \Rightarrow \left ( x\hat{i}+y\hat{j}+z\hat{k}-3\hat{i}-4\hat{j}-6\hat{k} \right ).\left (2\hat{i}+2\hat{j}+4\hat{k} \right )=0\\\Rightarrow \left ( x\hat{i}-3\hat{i}+y\hat{j}-4\hat{j}+z\hat{k}-6\hat{k} \right ).\left (2\hat{i}+2\hat{j}+4\hat{k} \right )=0\\\Rightarrow \left ( (x-3)\hat{i}+(y-4)\hat{j}+(z-6)\hat{k} \right ).\left (2\hat{i}+2\hat{j}+4\hat{k} \right )=0\\\Rightarrow 2\left ( x-3 \right )+2(y-4)+4(z-6)=0$
Upon further simplification,
$\Rightarrow 2x-6+2y-8+4z-24=0\\ \Rightarrow 2x+2y+4z-6-8-24=0\\ \Rightarrow 2x+2y+4z-38=0\\ \Rightarrow 2\left ( x+y+2z-19 \right )=0\\ \Rightarrow x+y+2z-19=0\\ \Rightarrow x+y+2z=19$
Therefore, the required equation of the plane is x + y + 2z = 19.

Question:8

Find the equation of a plane which is at a distance $3\sqrt{3}$ units from the origin and the normal to which is equally inclined to coordinate axes.

Given, the plane is at a distance of $3\sqrt{3}$ from the origin, and the normal is equally inclined to coordinate axes.
We need to find the equation of this plane.
We know, the vector equation of a plane located at a distance d from the origin is represented by:
$\vec{r}.\hat{n}=d\\ \Rightarrow \left ( x\hat{i}+y\hat{j}+z\hat{k} \right ).\left ( l\hat{i}+m\hat{j}+n\hat{k} \right )=d$
lx + my + nz = d ….(i) , where l, m and n are the direction cosines of the normal of the plane.
Since the normal is equally inclined to the coordinate axes,
$l = m = n \\ \cos \alpha =\cos\beta =\cos \gamma %u2026(ii)$
Also, we know,
$\cos^{2} \alpha =\cos^{2}\beta =\cos^{2} \gamma=1\\ \Rightarrow \cos^{2} \alpha =\cos^{2}\alpha =\cos^{2} \alpha =1 \: \: \left ( from(ii) \right )\\ \Rightarrow 3\cos^{2}\alpha=1\\ \Rightarrow \cos^{2}\alpha=\frac{1}{3}\\ \Rightarrow \cos \alpha =\frac{1}{\sqrt{3}}$
This means, $l=m=n =\frac{1}{\sqrt{3}}$
if we substitute the values of l, m and n in equation (i),
$\left (\frac{1}{\sqrt{3}} \right )x+\left (\frac{1}{\sqrt{3}} \right )y+\left (\frac{1}{\sqrt{3}} \right )z=d\: \: \left [where\: d=3\sqrt{3} \right ]$
So,
$\left (\frac{1}{\sqrt{3}} \right )x+\left (\frac{1}{\sqrt{3}} \right )y+\left (\frac{1}{\sqrt{3}} \right )z=3\sqrt{3} \\ \Rightarrow \frac{x+y+z}{\sqrt{3}}=3\sqrt{3}\\ \Rightarrow x+y+z=3\sqrt{3}\times \sqrt{3}\\ \Rightarrow x+y+z=3 \times 3=9$
Therefore, the required equation of the plane is x + y + z = 9.

Question:9

. If the line drawn from the point (-2, -1, -3) meets a plane at right angle at the point (1, -3, 3), find the equation of the plane.

Given: the line drawn from point (-2, -1, -3) meets a plane at 900 at the point (1, -3, 3). We must find the equation of the plane.
Any line perpendicular to the plane is the normal.
Let the points be P (-2, -1, -3) and Q (1, -3, 3), then the line PQ is a normal to the plane.
Hence, PQ = (1 + 2, -3 + 1, 3 + 3)=> PQ = (3, -2, 6)
=> Normal to the plane = $\vec{PQ}$
$\vec{PQ}=3\hat{i}-2\hat{j}+6\hat{k}$
The vector equation of a plane is represented by $\left (\vec{r}-\vec{a} \right ).\vec{n}=0$
Putting the obtained values in this equation,
$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\\ \vec{a}=\hat{i}-3\hat{j}+3\hat{k}\\ \vec{n}=3\hat{i}-2\hat{j}+6\hat{k}$
We get,
$\Rightarrow \left (\left (x\hat{i}+y\hat{j}+z\hat{k} \right )-\left (\hat{i}-3\hat{j}+3\hat{k} \right ) \right ).\left ( 3\hat{i}-2\hat{j}+6\hat{k} \right )=0\\ \Rightarrow \left ( x\hat{i}-y\hat{j}+z\hat{k}-\hat{i}+3\hat{j}-3\hat{k} \right ).\left ( 3\hat{i}-2\hat{j}+6\hat{k} \right )=0\\ \Rightarrow \left ( x\hat{i}-\hat{i}+y\hat{j}+3\hat{j}+z\hat{k}-3\hat{k} \right ).\left (3\hat{i}-2\hat{j}+6\hat{k} \right )=0\\ \Rightarrow \left ((x-1)\hat{i}+(y+3)\hat{j}+(z-3)\hat{k} \right ).\left (3\hat{i}-2\hat{j}+6\hat{k} \right )=0$
$\Rightarrow 3(x - 1) + (-2)(y + 3) + 6(z - 3) = 0\\ \Rightarrow3(x - 1) - 2(y + 3) + 6(z - 3) = 0\\ \Rightarrow 3x- 3 -2y - 6 + 6z - 18 = 0\\ \Rightarrow 3x - 2y +6z - 3 - 6 - 18 = 0\\ \Rightarrow 3x - 2y + 6z - 9 - 18 = 0\\ \Rightarrow 3x - 2y + 6z - 27 = 0\\ \Rightarrow 3x - 2y + 6z = 27$
Therefore, the required equation of the plane is 3x - 2y + 6z = 27.

Question:10

Find the equation of the plane through the points (2, 1, 0), (3, -2, -2) and (3, 1, 7).

Given points are (2, 1, 0), (3, -2, -2), and (3, 1, 7).
We know, equation of a line passing through 3 non-collinear points (x1 , y1 , z1 ), (x2 , y2 , z2 ) and (x3 , y3 , z3 ) is given as:
$\begin{vmatrix} x-x_{1} &y-y_{1} &z-z_{1} \\ x_{2}-x_{1}&y_{2}-y_{1} &z_{2}-z_{1} \\ x_{3}-x_{1}&y_{3}-y_{1} &z_{3}-z_{1} \end{vmatrix}=0$
Where, (x1 , y1 , z1 ) = (2, 1, 0)
(x2 , y2 , z2 ) = (3, -2, -2)
(x3 , y3 , z3 ) = (3, 1, 7)
Therefore, x1 = 2, y1 = 1, z1 = 0; x2 = 3, y2 = -2, z2 = -2; x3 = 3, y3 = 1, z3 = 7
Substituting these values in the line equation,
$\begin{vmatrix} x-2 &y-1 &z-0 \\ 3-2&-2-1 &-2-0 \\ 3-2&1-1 &7-0 \end{vmatrix}=0\\ \\ \\ \begin{vmatrix} x-2 &y-1 &z-0 \\ 1&-3 &-2 \\ 1&0 &7 \end{vmatrix}=0$
$\\ \\ \begin{vmatrix} x-2 &y-1 &z-0 \\ 1&-3 &-2 \\ 1&0 &7 \end{vmatrix}=\left ( x-2 \right )\left ( \left ( -3 \times 7 \right )-\left ( -2 \times 0 \right ) \right )$

$\begin{vmatrix} x-2 & y-1 &z \\ 1& -3& -2\\ 1& 0& 7 \end{vmatrix}=\left ( x-2 \right )\left ( -21-0 \right )-\left ( y-1 \right )\left ( 7-(-2) \right )+z\left ( 0-(-3) \right )$
$\begin{vmatrix} x-2 & y-1 &z \\ 1& -3& -2\\ 1& 0& 7 \end{vmatrix}=\left ( x-2 \right )\left ( -21 \right )-\left ( y-1 \right )\left ( 7+2 \right )+z\left ( 0+3 \right )$
$\begin{vmatrix} x-2 & y-1 &z \\ 1& -3& -2\\ 1& 0& 7 \end{vmatrix}=-21\left ( x-2 \right )-9\left ( y-1 \right )+3z$
$\begin{vmatrix} x-2 & y-1 &z \\ 1& -3& -2\\ 1& 0& 7 \end{vmatrix}=-21x+42-9y+9+3z$
$\begin{vmatrix} x-2 & y-1 &z \\ 1& -3& -2\\ 1& 0& 7 \end{vmatrix}=-21x-9y+3z+42+9$
$\begin{vmatrix} x-2 & y-1 &z \\ 1& -3& -2\\ 1& 0& 7 \end{vmatrix}=-21x-9y+3z+51$
Now, since
$\begin{vmatrix} x-2 & y-1 &z \\ 1& -3& -2\\ 1& 0& 7 \end{vmatrix}=0$
$\Rightarrow -21x -9y + 3z + 51 = 0\\ \Rightarrow -21x - 9y + 3z = -51\\ \Rightarrow -3(7x + 3y - z) = -3 \times 17\\ \Rightarrow 7x + 3y - z = 17$
Hence, the required equation of the plane is 7x + 3y - z = 17.

Question:11

Find the equations of the 2 lines through the origin which intersect the line $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}$ at angles of $\frac{\pi}{3}$ each.

Given the equation of the line, we need to find the equations of two lines through the origin which intersect the given line.
According to the theorem, equation of a line with direction ratios d1 = (b1 , b2 , b3 ) that passes through the point (x1 , y1 , z1 ) is expressed as:
$\frac{x-x_{1}}{b_{1}}=\frac{y-y_{1}}{b_{2}}=\frac{z-z_{1}}{b_{3}}$
We also know, the angle between two lines with direction ratios d1 and d2 respectively is given by:
$\theta = \cos^{-1}\left ( \frac{d_{1}d_{2}}{\left |d_{1} \right |\left |d_{2} \right |} \right )$
We use these theorems to find the equations of the two lines.
Let the equation of a line be:
$\theta = \cos^{-1}\left ( \frac{d_{1}d_{2}}{\left |d_{1} \right |\left |d_{2} \right |} \right )$
Given that it passes through the origin, (0, 0, 0)
Therefore, equation of both lines passing through the origin will be :
$\frac{x}{b_{1}}=\frac{y}{b_{2}}=\frac{z}{b_{3}}=\lambda \, \, .....(i)$
Let,
$\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}=\mu \, \, .....(ii)$
Direction ratio of the line = (2, 1, 1)
$\Rightarrow d_{1} = (2, 1, 1).... (iii)$
If we represent the direction ratio in terms of a position vector,
$d_{1}=2\hat{i}+\hat{j}+\hat{k} .....(iv)$

Any point on the line is given by (x, y, z). From (ii),
$\frac{x-3}{2}=\mu, \frac{y-3}{1}=\mu ,\frac{z}{1}=\mu$
$\\\text{take} \ \frac{x-3}{2}=\mu\\ \Rightarrow x-3=2\mu\\ \Rightarrow x=2\mu+3\\ \\ take \frac{y-3}{1}=\mu \\ \Rightarrow y-3=\mu\\ \Rightarrow y=\mu+3\\ \\ take \frac{z}{1}=\mu\\ \Rightarrow z=\mu$
Hence, any point on line (ii) is $P(2\mu + 3, \mu + 3, \mu)$
Since line (i) passes through the origin, we can say
$\left ( b_{1},b_{2},b_{3} \right )\equiv (2\mu + 3, \mu + 3, \mu)$
$\Rightarrow \: direction\: \: ratio\: of\; line(i)= (2\mu + 3, \mu + 3, \mu)\\ \Rightarrow d_{2}= (2\mu + 3, \mu + 3, \mu)....(v)$
We can represent the direction ratio in terms of position vector like:
$d_{2}= \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \: \: ....(vi)$
From the theorem, we know
$\cos \theta=\frac{d_{1}.d_{2}}{\left |d_{1} \right |\left |d_{2} \right |}$
If we substitute the values of d? and d? from (iv) and (vi) in the above equation, and putting $\theta=\frac{\pi}{3}$ from the question:
$\Rightarrow \cos \frac{\pi}{3}=\frac{\left ( 2\hat{i}+\hat{j}+\hat{k} \right )\left ( \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right )}{\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |}$
Solving the numerator,
$\left ( 2\hat{i}+\hat{j}+\hat{k} \right )\left ( \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right )= 2\left (2\mu + 3 \right )+1\left ( \mu + 3 \right )+1. \mu$
$\left ( 2\hat{i}+\hat{j}+\hat{k} \right )\left ( \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right )= 4\mu + 6+\mu + 3+ \mu$
$\left ( 2\hat{i}+\hat{j}+\hat{k} \right )\left ( \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right )= 4\mu +\mu + \mu+6+3$
$\left ( 2\hat{i}+\hat{j}+\hat{k} \right )\left ( \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right )= 6\mu+9$
Solving the denominator,
$\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{2^{2}+1^{2}+1^{2}}\sqrt{ \left (2\mu + 3 \right )^{2}+\left ( \mu + 3 \right )^{2}+ \mu^{2}}$
$\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{4+1+1}\sqrt{ \left (2\mu \right )^{2}+ 3^{2}+2\left ( 2\mu \right )\left ( 3 \right )+\left ( \mu \right )^{2}+3^{2}+ 2(\mu)(3)+\mu^{2}}$
$\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{6}\sqrt{4\mu^{2}+9+12\mu+u^{2}+9+6\mu+\mu^{2}}$
$\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{6}\sqrt{4\mu^{2}+u^{2}+\mu^{2}+12\mu+6\mu+9+9}$
$\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{6}\sqrt{6\mu^{2}+18\mu+18}$
$\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{6}\sqrt{6\left (\mu^{2}+3\mu+3 \right )}$
$\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=\sqrt{6}\sqrt{6}\sqrt{\mu^{2}+3\mu+3}$
$\left | 2\hat{i}+\hat{j}+\hat{k} \right |\left | \left (2\mu + 3 \right )\hat{i}+\left ( \mu + 3 \right )\hat{j}+ \mu\hat{k} \right |=6\sqrt{\mu^{2}+3\mu+3 }$
And cos π/3 = 1/2
Substituting the values, we get
$\Rightarrow \frac{1}{2}=\frac{6\mu+9}{6\sqrt{\mu^{2}+3\mu+3}}$
Performing cross multiplication,
$\Rightarrow 6\sqrt{\mu^{2}+3\mu+3}=2\left (6\mu+9 \right )\\ \Rightarrow 6\sqrt{\mu^{2}+3\mu+3}=2 \times 3\left (2\mu+3 \right )\\ \Rightarrow 6\sqrt{\mu^{2}+3\mu+3}=6\left (2\mu+3 \right )\\ \Rightarrow \sqrt{\mu^{2}+3\mu+3}=2\mu+3$
Squaring both sides,
$\Rightarrow \left (\sqrt{\mu^{2}+3\mu+3} \right )^{2}=\left (2\mu+3 \right )^{2}\\ \Rightarrow \mu^{2}+3\mu+3=(2\mu)^{2}+3^{2}+2(2\mu)(3)\left [ \because (a+b)^{2}=a^{2}+b^{2}+2ab \right ]\\ \Rightarrow \mu^{2}+3\mu+3=4\mu^{2}+9+12\mu\\ \Rightarrow 4\mu^{2}-\mu^{2}+12\mu-3\mu+9-3=0\\ \Rightarrow 3\mu^{2}+9\mu+6=0$
$\Rightarrow 3\left ( \mu^{2}+3\mu+2 \right ) =0\\ \Rightarrow \mu^{2}+3\mu+2=0\\ \Rightarrow\mu^{2}+2\mu+\mu+2=0\\ \Rightarrow \mu\left ( \mu+2 \right )+\left ( \mu+2 \right )=0\\ \Rightarrow \left ( \mu+1 \right )+\left ( \mu+2 \right )=0\\$
$\Rightarrow \left ( \mu+1 \right )=0 \: \: or\: \left ( \mu+2 \right )=0\\ \Rightarrow \mu=-1 \: \: \: or\: \: \: \mu=-2$
Therefore, from equation (v)
Direction ratio =$(2\mu+ 3, \mu + 3, \mu)$
Putting μ = -1:
Direction Ratio = (2(-1) + 3, (-1) + 3, -1)
⇒ Direction Ratio = (-2 + 3, -1 + 3, -1)
⇒ Direction Ratio = (1, 2, -1) …(vi)
Now putting μ = -2:
Direction Ratio = (2(-2) + 3, (-2) + 3, -2)
⇒ Direction Ratio = (-4 + 3, -2 + 3, -2)
⇒ Direction Ratio = (-1, 1, -2) …(vii)
Using the direction ratios in (vi) and (vii) in equation (i);
$\frac{x}{b_{1}}=\frac{y}{b_{2}}=\frac{z}{b_{3}}=\lambda\\ \\ \\ \frac{x}{1}=\frac{y}{2}=\frac{z}{-1}=\lambda$
And,
$\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}=\lambda$
Therefore, the two required lines are $\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}=\lambda$ and $\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}=\lambda$

Question:12

Find the angle between the lines whose direction cosines are given by equations l + m + n = 0, l2 + m2 - n2 = 0.

Given, two lines whose direction cosines are l + m + n = 0 - (i); and l² + m² - n² = 0 - (ii). We need to find the angle between these lines.
First, we must find the values of l, m and n.
From equation (i), l + m + n = 0
=> n = - l - m
=> n = -(l + m) …(iii)
If we substitute the value of n from (i) in (ii),
$l^{2}+m^{2}-n^{2}=0\\ \Rightarrow l^{2}+m^{2}-\left (-\left ( l+m \right ) \right )^{2}=0\\ \Rightarrow l^{2}+m^{2}-\left ( l+m \right ) ^{2}=0\\ \Rightarrow l^{2}+m^{2}-\left ( l^{2}+m^{2}+2lm \right )=0\\ \Rightarrow l^{2}+m^{2} -l^{2}-m^{2}-2lm=0\\ \Rightarrow l^{2}-l^{2}+m^{2}-m^{2}-2lm=0\\ \Rightarrow -2lm=0\\ \Rightarrow lm=0$
⇒ l = 0 or m = 0
Putting l = 0 in equation (i),
=> 0 + m + n = 0
=> m + n = 0
=> m = -n
If m = $\lambda$, then
n = -m = -$\lambda$
Hence, direction ratios (l, m, n) = (0, $\lambda$, -$\lambda$)
=> Position vector parallel to these given lines = $0\hat{i}+\lambda\hat{j}-\lambda \hat{k}$
$\Rightarrow d_{1}=\lambda\hat{j}-\lambda \hat{k}$
Now, putting m = 0 in equation (i),
=> l + 0 + n = 0
=> l + n = 0
=> l = -n
If n = $\lambda$, then
l = -n = -$\lambda$
Hence, direction ratios (l, m, n) = (-$\lambda$, 0, $\lambda$)
=> Position vector parallel to these given lines = $-\lambda \hat{i}+\0\hat{j}+\lambda \hat{k}$
$\Rightarrow d_{2}=-\lambda\hat{i}+\lambda \hat{k}$
From the theorem, we get the angle between the two lines whose direction ratios are d1 and d2 as:
$\theta=\cos^{-1}\left ( \frac{\left | d_{1}.d_{2} \right |}{\left | d_{1}\right |\left |d_{2} \right |} \right )$
If we substitute the values of d1 and d2, we get
$\theta=\cos^{-1}\left ( \frac{\left| \left ( \lambda \hat{j}-\lambda \hat{k} \right )\left (- \lambda \hat{i}+\lambda \hat{k} \right ) \right |}{\left | \left ( \lambda \hat{j}-\lambda \hat{k} \right )\right |\left |\left (- \lambda \hat{i}+\lambda \hat{k} \right ) \right |} \right )$
Solving the numerator,
$\left ( \lambda \hat{j}-\lambda \hat{k} \right )\left (- \lambda \hat{i}+\lambda \hat{k} \right ) =0+0+\left ( -\lambda \right )\left ( \lambda \right )\\ \Rightarrow \left ( \lambda \hat{j}-\lambda \hat{k} \right )\left (- \lambda \hat{i}+\lambda \hat{k} \right )= -\lambda^{2}$
Solving the denominator,
$\left |\left ( \lambda \hat{j}-\lambda \hat{k} \right ) \right |\left |\left (- \lambda \hat{i}+\lambda \hat{k} \right ) \right |=\sqrt{ \lambda ^{2}\left (-\lambda \right )^{2}} \sqrt{\left (- \lambda \right )^{2}+\lambda ^{2}}$
$\left |\left ( \lambda \hat{j}-\lambda \hat{k} \right ) \right |\left |\left (- \lambda \hat{i}+\lambda \hat{k} \right ) \right |=\sqrt{ \lambda ^{2}+\lambda^{2}} \sqrt{ \lambda^{2}+\lambda ^{2}}$
$\left |\left ( \lambda \hat{j}-\lambda \hat{k} \right ) \right |\left |\left (- \lambda \hat{i}+\lambda \hat{k} \right ) \right |= \lambda ^{2}+\lambda^{2}$
$\left |\left ( \lambda \hat{j}-\lambda \hat{k} \right ) \right |\left |\left (- \lambda \hat{i}+\lambda \hat{k} \right ) \right |=2 \lambda ^{2}$
Substituting the values in θ,
$\theta=\cos^{-1}\left ( \frac{\left | -\lambda^{2} \right |}{\lambda^{2}} \right )\\ \Rightarrow \theta=\cos^{-1}=\frac{1}{2}\\ \Rightarrow \theta=\frac{\pi}{3}\left [ \because \cos\frac{\pi}{3}=\frac{1}{2} \right ]$
Therefore, the required angle between the lines is π/3.

Question:13

If a variable line in two adjacent positions has direction cosines l, m, n and $l+\delta l,\: m+\delta m,\: n+\delta n,$ show that the small angle $\delta \theta$ between the two positions is given by $\delta \theta^{2}=\delta l^{2}+\delta m^{2}+\delta n^{2}$

Given: direction cosines of a variable line in two adjacent positions are l, m, n and $l+\delta l, m+\delta m,n+\delta n,$
We have to prove that the small angle $\delta \theta$ between the two positions is given by $\delta \theta^{2}=\delta l^{2}+\delta m^{2}+\delta n^{2}$
We know, the relationships between direction cosines is given as
$l^{2}+ m^{2}+ n^{2}=1 ....(1)$
Also, $\left (l+\delta l \right )^{2}+ \left (m+\delta m \right )^{2}+ \left (n+\delta n \right )^{2}=1$
$\Rightarrow l^{2}+(\delta l)^{2}+2(l)(\delta l)+m^{2}+(\delta m)^{2}+2(m)(\delta m)+n^{2}+(\delta n)^{2}+2(n)(\delta n)=1\\ \Rightarrow l^{2}+m^{2}+n^{2}+\left (\delta l \right )^{2}+\left (\delta m \right )^{2}+\left (\delta n \right )^{2}+2l\delta l+2m\delta m+2n\delta n=1\\ \Rightarrow 1+\delta l^{2}+\delta m^{2}+ \delta n^{2}+2l\delta l+2m\delta m+2n\delta n=1\: \: \: \left [ from (i) \right ] \\ \Rightarrow 2l\delta l+2m\delta m+2n\delta n+\delta l^{2}+\delta m^{2}+ \delta n^{2}=1-1 \\ \Rightarrow 2 \left (l\delta l+m\delta m+n\delta n \right )=-\left (\delta l^{2}+\delta m^{2}+ \delta n^{2} \right )\\ \Rightarrow l\delta l+m\delta m+n\delta n =-\frac{1}{2}\left (\delta l^{2}+\delta m^{2}+ \delta n^{2} \right ).......(iii)$
Let
$\vec{a}=l\hat{i}+m\hat{j}+n\hat{k}\\ \Rightarrow \vec{b}=\left ( l+\delta l \right )\hat{i}+\left ( m+\delta m \right )\hat{j}+\left ( n+\delta n \right )\hat{k}$
We know, angle between two lines = $\cos \theta=\vec{a}. \vec{b}$
Here, the angle is very small because the line is variable in different although adjacent positions. According to the question, this small angle is $\delta \theta$
Therefore,
$\cos \delta \theta=\vec{a}. \vec{b}$
Substituting the values of the two vectors, we get
$\Rightarrow \cos \delta \theta = \left (l\hat{i}+m\hat{j}+n\hat{k} \right ).\left (\left ( l+\delta l \right )\hat{i}+\left ( m+\delta m \right )\hat{j}+\left ( n+\delta n \right )\hat{k} \right )$
The dot product of 2 vectors is calculated by obtaining the sum of the product of the coefficients of $\hat{i},\hat{j}\; and \; \hat{k}$
$\Rightarrow \cos \delta \theta =l \left ( l+\delta l \right )+m \left ( m+\delta m \right )+n \left ( n+\delta n \right )\\ \Rightarrow \cos \delta \theta = l^{2}+l \delta l+m^{2}+m \delta m+n^{2}+n \delta n\\ \Rightarrow \cos \delta \theta =l^{2}+m^{2}+n^{2}+l \delta l+m \delta m+n \delta n\\ \Rightarrow \cos \delta \theta =1+l \delta l+m \delta m+n \delta n\: \: \left [ from(i) \right ]\\ \Rightarrow \cos \delta \theta=1-\frac{1}{2}\left ( \delta l^{2}+\delta m^{2}+\delta n^{2} \right )\: \: \: \: \left [ \because from(ii) \right ]\\ \Rightarrow \frac{1}{2}\left ( \delta l^{2}+\delta m^{2}+\delta n^{2} \right )=1-\cos \delta \theta$
Or,
$\Rightarrow 1-\cos \delta \theta= \frac{1}{2}\left ( \delta l^{2}+\delta m^{2}+\delta n^{2} \right )$
We know, $1 -\cos 2 \theta = 2\sin^{2} \theta$
On the left-hand side, the angle is $2 \theta$. On the right hand side, it becomes half, that is, $\frac{ 2 \theta}{2} =\theta$.
Similarly replacing $2 \theta$ by $\delta \theta$ in LHS, then making the angle on the RHS half,
We get:
$1 -\cos \delta \theta = 2\sin^{2} \frac{\delta \theta}{2}$
$\Rightarrow 2\sin^{2} \frac{\delta \theta}{2}=\frac{1}{2}\left ( \delta l^{2}+\delta m^{2}+\delta n^{2} \right )\\ \Rightarrow 2 \times 2\sin^{2} \frac{\delta \theta}{2}= \delta l^{2}+\delta m^{2}+\delta n^{2} \\ \Rightarrow 4\sin^{2} \frac{\delta \theta}{2}= \delta l^{2}+\delta m^{2}+\delta n^{2} \\ \Rightarrow 4\left ( \sin \frac{\delta \theta}{2} \right )^{2}= \delta l^{2}+\delta m^{2}+\delta n^{2} \\$
Since $\delta \theta$ is a very small angle, $\frac{\delta \theta}{2}$ will be much smaller. Hence $\sin \frac{\delta \theta}{2}$ will also be very small in value.
$\Rightarrow \sin \frac{\delta \theta}{2}=\frac{\delta \theta}{2}\\ \\ \Rightarrow 4\left ( \frac{\delta \theta}{2} \right )^{2}=\delta l^{2}+\delta m^{2}+\delta n^{2}\\ \Rightarrow 4 \frac{\delta \theta^{2}}{4} =\delta l^{2}+\delta m^{2}+\delta n^{2}\\ \\ \Rightarrow \delta \theta^{2} =\delta l^{2}+\delta m^{2}+\delta n^{2}\\$
Hence, proved.

Question:14

O is the origin and A is (a, b, c). Find the direction cosines of the line OA and the equation of the plane through A at right angle to OA.

We have the points O (0, 0, 0) and A (a, b, c) where a, b, and c are direction ratios. We need to find the direction cosines of line OA and the equation of the plane through A at right angle to OA.
To begin with,
$\vec{OA}=Position\: vector\: of\: A-Position\: vector\: of\:O\\ \Rightarrow \vec{OA}=\left ( a\hat{i}+b\hat{j}+c\hat{k} \right )-\left ( 0\hat{i}+0\hat{j}+0\hat{k} \right ) \\ \Rightarrow \vec{OA}= a\hat{i}-0\hat{i}+b\hat{j}-0\hat{j}+c\hat{k}-0\hat{k}\\ \Rightarrow \vec{OA}= a\hat{i}+b\hat{j}+c\hat{k}$
We know, if (a, b, c) are the direction ratios of a given vector, then its direction cosines will be:
$\left ( \frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}},\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}},\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}} \right )$
According to the question, the direction ratios are (a, b, c), therefore the direction cosines of the vector OA are the same as the above formula, that is,
$\left ( \frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}},\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}},\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}} \right )$
Given, the plane is perpendicular to OA. We know, a normal is a line or vector which is perpendicular to a given object. Therefore, we can say:
$\vec{n}=\vec{OA}\\ \Rightarrow \vec{n}=a\hat{i}+b\hat{j}+c\hat{k}\\ \left [ \because \vec{OA}=a\hat{i}+b\hat{j}+c\hat{k} \right ]$
Also, the vector equation of a plane where the normal is passing through the plane and passing through is,
$\left ( \vec{r}-\vec{a} \right ).\vec{n}=0$
Where
$\vec{r}-\vec{a} = vector\: from\: \vec{A}\: to\: \vec{R} \\ \vec{a}=Position\: vector\: of\: the\: given\: point\: in\: the\: plane\\ \vec{n}=normal\: vector\: to\: the\: plane$
Here, the given point in the plane is A (a, b, c).
$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\\ \vec{a}=a\hat{i}+b\hat{j}+c\hat{k}\\ \vec{n}=a\hat{i}+b\hat{j}+c\hat{k}\\$
Substituting the vectors respectively, we get:
$\left (\left (x\hat{i}+y\hat{j}+z\hat{k} \right )-\left (a\hat{i}+b\hat{j}+c\hat{k} \right ) \right ).\left (a\hat{i}+b\hat{j}+c\hat{k} \right )=0\\ \Rightarrow \left (x\hat{i}+y\hat{j}+z\hat{k} -a\hat{i}-b\hat{j}-c\hat{k} \right ).\left (a\hat{i}+b\hat{j}+c\hat{k} \right )=0\\ \Rightarrow \left (\left (x-a \right )\hat{i}+\left (y-b \right )\hat{j}+\left (z-c \right )\hat{k}\right ).\left (a\hat{i}+b\hat{j}+c\hat{k} \right )=0\\ \Rightarrow a\left (x-a \right )+b\left (y-b \right )+c\left (z-c \right )=0\\$
Upon simplifying this, we get:
$\Rightarrow ax - a^{2} + by-b^{2} + cz - c^{2} =0\\ \Rightarrow ax + by + cz - a^{2}-b^{2}- c^{2} =0\\ \Rightarrow a^{2}+b^{2}+ c^{2} =ax + by + cz$
Hence, the required equation of the plane is a² + b² + c² = ax + by + cz.

Question:15

Two systems of rectangular axis have the same origin. If a plane cuts them at distances a, b, c and a’, b’, c’ respectively from the origin, prove that:
$\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{a'^{2}}+\frac{1}{b'^{2}}+\frac{1}{c'^{2}}$

Given, we have 2 systems of rectangular axes. Both the systems have the same origin, and there is a plane that cuts both systems.
One system is cut at a distance of a, b, c.
The other system is cut at a distance of a’, b’, c’.
To prove:
$\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{a'^{2}}+\frac{1}{b'^{2}}+\frac{1}{c'^{2}}$
Proof: Since a plane intersects both the systems at distances a, b, c, and a’, b’, c’ respectively, this plane will have different equations in the two different systems.
Let us consider the equation of the plane in the system with distances a, b, c to be:
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Let us consider the equation of the plane in the system with distances a’, b’, c’ be:
$\frac{x}{a'}+\frac{y}{b'}+\frac{z}{c'}=1$
According to the question, the plane cuts both the systems from the origin. We know, the perpendicular distance of a plane $ax + by + cz + d =0$ from the origin is given by:
$\left | \frac{d}{\sqrt{a^{2}+b^{2}+c^{2}}} \right |$
(where not all a, b, and c are zero)
Therefore, the perpendicular distance from the origin of the first plane is:
$\left | \frac{-1}{\sqrt{\left (\frac{1}{a} \right )^{2}+\left (\frac{1}{b} \right )^{2}+\left (\frac{1}{c} \right )^{2}}} \right |$
And, the perpendicular distance from the origin of the second plane:
$\left | \frac{-1}{\sqrt{\left (\frac{1}{a'} \right )^{2}+\left (\frac{1}{b'} \right )^{2}+\left (\frac{1}{c'} \right )^{2}}} \right |$
We also know, if two systems of lines have the same origin, their perpendicular distances from the origin to the plane in both systems are equal.
Therefore,
$\left | \frac{-1}{\sqrt{\left (\frac{1}{a} \right )^{2}+\left (\frac{1}{b} \right )^{2}+\left (\frac{1}{c} \right )^{2}}} \right |=\left | \frac{-1}{\sqrt{\left (\frac{1}{a'} \right )^{2}+\left (\frac{1}{b'} \right )^{2}+\left (\frac{1}{c'} \right )^{2}}} \right |$
$\Rightarrow \frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}} =\frac{1}{\sqrt{\frac{1}{a'^{2}}+\frac{1}{b'^{2}}+\frac{1}{c'^{2}}}}$
Cross-multiplying,
$\Rightarrow \sqrt{\frac{1}{a'^{2}}+\frac{1}{b'^{2}}+\frac{1}{c'^{2}}}=\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}$
Squaring both sides,
$\Rightarrow \sqrt{\frac{1}{a'^{2}}+\frac{1}{b'^{2}}+\frac{1}{c'^{2}}}=\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}$
$\Rightarrow \frac{1}{a'^{2}}+\frac{1}{b'^{2}}+\frac{1}{c'^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}$

Or
$\Rightarrow \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{a'^{2}}+\frac{1}{b'^{2}}+\frac{1}{c'^{2}}$
Hence, proved.

Question:16

Find the foot of the perpendicular from the point (2, 3, -8) to the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$
Also, find the perpendicular distance from the given point to the line.

Given, the perpendicular from the point (let) C (2, 3, -8) to the line of which the equation is,
$\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$
This can be re-written as,
$\frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}$
Hence, the vector equation of the line is, $-2\hat{i}+6\hat{j}-3\hat{k}$
We must find the foot of the perpendicular from the point C (2, 3, -8) to given line, as well as the perpendicular distance from the given point C to the line.
To start with, let us locate the point of intersection between the point and the line.
Let us take,
$\frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}=\lambda$
$\frac{x-4}{-2}=\lambda,\frac{y}{6}=\lambda,\frac{z-1}{-3}=\lambda$
$from\frac{x-4}{-2}=\lambda\\\Rightarrow x-4=-2\lambda \\\Rightarrow x=4-2\lambda\\ \\ from \frac{y}{6}=\lambda\\ \Rightarrow y=6\lambda\\ \\ from \frac{z-1}{-3}=\lambda\\ \Rightarrow z-1=-3\lambda\\\Rightarrow z=1-3\lambda$
We have,$x = 4 - 2\lambda , y = 6\lambda, z = 1 - 3\lambda$
Therefore, the coordinates of any point on the given line is $\left ( 4 - 2\lambda , 6\lambda, 1 - 3\lambda \right )$

Let us consider the foot of the perpendicular from C(2, 3, -8) on line to be$L\left ( 4 - 2\lambda , 6\lambda, 1 - 3\lambda \right )$
Therefore, the direction ratios of $CL\left ( 4 - 2\lambda-2 , 6\lambda-3, 1 - 3\lambda-(-8) \right )$
$=\left ( 4 - 2\lambda-2 , 6\lambda-3, 1+8 - 3\lambda \right )\\ =\left ( 2-2\lambda , 6\lambda-3, 9 - 3\lambda \right )$

Also, the direction ratio of the line is,$\frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}$ (-2, 6, -3).
Since L is the foot of the perpendicular on the line,
Sum of the product of these direction ratios $\left ( 2-2\lambda , 6\lambda-3, 9 - 3\lambda \right )$ and (-2, 6, -3) = 0.
$-2\left ( 2-2\lambda\right ) +6\left (6\lambda-3 \right )+(-3)\left ( 9 - 3\lambda \right )\\\\ \Rightarrow -4+4\lambda+36\lambda-18-27+9\lambda=0\\ \Rightarrow \left ( 4\lambda+36\lambda+9\lambda \right )+\left ( -4-18-27 \right )=0\\ \Rightarrow 49\lambda-49=0\\ \Rightarrow 49\lambda=49\\ \Rightarrow \lambda= \frac{49}{49}\\Hence \: \: \lambda=1$
If we substitute this value of λ in $L\left ( 4 - 2\lambda , 6\lambda, 1 - 3\lambda \right )$, we get
$\Rightarrow L\left ( 4 - 2\lambda , 6\lambda, 1 - 3\lambda \right )=L(4 - 2(1), 6(1), 1 - 3(1))$
$\Rightarrow L\left ( 4 - 2\lambda , 6\lambda, 1 - 3\lambda \right )=L(4 - 2, 6, 1 - 3)$
$\Rightarrow L\left ( 4 - 2\lambda , 6\lambda, 1 - 3\lambda \right )=L(2, 6, -2)$
Now, we must calculate the perpendicular distance of point C from the line, that is point L.
In other words, we need to find $\left | \vec{CL} \right |$
We know, $\vec{CL} =\left ( 2-2\lambda,6\lambda-3,9-3\lambda \right )$
Substituting λ = 1,
$\vec{CL} =\left ( 2-2(1),6(1)-3,9-3(1) \right )\\ \Rightarrow \vec{CL} =\left ( 2-2,6-3,9-3 \right )\\ \Rightarrow \vec{CL} =\left ( 0,3,6 \right )$
To find $\left | \vec{CL} \right |$
$\left | \vec{CL} \right |=\sqrt{0^{2}+3^{2}+6^{2}}\\ \Rightarrow \left | \vec{CL} \right |=\sqrt{0+9+36} \Rightarrow \left | \vec{CL} \right |=\sqrt{45}\\ \Rightarrow \left | \vec{CL} \right |=3\sqrt{5}$
Therefore, the foot of the perpendicular from the point C to the given line is (2, 6, -2) and the perpendicular distance is $3\sqrt{5}$ units.

Question:17

Given, the point P (2, 4, -1), the equation of the line is $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$
We must find the distance of point P from this line.
Note, to find the distance between a point and a line, we should get foot of the perpendicular from the point on the line.
Let, P(2, 4, -1) be the given point and be $L:\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}=\lambda$ the given line.
Direction ratio of the line L is (1, 4, -9) …(i)
Let us find any point on this line.
Taking L,
$\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}=\lambda$
$\frac{x+5}{1}=\lambda,\frac{y+3}{4}=\lambda,\frac{z-6}{-9}=\lambda$
$Take \frac{x+5}{1}=\lambda\\ \Rightarrow x+5=\lambda\\ \Rightarrow x=\lambda - 5\\\\ Take \frac{y+3}{4}=\lambda\\ \Rightarrow y+3=4\lambda\\ \Rightarrow y=4\lambda-3\\ \\ Take \frac{z-6}{-9}=\lambda\\ \Rightarrow z-6=-9\lambda\\ \Rightarrow z=6-9\lambda$
Therefore, any point on the line L is $(\lambda - 5, 4\lambda - 3, 6 - 9\lambda)$
Let this point be $Q(\lambda - 5, 4\lambda - 3, 6 - 9\lambda)$, the foot of the perpendicular from the point P (2, 4, -1) on the line L.
Hence, the direction ratio of PQ is given by
$(\lambda - 5-2, 4\lambda - 3-4, 6 - 9\lambda-(-1))$
=> Direction ratio of PQ= $(\lambda - 7, 4\lambda - 7, 7 - 9\lambda)$ …(ii)
Also, we know, if two lines are perpendicular to each other, then the dot product of their direction ratios should be 0.
Here, PQ is perpendicular to L. We have, from (i) and (ii),
Direction ratio of L = (1, 4, -9)
Direction ratio of PQ = $(\lambda - 7, 4\lambda - 7, 7 - 9\lambda)$
Therefore,
$(1, 4, -9).(\lambda - 7, 4\lambda - 7, 7 - 9\lambda) = 0\\ \Rightarrow 1 (\lambda- 7) + 4 (4\lambda - 7) + (-9) (7 - 9\lambda) = 0\\ \Rightarrow \lambda - 7 + 16\lambda - 28 -63 + 81\lambda = 0\\ \Rightarrow \lambda + 16\lambda + 81\lambda - 7 - 28- 63 = 0\\ \Rightarrow 98\lambda - 98 = 0\\ \Rightarrow 98\lambda = 98\\ \Rightarrow \lambda = 1$
Hence, the coordinate of Q, i.e. the foot of the perpendicular from the point on the given line is,
$Q (\lambda - 5, 4\lambda - 3, 6 - 9\lambda) = Q (1 - 5, 4(1) - 3, 6 - 9)\\ \Rightarrow Q (\lambda - 5, 4\lambda - 3, 6 - 9\lambda) = Q (1 - 5, 4 - 3, 6 - 9)\\ \Rightarrow => Q(\lambda - 5, 4\lambda - 3, 6 - 9\lambda) = (-4, 1, -3)$
Now, to find the perpendicular distance from P to the line, that is point Q,
That is, to find $\left | \vec{PQ} \right |$
We know,
$\left | \vec{PQ} \right |=(\lambda - 7, 4\lambda - 7, 7 - 9\lambda)$
Substituting $\lambda=1$
$\vec{PQ}=(1 - 7, 4(1)- 7, 7 - 9(1))\\ \Rightarrow \vec{PQ}=\left ( -6,4-7,7-9 \right )\\ \Rightarrow \vec{PQ}=\left ( -6,-3,-9 \right )$
Now, to find
$\left | \vec{PQ} \right |= \sqrt{(-6)^{2}+(-3)^{2}+(-2)^{2} }\\ \Rightarrow \left | \vec{PQ} \right |= \sqrt{36+9+4}\\ \Rightarrow \left | \vec{PQ} \right |= \sqrt{49}\\ \Rightarrow \left | \vec{PQ} \right |= 7$
Therefore, the distance from the given point to the given line = 7 units.

Question:18

Find the length and the foot of the perpendicular from the point (1, 3/2, 2) to the plane 2x - 2y + 4z + 5 = 0.

Given, point P (1, 3/2, 2)
The plane is 2x - 2y + 4z + 5 = 0
We must find the foot of the perpendicular from the point P to the equation of the given plane.
Also, we must find the distance from the point P to the plane.
Let us consider the foot of the perpendicular from point P to be Q.
Let Q be Q (x1 , y1 , z1)
So, the direction ratio of PQ is given by
(x1 - 1, y1 - 3/2, z1 - 2)
Now, let us consider the normal to the plane 2x - 2y + 4z + 5 = 0:
It is obviously parallel to PQ, since a normal is a line or vector that is perpendicular to a given object. The direction ratio simply states the number of units to move along each axis.
For any plane, ax + by + cz = d, where, a, b, and c are normal vectors to the plane.
Hence, the direction ratios are (a, b, c).
Therefore, the direction ratio of the normal = (2, -2, 4) for plane 2x - 2y + 4z + 5 = 0.
The Cartesian equation of the line PQ, where P(1, 3/2, 2) and Q (x1 , y1 , z1) is:
$\frac{x_{1}-1}{2}=\frac{y_{1}-\frac{3}{2}}{-2}=\frac{z_{1}-2}{4}=\lambda(say)$
To find any point on this line,
$\frac{x_{1}-1}{2}=\lambda,\frac{y_{1}-\frac{3}{2}}{-2}=\lambda,\frac{z_{1}-2}{4}=\lambda$
$from \frac{x_{1}-1}{2}=\lambda\\\Rightarrow x_{1}-1=2\lambda\\\Rightarrow x_{1}=2\lambda+1\\ \\ \\ from \frac{y_{1}-\frac{3}{2}}{-2}=\lambda\\ \Rightarrow y_{1}-\frac{3}{2}=-2\lambda\\\Rightarrow y_{1}=\frac{3}{2}-2\lambda\\ \\ \\ from \frac{z_{1}-2}{4}=\lambda\\\Rightarrow z_{1}-2=4\lambda \\\Rightarrow z_{1}=4\lambda+2$
Any point on the line is $(2\lambda+ 1, \frac{3}{2} - 2\lambda, 4\lambda + 2).$
This point is Q.
$Q\left ( x_{1},y_{1},z_{1} \right )=Q(2\lambda+ 1, \frac{3}{2} - 2\lambda, 4\lambda + 2)....(i)$
And, it was assumed that is lies on the given plane. Substituting x1, y1, and z1 in the plane equation, we get:
2x1 - 2y1 + 4z1 + 5 = 0
$\Rightarrow 2\left (2\lambda+ 1 \right )-2\left ( \frac{3}{2} - 2\lambda \right )+4\left (4\lambda + 2 \right )+5=0$
Simplifying to find the value of $\lambda$
$\Rightarrow 4\lambda + 2 - 3 + 4\lambda + 16\lambda + 8 + 5 = 0\\ \Rightarrow 4\lambda + 4\lambda + 16\lambda + 2 - 3 + 8 + 5 = 0\\ \Rightarrow 24\lambda + 12 = 0\\ \Rightarrow 24\lambda = -12\\ \Rightarrow \lambda =-\frac{12}{24}\\ \Rightarrow \lambda =-\frac{1}{2}$
Since Q is the foot of the perpendicular from the point P,
We substitute the value of $\lambda$ in equation (i) to get:
$Q\left ( x_{1},y_{1},z_{1} \right )=Q(2(-\frac{1}{2})+ 1, \frac{3}{2} - 2(-\frac{1}{2}), 4(-\frac{1}{2}) + 2)\\ \Rightarrow Q\left ( x_{1},y_{1},z_{1} \right )=Q\left ( -1+1,\frac{3}{2}+1,-2+2 \right )\\ \Rightarrow Q\left ( x_{1},y_{1},z_{1} \right )=Q\left ( 0,\frac{5}{2},0 \right )$
Then, to find $\left | \vec{PQ} \right |$
Where, P = (1, 3/2, 2) and Q = (0, 5/2, 0)
$\left | \vec{PQ} \right |=\sqrt{\left (0-1 \right )^{2}+\left ( \frac{5}{2}-\frac{3}{2} \right )^{2}+\left ( 0-1 \right )^{2}}\\ \Rightarrow \left | \vec{PQ} \right |=\sqrt{(-1)^{2}+(1)^{2}+(-2)^{2}}\\ \Rightarrow \left | \vec{PQ} \right |=\sqrt{1+1+4}\\ \Rightarrow \left | \vec{PQ} \right |=\sqrt{6}$
Thus, the foot of the perpendicular from the given point to the plane is (0, 5/2, 0) and the distance is $\sqrt{6}$ units.

Question:19

Find the equations of the line passing through the point (3, 0, 1) and parallel to the planes x + 2y = 0 and 3y - z = 0.

Given, a line passes through a point P (3, 0, 1) and is parallel to the planes x + 2y = 0 and 3y - z = 0.
We must find the equation of this line.
Let the position vector of point P be
$\vec{a}=3\hat{i}+0\hat{j}+\hat{k}$
Or,
$\vec{a}=3\hat{i}+\hat{k}.....(i)$
Let us consider the normal to the given planes, that is, perpendicular to the normal of the plane x + 2y = 0 and 3y - z = 0
Normal to the plane x + 2y = 0 can be given as $\vec{n_{1}}=\hat{i}+2\hat{j}$
Normal to the plane 3y - z = 0 can be given as $\vec{n_{2}}=3\hat{j}-\hat{k}$
So, $\vec{n}$ is perpendicular to both these normals.
So,
$\vec{n}=\vec{n_{1}}\times \vec{n_{2}}$
$\Rightarrow \vec{n}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 0 \\ 0 & 3 & -1 \end{vmatrix}$
Taking the 1st row and the 1st column, we multiply the 1st element of the row $\left (a_{11} \right )$ with the difference of products of the opposite elements $\left (a_{22}\times a_{33}-a_{23} \times a_{32} \right )$, excluding 1st row and 1st column
$\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21}& a_{22} & a_{23} \\ a_{31}& a_{32} & a_{33} \end{vmatrix}=a_{11}\left ( a_{22} \times a_{33}-a_{23} \times a_{32} \right )$
Here,
$\begin{vmatrix} \hat{i}& \hat{j} & \hat{j} \\ 1& 2 & 0 \\ 0& 3 & -1 \end{vmatrix}=\hat{i}\left ( \left ( 2 \times -1 \right )-\left (0 \times 3 \right ) \right )$
Now, we take the 2nd column and 1st row, and multiply the 2nd element of the row (a??) with the difference of the product of opposite elements $\left (a_{21}\times a_{33}-a_{23} \times a_{31} \right )$
$\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21}& a_{22} & a_{23} \\ a_{31}& a_{32} & a_{33} \end{vmatrix}=a_{11}\left ( a_{22} \times a_{33}-a_{23} \times a_{32} \right )-a_{12}\left ( a_{21} \times a_{33}-a_{23} \times a_{31} \right )$
Here
$\begin{vmatrix} \hat{i}& \hat{j} & \hat{j} \\ 1& 2 & 0 \\ 0& 3 & -1 \end{vmatrix}=\hat{i}\left ( \left ( 2 \times -1 \right )-\left (0 \times 3 \right ) \right )-\hat{j}\left (\left ( 1 \times -1 \right )-\left ( 0 \times 0 \right ) \right )$
Finally, taking the 1st row and 3rd column , we multiply the 3rd element of the row (a??) with the difference of the product of opposite elements $\left (a_{22}\times a_{33}-a_{23} \times a_{32} \right )$ excluding the 1st row and 3rd column.
$\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21}& a_{22} & a_{23} \\ a_{31}& a_{32} & a_{33} \end{vmatrix}=a_{11}\left ( a_{22} \times a_{33}-a_{23} \times a_{32} \right )-a_{12}\left ( a_{21} \times a_{33}-a_{23} \times a_{31} \right )+a_{13}\left ( a_{21} \times a_{32}-a_{22} \times a_{31} \right )$
Here
$\begin{vmatrix} \hat{i}& \hat{j} & \hat{j} \\ 1& 2 & 0 \\ 0& 3 & -1 \end{vmatrix}=\hat{i}\left ( \left ( 2 \times -1 \right )-\left (0 \times 3 \right ) \right )-\hat{j}\left (\left ( 1 \times -1 \right )-\left ( 0 \times 0 \right ) \right )+\hat{k}\left ( \left ( 1 \times 3 \right )-\left ( 2 \times 0 \right ) \right )$
Futher simplifying it,
$\Rightarrow \begin{vmatrix} \hat{i}& \hat{j} & \hat{j} \\ 1& 2 & 0 \\ 0& 3 & -1 \end{vmatrix}=\hat{i}\left ( -2-0 \right )-\hat{j}\left ( -1-0 \right )+\hat{k}\left ( 3-0 \right )\\ \\ \\ \Rightarrow \begin{vmatrix} \hat{i}& \hat{j} & \hat{j} \\ 1& 2 & 0 \\ 0& 3 & -1 \end{vmatrix}=-2\hat{i}+\hat{j}+3\hat{k}\\ \\ \\ \rightarrow \vec{n}=-2\hat{i}+\hat{j}+3\hat{k}$
Therefore, the direction ratio is (-2, 1, 3) …(iii)
We know, vector equation of any line passing through a point and parallel to a vector is $\vec{r}=\vec{a}+\lambda \vec{b}$ where $\lambda \epsilon \mathbb{R}$
Hence, from (i) and (ii),
$\vec{a}=3\hat{i}+\hat{k}\\ \vec{n}=-2\hat{i}+\hat{j}+3\hat{k}$

Putting these vectors in the equation $\hat{r}=\hat{a}+\lambda \hat{n}\\$

We get
$\hat{r}=\left ( 3\vec{i}+\vec{k} \right )+\lambda \left ( -2\hat{i}+\hat{j}+3\hat{k} \right )$
But we know,
$\hat{r}=x\vec{i}+y\hat{j}+z\vec{k}$
Substituting this,
$\left (x\vec{i}+y\hat{j}+z\vec{k} \right )=\left ( 3\hat{i}+\hat{k} \right )+\lambda\left ( -2\hat{i}+\hat{j}+3\hat{k} \right )\\ \\ \Rightarrow \left (x\vec{i}+y\hat{j}+z\vec{k} \right )-\left ( 3\hat{i}+\hat{k} \right )=\lambda\left ( -2\hat{i}+\hat{j}+3\hat{k} \right )\\ \\ \Rightarrow x\hat{i}+y\hat{j}+z\hat{k}-3\hat{i}-\hat{k}=\lambda \left ( -2\hat{i}+\hat{j}+3\hat{k} \right )\\ \\ \Rightarrow \left ( x-3 \right )\hat{i}+y\hat{j}+\left ( z-1 \right )\hat{k}=\lambda \left ( -2\hat{i}+\hat{j}+3\hat{k} \right )\\ \\$
Thus, the required equation of the line is $\left ( x-3 \right )\hat{i}+y\hat{j}+\left ( z-1 \right )\hat{k}=\lambda \left ( -2\hat{i}+\hat{j}+3\hat{k} \right )\\ \\$

Question:20

Find the equation of the plane through the points (2, 1, -1) and (-1, 3, 4) and perpendicular to the plane x - 2y + 4z = 10.

Given, a plane passes through the points (2, 1, -1) and (-1, 3, 4) and is perpendicular to the plane x - 2y + 4z = 10.
We want to find the equation of this plane.
We know, the Cartesian equation of a plane passing through (x1, y1, z1)
with direction ratios perpendicular to a, b, c for its normal is given as:
a (x - x1) +b (y - y1) + c (z - z1) = 0
Hence,
Let us consider the equation of the plane passing through (2, 1, -1) to be
a(x – 2) + b(y – 1) + c(z – (-1)) = 0
⇒ a(x – 2) + b(y – 1) + c(z + 1) = 0 …(i)
Since it also passes through point (-1, 3, 4) we just replace x, y, z by -1, 3, and 4 respectively.
⇒ a(-1 – 2) + b(3 – 1) + c(4 + 1) = 0
⇒ -3a + 2b + 5c = 0 …(ii)
Since a, b, and c are direction ratios and this plane is perpendicular to the plane x - 2y + 4z = 10, we just replace x, y, and z with a, b, and c respectively (neglecting 10) and we equate this to 0.
=> a - 2b + 4c = 0 …(iii)
To solve two equations x1a + y1b + z1c = 0 and x2a + y2b + z2c = 0, we use the formula
$\frac{a}{\begin{vmatrix} y_{1} &z_{1} \\ y_{2}&z_{2} \end{vmatrix}}=\frac{b}{\begin{vmatrix} z_{1} &x_{1} \\ z_{2}&x_{2} \end{vmatrix}}=\frac{c}{\begin{vmatrix} x_{1} &y_{1} \\ x_{2}&y_{2} \end{vmatrix}}$
Similarly, to solve for equations (ii) and (iii):
$\frac{a}{\begin{vmatrix}2 &5 \\ -2&4 \end{vmatrix}}=\frac{b}{\begin{vmatrix} 5 &-3 \\ 4&1 \end{vmatrix}}=\frac{c}{\begin{vmatrix}-3 & 2 \\ 1&-2 \end{vmatrix}}$
$\Rightarrow \frac{a}{\left ( 2 \times 4 \right )-\left ( 5 \times -2 \right )}=\frac{b}{\left ( 5 \times 1 \right )-\left ( -3 \times 4 \right )}=\frac{c}{\left ( -3 \times -2 \right )-\left ( 2 \times 1 \right )}$
$\Rightarrow \frac{a}{8+10}=\frac{b}{5+12}=\frac{c}{6-2}$
$\Rightarrow \frac{a}{18}=\frac{b}{17}=\frac{c}{4}=\lambda$
$\Rightarrow \frac{a}{18}=\lambda, \frac{b}{17}=\lambda, \frac{c}{4}=\lambda$
That is,
$\Rightarrow \frac{a}{18}=\lambda\\ \Rightarrow a=18 \lambda\\ \\ \Rightarrow \frac{b}{17}=\lambda\\ \Rightarrow b=17 \lambda\\ \\ \Rightarrow \frac{c}{4}=\lambda\\ \Rightarrow c=4 \lambda\\ \\$
Substituting these values of a, b, and c in equation (i), we get
$a(x - 2) + b(y - 1) + c(z + 1) = 0\\ \Rightarrow 18\lambda(x - 2) + 17\lambda(y - 1) + 4\lambda(z + 1) = 0\\ \Rightarrow \lambda[18(x - 2) + 17(y - 1) + 4(z + 1)] = 0\\ \Rightarrow 18(x - 2) + 17(y - 1) + 4(z + 1) = 0\\ \Rightarrow 18x - 36 + 17y - 17 + 4z + 4 = 0\\ \Rightarrow 18x + 17y + 4z - 36 - 17 + 4 = 0\\ \Rightarrow 18x + 17y + 4z - 49 = 0\\ \Rightarrow 18x + 17y + 4z = 49\\$
Therefore, the required equation of the plane is 18x + 17y + 4z = 49.

Question:21

Find the shortest distance between the lines given by $r=\left ( 8+3\lambda \right )\hat{i}-\left ( 9+16\lambda \right )\hat{j}+\left ( 10+7\lambda \right )\hat{k}$ and $r=15\hat{i}+29\hat{j}+5\hat{k}+\mu\left ( 3\hat{i}+8\hat{j}-5\hat{k} \right )$

Given two lines,
$\vec{r}=\left ( 8+3\lambda \right )\hat{i}-\left ( 9+16\lambda \right )\hat{j}+\left ( 10+7\lambda \right )\hat{k}...........(i)\\ \\ \vec{r}=15\hat{i}+29\hat{j}+5\hat{k}+\mu\left ( 3\hat{i}+8\hat{j}-5\hat{k} \right )...........(ii)$
Taking equation (i),
$\vec{r}=\left ( 8+3\lambda \right )\hat{i}-\left ( 9+16\lambda \right )\hat{j}+\left ( 10+7\lambda \right )\hat{k} \\ \Rightarrow \vec{r}= 8\hat{i}+3\lambda \hat{i}- 9\hat{j}+16\lambda \hat{j})+ 10 \hat{k}+7\lambda \hat{k} \\ \Rightarrow \vec{r}=8\hat{i}-9\hat{j}+10\hat{k}+3 \lambda \hat{i}-16 \lambda \hat{j}+7 \lambda \hat{k}\\ \Rightarrow \vec{r}=8\hat{i}-9\hat{j}+10\hat{k}+\lambda \left(3 \hat{i}-16 \hat{j}+7 \hat{k} \right ).............(iii)$
We know, the vector equation of a line passing through a point and parallel to a vector is where $\lambda\epsilon \mathbb{R}$
$\vec{a}$ = Position vector of the point through which line passes
$\vec{b}$ = Normal to the line
Comparing this with equation (iii), we get
$\vec{a_{1}}=8\hat{i}-9\hat{j}+10\hat{k}\\ \vec{b_{1}}=3\hat{i}-16\hat{j}+7\hat{k}\\$
Now take equation (ii)
$\vec{r}=15\hat{i}+29\hat{i}+5\hat{k}+\mu\left ( 3\hat{i}+8\hat{j}-5\hat{k} \right ) \\ \vec{r}=\left (15\hat{i}+29\hat{i}+5\hat{k} \right )+\mu\left ( 3\hat{i}+8\hat{j}-5\hat{k} \right )..........(iv)$
Similarly from (iv)
$\vec{a_{2}}=\left (15\hat{i}+29\hat{i}+5\hat{k} \right )\\ \vec{b_{2}}=\left ( 3\hat{i}+8\hat{j}-5\hat{k} \right )$
So, the shortest distance between two lines can be represented as:
$d=\left |\frac{\left ( \vec{b_{1}} \times \vec{b_{2}} \right ).\left ( \vec{a_{2}} - \vec{a_{1} }\right ) } {\left | \vec{b_{1}} \times \vec{b_{2}} \right |}\right |$
solve $\vec{b_{1}} \times \vec{b_{2}}$
$\vec{b_{1}} \times \vec{b_{2}}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}$
Taking 1st row and 1st column, we multiply the 1st element of the row (a??) with the difference of the product of the opposite elements $\left ( a_{22}\times a_{33}-a_{23} \times a_{32}\right )$, excluding the 1st row and the 1st column;
$\begin{vmatrix} a_{11} &a_{12} & a_{13} \\ a_{21} &a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}=a_{11}\left ( a_{22}\times a_{33}-a_{23} \times a_{32} \right )$
Here
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}=\hat{i}\left (\left ( -16 \times -5 \right )-\left ( 7 \times 8 \right ) \right )$
Now, we take the 2nd column and 1st row, and multiply the 2nd element of the row (a??) with the difference of the product of opposite elements (a?? x a?? - a?? x a??)
$\begin{vmatrix} a_{11} &a_{12} & a_{13} \\ a_{21} &a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}=a_{11}\left ( a_{22}\times a_{33}-a_{23} \times a_{32} \right )-a_{12}\left ( a_{21}\times a_{33}-a_{23} \times a_{31} \right )$
Here
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}=\hat{i}\left (\left ( -16 \times -5 \right )-\left ( 7 \times 8 \right ) \right )-\hat{j}\left (\left ( 3 \times -5 \right )-\left ( 7 \times 3 \right ) \right )$
Finally, taking the 1st row and 3rd column , we multiply the 3rd element of the row (a??) with the difference of the product of opposite elements (a?? x a?? - a?? x a??), excluding the 1st row and 3rd column.
$\begin{vmatrix} a_{11} &a_{12} & a_{13} \\ a_{21} &a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}=a_{11}\left ( a_{22}\times a_{33}-a_{23} \times a_{32} \right )-a_{12}\left ( a_{21}\times a_{33}-a_{23} \times a_{31} \right )+a_{13}\left ( a_{21}\times a_{32}-a_{22} \times a_{31} \right )$
Here
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}=\hat{i}\left (\left ( -16 \times -5 \right )-\left ( 7 \times 8 \right ) \right )-\hat{j}\left (\left ( 3 \times -5 \right )-\left ( 7 \times 3 \right ) \right )+\hat{k}\left ( \left ( 3 \times 8 \right )-\left ( -16 \times 3 \right ) \right )$
Further simplifying it.
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}=\hat{i}\left (80-56 \right )-\hat{j}\left (-15-21 \right )+\hat{k}\left (24+48 \right )$
$\Rightarrow \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}=24\hat{i}+36\hat{j}+72\hat{k}\\ \\ \\ \Rightarrow \vec{b}\times \vec{b}= 24\hat{i}+36\hat{j}+72\hat{k}........(v)$
And,
$\left | \vec{b}\times \vec{b} \right |= \left |24\hat{i}+36\hat{j}+72\hat{k} \right |\\ \\ \Rightarrow \left | \vec{b}\times \vec{b} \right |=\sqrt{24^{2}+36^{2}+72^{2}} \\ \Rightarrow \left | \vec{b}\times \vec{b} \right |=12\sqrt{2^{2}+3^{2}+6^{2}} \\ \Rightarrow \left | \vec{b}\times \vec{b} \right |=12 \sqrt{4+9+36}\\ \Rightarrow \left | \vec{b}\times \vec{b} \right |=12\sqrt{49}\\ \Rightarrow \left | \vec{b}\times \vec{b} \right |=12 \times 7\\ \Rightarrow \left | \vec{b}\times \vec{b} \right |=84...........(vi)$
$Now \; \; solving \: \: \vec{a_{2}}- \vec{a_{1}}\\ \\ \vec{a_{2}}- \vec{a_{1}} =\left (15\hat{i}+29\hat{i}+5\hat{k} \right )-\left ( 8\hat{i}-9\hat{j}+10\hat{k} \right )\\ \Rightarrow \vec{a_{2}}- \vec{a_{1}} = 15\hat{i}-8\hat{i}+29\hat{j}+9\hat{j}+5\hat{k}-10\hat{k}\\ \Rightarrow \vec{a_{2}}- \vec{a_{1}} = 7\hat{i}+38\hat{j}-5\hat{k}.....(vii)$
Substituting the values from (v), (vi) and (vii) in d, we get
$d=\left |\frac{\left ( \vec{b_{1}} \times \vec{b_{2}} \right ).\left ( \vec{a_{2}} - \vec{a_{1} }\right ) } {\left | \vec{b_{1}} \times \vec{b_{2}} \right |}\right |$
$\Rightarrow d =\left | \frac{\left ( 24\hat{i}+36\hat{j}+72\hat{k} \right ).\left ( 7\hat{i}+38\hat{j}-5\hat{k} \right )}{84} \right |\\ \Rightarrow d =\left | \frac{24 \times 7 +36 \times 38+72 \times -5}{84} \right |\\ \Rightarrow d =\left | \frac{168+1368-360}{84} \right |\\ \Rightarrow d =\left | \frac{1176}{84} \right |\\ \Rightarrow d =14\\$
Thus, the shortest distance between the lines is 14 units.

Question:22

Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 , which contains the line of intersection of the planes x + 2y + 3z - 4 = 0 and 2x + y -z + 5 = 0.

Given, a plane is perpendicular to another plane 5x + 3y + 6z + 8 = 0,and also contains line of intersection of the planes x + 2y + 3z - 4 = 0 and 2x + y -z + 5 = 0.
We must find the equation of this plane.
We know, the equation of a plane passing through the line of intersection of the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is given as,
$( a_{1}x + b_{1}y + c_{1}z + d_{1}) +\lambda(a_{2}x + b_{2}y + c_{2}z + d_{2}) = 0$
Similarly, the equation of a plane through the line of intersection of the planes x + 2y + 3z - 4 = 0 and 2x + y -z + 5 = 0. is given by,
$(x + 2y + 3z - 4) + \lambda(2x + y - z + 5) = 0 \\ \Rightarrow x + 2y + 3z - 4 + 2\lambda x + \lambda - \lambda z + 5\lambda = 0\\ \Rightarrow x + 2 \lambda x + 2y + \lambda y + 3z - \lambda z - 4 + 5 \lambda = 0\\ \Rightarrow (1 + 2 \lambda)x + (2 + \lambda)y + (3 - \lambda)z - 4 + 5 \lambda = 0 ....(i)$
Thus, the direction ratio of plane in (i) is,
$(1 + 2\lambda, 2 + \lambda, 3 - \lambda)$
Since the plane in equation (i) is perpendicular to the plane 5x + 3y + 6z + 8 = 0;
we can replace x, y, z with (1 + 2λ), (2 + λ) and (3 - λ) respectively in the plane 5x + 3y + 6z + 8 = 0 (neglecting 8) and equating to 0.
This gives us,
$5(1 + 2\lambda) + 3(2 + \lambda) + 6(3 - \lambda) = 0\\ \Rightarrow 5 + 10\lambda + 6 + 3\lambda + 18 - 6\lambda = 0\\ \Rightarrow 10\lambda + 3\lambda- 6\lambda + 5 + 6 + 18 = 0\\ \Rightarrow 7\lambda + 29 = 0\\ \Rightarrow 7\lambda = -29\\ \Rightarrow \lambda=- \frac{29}{7}$
Substituting this value of $\lambda$ in equation (i) we get
$\left ( 1+2\left (-\frac{29}{7} \right ) \right )x+\left ( 2-\frac{29}{7} \right )y+\left ( 3+\frac{29}{7} \right )z-4+5\left (-\frac{29}{7} \right )=0\\ \\ \\ \Rightarrow \left ( 1+\frac{58}{7} \right )x+\left ( 2-\frac{29}{7} \right )y+\left ( 3+\frac{29}{7} \right )z-4-\frac{145}{7}=0\\ \\ \\ \Rightarrow \left ( \frac{7-58}{7} \right )x+\left ( \frac{14-29}{7} \right )y+\left ( \frac{21+29}{7} \right )z+\left ( \frac{-28-145}{7} \right )=0 \\ \\ \\ \Rightarrow -\frac{51}{7}x-\frac{15}{7}y+\frac{50}{7}z-\frac{173}{7}=0\\ \\ \\ \Rightarrow -51x - 15y + 50z - 173 = 0 \\ \\ \Rightarrow 51x + 15y - 50z + 173 = 0$
Thus, the required equation of the plane is 51x + 15y - 50z + 173 = 0.

Question:23

The plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle α. Prove the equation of the plane in its new position is $ax+by\pm \left ( \sqrt{a^{2}+b^{2}} \tan \alpha \right )z=0$

Given, the plane ax + by = 0 is rotated about its line of intersection with z = 0 by an angle $\alpha$
To prove: equation of the plane in its new position is
$ax+by\pm z\sqrt{a^{2}+b^{2}}\tan \alpha=0$
Proof: Two planes are given, ax + by = 0 …(i) and z = 0 …(ii)
We know, the equation of the plane passing through the line of intersection of the planes (i) and (ii) is
$ax + by + \lambda z = 0...(iii)$
where, $\lambda \epsilon \mathbb{R}$
The angle between the new plane and plane (i) is given as $\alpha$
Since the angle between two planes is equivalent to the angle between their normals, the direction ratio of normal to ax + by = 0 or ax + by +0z = 0 is (a, b, 0).
$\Rightarrow \vec{l}=a\hat{i}+b\hat{j}$
And, the direction ratio of normal to $ax + by + \lambda z = 0$ is (a, b, λ).
$\Rightarrow \vec{m}=a\hat{i}+b\hat{j}+\lambda \hat{k}$
Also, we know, angle between 2 normal vectors of the two given planes can be given as;
$\cos \alpha=\frac{\vec{l}\vec{m}}{\left |\vec{l} \right |\left |\vec{m} \right |}$
If we substitute the values of these vectors, we get
$\cos \alpha=\frac{\left (a\hat{i}+b\hat{j} \right )\left ( a\hat{i}+b\hat{j}+\lambda \hat{k} \right )}{\left |\left (a\hat{i}+b\hat{j} \right ) \right |\left |\left ( a\hat{i}+b\hat{j}+\lambda \hat{k} \right ) \right |}\\$
$\Rightarrow \cos \alpha=\frac{a.a+b.b+0.\lambda}{\sqrt{a^{2}+b^{2}}\sqrt{a^{2}+b^{2}+\lambda^{2}}}\\ \Rightarrow \cos \alpha=\frac{a^{2}+b^{2}}{\sqrt{a^{2}+b^{2}}\sqrt{a^{2}+b^{2}+\lambda^{2}}}$
We then multiply $\sqrt{a^{2}+b^{2}}$ by the numerator and denominator on the right hand side of the equation to get
$\Rightarrow \cos \alpha=\frac{a^{2}+b^{2}}{\sqrt{a^{2}+b^{2}}\sqrt{a^{2}+b^{2}+\lambda^{2}}}\times \frac{\sqrt{a^{2}+b^{2}}}{\sqrt{a^{2}+b^{2}}}\\ \\ \Rightarrow \cos \alpha=\frac{\left (a^{2}+b^{2} \right )\sqrt{a^{2}+b^{2}}}{\left (a^{2}+b^{2} \right )\sqrt{a^{2}+b^{2}+\lambda^{2}}}\\ \\ \Rightarrow \cos \alpha=\frac{\sqrt{a^{2}+b^{2}}}{\sqrt{a^{2}+b^{2}+\lambda^{2}}}\\ \\$
Applying square on both sides,
$\Rightarrow \cos^{2} \alpha=\left (\frac{\sqrt{a^{2}+b^{2}}}{\sqrt{a^{2}+b^{2}+\lambda^{2}}} \right )^{2}\\ \\ \Rightarrow \cos^{2} \alpha=\frac{a^{2}+b^{2}}{a^{2}+b^{2}+\lambda ^{2}}$
$\Rightarrow (a^{2} + b^{2} + \lambda^{2}) cos^{2} \alpha = a^{2} + b^{2}\\ \Rightarrow a^{2} \cos^{2} \alpha + b^{2} \cos^{2} \alpha + \lambda^{2} \cos^{2} \alpha = a^{2} + b^{2}\\ \Rightarrow \lambda^{2} \cos^{2} \alpha = a^{2} + b^{2} - a^{2} \cos^{2} \alpha -b^{2}\cos^{2} \alpha\\ \Rightarrow \lambda^{2} \cos^{2} \alpha = a^{2} - a^{2} \cos^{2} \alpha + b^{2} -b^{2} \cos^{2} \alpha\\ \Rightarrow \lambda^{2} \cos^{2} \alpha = a^{2}(1 -\cos^{2} \alpha) + b^{2}(1 - \cos^{2} \alpha)\\ \Rightarrow \lambda^{2} \cos^{2} \alpha = (a^{2} + b^{2})(1 - \cos^{2} \alpha)\\ \Rightarrow \lambda^{2} \cos^{2} \alpha = (a^{2} + b^{2}) \sin^{2} \alpha [since, \sin^{2} \alpha + \cos^{2} \alpha = 1]$
$\Rightarrow \lambda^{2} = \frac{(a^{2} + b^{2}) \sin^{2} \alpha }{\cos^{2} \alpha}\\ Since \frac{sin^{2}\alpha}{\cos^{2}\alpha}=\tan^{2}\alpha\\ \Rightarrow \lambda^{2}=\left ( a^{2}+b^{2} \right )tan^{2}\alpha\\ \Rightarrow \lambda =\pm \sqrt{\left ( a^{2}+b^{2} \right )tan^{2}\alpha}\\ \Rightarrow \lambda =\pm \sqrt{\left ( a^{2}+b^{2} \right )}tan^{2}\alpha\\$
Substituting the value of $\lambda$ in equation (iii) to find the plane equation,
ax + by + λz = 0
$\lambda =\pm \sqrt{\left ( a^{2}+b^{2} \right )}tan^{2}\alpha\\$
Hence proved.

Question:24

Find the equation of the plane through the intersection of the planes $r.\left ( \hat{i}+3\hat{j} \right )-6=0$and $r.\left ( 3\hat{i}-\hat{j}-4\hat{k} \right )=0$whose perpendicular distance from origin is unity.

Given two planes,
$\vec{r}.\left ( \hat{i}+3\hat{j} \right )-6=0\\ \vec{r}.\left(3\hat{i}-\hat{j}-4\hat{k} \right )=0$
Also given, the perpendicular distance of the plane from the origin = 1.
We must find the equation of this plane.
We know,
$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$
Simplifying the planes,
$\vec{r}.\left ( \hat{i}+3\hat{j} \right )-6=0\\ \Rightarrow \left ( x\hat{i}+y\hat{j}+z\hat{k} \right ).\left ( \hat{i}+3\hat{j} \right )-6=0\\ \Rightarrow x+3y-6=0........(i)$
Also, for
$\vec{r}.\left(3\hat{i}-\hat{j}-4\hat{k} \right )=0\\ \Rightarrow \left ( x\hat{i}+y\hat{j}+z\hat{k} \right ).\left(3\hat{i}-\hat{j}-4\hat{k} \right )=0\\ \Rightarrow 3x-y-4z=0$
The equation of a plane through the line of intersection of x + 3y - 6 = 0 and 3x - y - 4z = 0 can be given as
$(x + 3y - 6) + \lambda(3x - y - 4z) = 0\\ \Rightarrow x + 3y - 6 + 3\lambda x - \lambda y - 4\lambda z = 0\\ \Rightarrow x + 3\lambda x + 3y - \lambda y - 6 - 4\lambda z = 0\\ \Rightarrow (1 + 3\lambda)x + (3 - \lambda)y - 4\lambda z - 6 = 0 %u2026(iii)$
Also, we know, the perpendicular distance of a plane, ax + by + cz + d = 0 from the origin, let’s say P, is given by
$P=\left | \frac{d}{\sqrt{a^{2}+b^{2}+c^{2}}} \right |$
Similarly, the perpendicular distance of the plane in equation (iii) from the origin (=1 according to the question) is:
$1=\left | \frac{-6}{\sqrt{\left ( 1+3 \lambda \right )^{2}+\left ( 3-\lambda \right )^{2}+\left ( -4\lambda \right )^{2}}} \right |\\ \\ \Rightarrow \sqrt{\left ( 1+3 \lambda \right )^{2}+\left ( 3-\lambda \right )^{2}+\left ( -4\lambda \right )^{2}}=6$
Taking the square of both sides,
$\Rightarrow \left (\sqrt{\left ( 1+3 \lambda \right )^{2}+\left ( 3-\lambda \right )^{2}+\left ( -4\lambda \right )^{2}} \right )^{2}=6^{2}$
$\Rightarrow (1 + 3\lambda)^{2} + (3 - \lambda)^{2} + (-4\lambda)^{2} = 36\\ \Rightarrow 1 + (3\lambda)^{2} + 2(1)(3\lambda) + (3)^{2} + \lambda^2 - 2(3)(\lambda) + 16\lambda^{2} = 36\\ \Rightarrow 1 + 9\lambda^{2} + 6\lambda + 9 + \lambda^{2} - 6\lambda + 16\lambda^{2} = 36\\ \Rightarrow 9\lambda^{2} + 16\lambda^{2} + \lambda^{2} + 6\lambda - 6\lambda = 36 - 1 - 9\\ \Rightarrow 26\lambda^{2} + 0 = 26\\ => \lambda^{2} = 26/26\\ => \lambda^{2} = 1\\ => \lambda = \pm 1$
First, we subsitute $\lambda=1$ in eq (iii) to find the plane equation
$(1 + 3\lambda)x + (3 - \lambda)y - 4\lambda z - 6 = `0\\ \Rightarrow (1 + 3(1))x - (3 - 1)y - 4(1)z - 6 = 0\\ \Rightarrow 4x - 2y - 4z- 6 = 0$
Now, we substitute λ= -1 in eq (iii) to find the plane equation
$(1 + 3\lambda)x + (3 - \lambda)y - 4 \lambda z - 6 = 0\\ \Rightarrow (1 + 3(-1))x + (3 - (-1))y - 4(-1)z - 6 = 0\\ \Rightarrow (1 - 3)x + (3 + 1)y + 4z - 6 = 0\\ \Rightarrow -2x + 4y + 4z - 6 = 0$
Therefore, the equation of the required plane is -2x + 4y + 4z – 6 = 0 and 4x – 2y – 4z – 6 = 0.

Question:25

Show that the points $\hat{i}-\hat{j}+3\hat{k}$ and $3\left ( \hat{i}+\hat{j}+\hat{k} \right )$ are equidistant from the plane $r.\left ( 5\hat{i}+2\hat{j}-7\hat{k} \right )+9=0$and lies on the opposite of it.

Given two points,
$\\\vec{A}=\hat{i}-\hat{j}+3\hat{k}\\ \vec{B}=3\left ( \hat{i}+\hat{j}+\hat{k} \right )=3\hat{i}+3\hat{j}+\hat{k}\\ \vec{r}.\left ( 5\hat{i}+2\hat{j}-7\hat{k} \right )+9=0$
Also,
$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\\\\$
Where,
Therefore,
$\left (x\hat{i}+y\hat{j}+z\hat{k} \right ).\left (5\hat{i}+2\hat{j}-7\hat{k} \right )+9=0\\ \\ \Rightarrow 5x + 2y - 7z + 9 = 0$
We must show that the points A and B are equidistant from the plane
5x + 2y - 7z + 9 = 0
We also need to show that the points lie on the opposite side of the plane.
Normal of the plane is, $\vec{N=} 5{i} + 2\hat{j} - 7\hat{k}$
We know, the perpendicular distance of the position vector of a point
to the plane, p: ax + by + cz + d = 0 is given as:$D=\left | \frac{p(l,m,n)}{\left |\vec{N} \right |} \right |$
Where $\left |\vec{N} \right |=Normal \: vector\: of\: the\: plane$
$\vec{N} =a\vec{i}+b\vec{j}+c\vec{k}$
Thus, the perpendicular distance of the point $\vec{A} =\vec{i}-\vec{j}+3\vec{k}=A(1,-1,3)$ to the plane 5x + 2y - 7z + 9 = 0 having normal $\vec{N} =5\vec{i}+2\vec{j}-7\vec{k}$ is given by,
$\left | D_{1} \right |=\left |\frac{5(1)+2(-1)-7(3)+9}{|5\hat{i}+2\hat{j}-7\hat{k}|} \right |\\ \Rightarrow \left | D_{1} \right |=\left |\frac{5-2-21+9}{\sqrt{5^{2}+2^{2}+(-7)^{2}}} \right |\\ \Rightarrow \left | D_{1} \right |=\left |\frac{-9}{\sqrt{25+4+49}} \right |\\ \Rightarrow \left | D_{1} \right |=\left |\frac{9}{\sqrt{78}} \right |\\$
Hence, the perpendicular distance of the point $\vec{B}=3\hat{i}+3\hat{j}+3\hat{k}=B(3,3,3)$ to the plane 5x + 2y - 7z + 9 = 0 having normal $\vec{N}=5\hat{i}+2\hat{j}-7\hat{k}$
$\left | D_{2} \right |=\left |\frac{5(3)+2(3)-7(3)+9}{|5\hat{i}+2\hat{j}-7\hat{k}|} \right |\\ \Rightarrow \left | D_{2} \right |=\left |\frac{15+6-21+9}{\sqrt{5^{2}+2^{2}+(-7)^{2}}} \right |\\ \Rightarrow \left | D_{2} \right |=\left |\frac{9}{\sqrt{25+4+49}} \right |\\ \Rightarrow \left | D_{2} \right |=\left |\frac{9}{\sqrt{78}} \right |\\$
Therefore, |D1| = |D2|
However, D1 and D2 have different signs.
Therefore, the points A and B will lie on opposite sides of the plane.
Hence, we have successfully shown that the points are equidistant from the plane and lie on opposite sides of the plane.

Question:26

$AB=3\hat{i}-\hat{j}+\hat{k}\; \; and\; \; AB=-3\hat{i}+2\hat{j}+4\hat{k}$are two vectors. The positions vectors of the points A and C are $6\hat{i}+7\hat{j}+4\hat{k}\; \; and\; \;-9\hat{j}+2\hat{k}$ respectively. Find the position vector of a point P on the line AB and a point Q on the line CD, such that PQ is perpendicular to AB and CD both.

Given,
$\vec{AB}=3\hat{i}-\hat{j}+\hat{k}\\ \vec{CD}=-3\hat{i}+2\hat{j}+4\hat{k}$
And the position vectors
$\vec{OA}=6\hat{i}+7\hat{j}+4\hat{k}\\ \vec{OC}=-9\hat{j}+2\hat{k}$
Therefore, the line passing through A and along AB will have the equation:
$\vec{r}=6\hat{i}+7\hat{j}+4\hat{k}+\lambda\left ( 3\hat{i}-\hat{j}+\hat{k} \right )\\ \Rightarrow \vec{r}=\left ( 6+3\lambda \right )\hat{i}+\left ( 7-\lambda \right )\hat{j}+\left ( 4+\lambda \right )\hat{k}$
and the line passing through C and along CD will have equation
$\vec{r}=-9\hat{j}+2\hat{k}+\mu \left ( -3\hat{i}+2\hat{j}+4\hat{k} \right )\\ \Rightarrow \vec{r}=-3\mu \hat{i}+\left ( 2\mu-9 \right )\hat{j}+\left ( 2+4\mu \right )\hat{k}$
Now, PQ is a vector perpendicular to both AB and CD, such that Q lies on CD and P lies on AB. Thus, coordinates of P and Q will be of the form
$P (6 + 3\lambda, 7 - \lambda, 4 + \lambda)....(i)\\ Q (-3\mu, 2\mu - 9, 2 + 4\mu) .....(ii)$
Hence, the vector PQ will be given as
$\vec{PQ}=\left ( -3\mu-6-3\lambda \right )\hat{i}+\left ( 2\mu-16+\lambda \right )\hat{j}+\left ( 4\mu-2-\lambda \right )\hat{k}$
Now, since PQ is perpendicular to both, hence the dot products of AB.PQ and CD.PQ will be equal to 0.
AB. PQ = 0 and CD. PQ = 0
$AB. PQ = 3(-3\mu - 6 - 3\lambda) - (2\mu - 16 + \lambda) + (4\mu - 2 - \lambda)\\ \Rightarrow 0 = -9\mu - 18 - 9\lambda - 2\mu + 16 - \lambda + 4\mu - 2 - \lambda\\ \Rightarrow -7\mu - 11\lambda - 4 = 0.....(iii)\\ CD.PQ = 3(-3\mu - 6 - 3\lambda) + 2(2\mu - 16 + \lambda) + 4(4\mu - 2 - \lambda)\\ \Rightarrow 0 = -9\mu - 18 -9\lambda + 4\mu - 32 + 2\lambda +16 \mu -8 - 4\lambda\\ \Rightarrow 29\mu + 7\lambda - 22 = 0 ....(iv)$
Solving (iii) and (iv), we get
$\lambda = -1\; \; and \: \: \mu = 1$
Putting the value of $\lambda$ in (i) we get,
$P (6 + 3(-1), 7 - (-1), 4 + (-1))\\ P (6-3,7+1,4-1)\\ P (3,8,3)$
Putting the value of $\mu$ in (ii) we get,
$Q (-3(1), 2(1) - 9, 2 + 4(1))\\ Q (-3, 2 - 9, 2 +4)\\ Q (-3, -7, 6)$
Hence, position vector of P and Q will be
$\\\vec{OP}=3\hat{i}+8\hat{j}+3\hat{k}\\ \vec{OQ}=-3\hat{i}-7\hat{j}+6\hat{k}$

Question:27

Show that the straight lines whose direction cosines are given by 2l + 2m - n = 0 and mn + nl + lm = 0 are at right angles.

We have given
$2l + 2m - n = 0...(i)\\ \Rightarrow n = 2(l + m)...(ii)$
, and
$mn + nl + lm = 0\\ \Rightarrow 2m(l + m) + 2(l + m)l + lm = 0\\ \Rightarrow 2lm + 2m^{2} + 2l^{2} + 2lm + lm = 0\\ \Rightarrow 2m^{2} + 5lm + 2l^{2} = 0\\ \Rightarrow 2m^{2} + 4lm + lm + 2l^{2} = 0\\ \Rightarrow (2m + l)(m + 2l) = 0$
Thus, we get two cases:
$l = -2m$
=> -4m + 2m - n = 0 [from (i)]
=> n = 2m
, and
m = -2l
=> 2l + 2(-2l) - n = 0
=> 2l - 4l = n
=> n = -2l
Hence, the direction ratios of one line is proportional to -2m, m or -2m or direction ratios are (-2, 1, -2) and the direction ratios of another line is proportional to l, -2l, -2l, or direction ratios are (1, -2, -2)
Thus, the direction vectors of two lines are $b_{1}=-2\hat{i}+\hat{j}-2\hat{k} \: and \: b_{2}=\hat{i}-2\hat{j}-2\hat{k}$
Also, the angle between the two lines $\vec{r}=\vec{a_{1}}+\lambda \vec{b_{1}} \: and \: \vec{r}=\vec{a_{2}}+\mu \vec{b_{2}}$ is given by:
$\cos \theta=\left |\frac{\vec{b_{1}}.\vec{b_{2}}}{\left |\vec{b_{1}} \right |\left |\vec{b_{2}} \right |} \right |$
Now,
$\vec{b_{1}}.\vec{b_{2}}=2(1)+1(-2)+(-2)(-2)\\ =2-2+4\\=0\\ \Rightarrow \cos\theta=0\\ \Rightarrow \theta=90^{\circ}$
Therefore, the lines have a 900 angle between them.

Question:28

If l1, m1, n1; l2, m2, n2; l3, m3, n3 are the direction cosines of 3 mutually perpendicular lines, prove that the line whose direction cosines are proportional to l1 + l2 + l3 , m1 + m2 + m3 , n1 + n2 + n3 makes equal angles with them.

Let the direction vector of the 3 mutually perpendicular lines be
$\vec{a}=l_{1}\hat{i}+m_{1}\hat{j}+n_{1}\hat{k}\\ \\ \vec{b}=l_{2}\hat{i}+m_{2}\hat{j}+n_{2}\hat{k}\\ \\ \vec{c}=l_{3}\hat{i}+m_{3}\hat{j}+n_{3}\hat{k}$
Let the direction vectors associated with direction cosines $l_{1} + l_{2} + l_{3} , m_{1} + m_{2} + m_{3} , n_{1} + n_{2} + n_{3}$ be
$\vec{p}=\left (l_{1} + l_{2} + l_{3} \right )\hat{i}+\left (m_{1} + m_{2} + m_{3} \right )\hat{j}+\left ( n_{1} + n_{2} + n_{3} \right )\hat{k}$
Since the lines associated with the direction vectors a, b, and c are mutually perpendicular, we get
$\vec{a}.\vec{b}=0$ (Since the dot product of two perpendicular vectors is 0)
=>$l_{1}l_{2} + m_{1}m_{2}+n_{1}n_{2}=0$ …(1)
Similarly,
$l_{1}l_{3} + m_{1}m_{3}+n_{1}n_{3}=0$ …(2)
Finally, $\vec{b}.\vec{c}=0$
=>$l_{2}l_{3} + m_{2}m_{3}+n_{2}n_{3}=0$ …(3)
Now, let us consider x, y and z as the angles made by direction vectors a, b, and c respectively with p.
Then,
$\cos x=\vec{a}.\vec{p}$
$\Rightarrow \cos x =l_{1}\left ( l_{1}+l_{2}+l_{3} \right )+m_{1}\left ( m_{1}+m_{2}+m_{3} \right )+n_{1}\left ( n_{1}+n_{2}+n_{3} \right )\\ \Rightarrow \cos x=l_{1}^{2}+l_{1}l_{2}+l_{1}l_{3}+m_{1}^{2}+m_{1}m_{2}+m_{1}m_{3}+n_{1}^{2}+n_{1}n_{2}+n_{1}n_{3}\\ \Rightarrow cos x= l_{1}^{2}+m_{1}^{2}+n_{1}^{2}+\left ( l_{1}l_{2}+m_{1}m_{2}+ n_{1}n_{2} \right )+\left ( l_{1}l_{3}+m_{1}m_{3}+ n_{1}n_{3} \right )\\$
We know, $l_{1}^{2}+m_{1}^{2}+n_{1}^{2}=1$ [since the sum of squares of direction cosines of a line = 1]
$\Rightarrow \cos x=1+0=1$ [from (1) and (2)]
Then, $\cos y=1$ and $\cos z=1$
=> x = y = z = 0.
Therefore, the vector p makes equal angles with the vectors a, b and c.

Question:29

Distance of the point (α, β, y) is:
A. β B. |β| C. |β| + |y| D. √(α² + y²)

Drawing a perpendicular from (α, β, y) to the y-axis gives us a foot of $\left ( \alpha ,\beta ,\gamma \right )$ perpendicular with coordinates (0, β, 0).
Also, using the distance formula, we can calculate the distance between two points as:
$AB=\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}+\left ( z_{2}-z_{1} \right )^{2}}$
Thus, the required distance $=\sqrt{\left ( \alpha-0 \right )^{2}+\left ( \beta-\beta \right )^{2}+\left(\gamma-0 \right )^{2}}=\sqrt{\alpha^{2}+\gamma^{2}}$
(Option D)

Question:30

If the direction cosines of a line are k, k, k, then:
A. k > 0
B. 0 < k < 1
C. k = 1
D. k = 1/√3 or -1/√3

We know that the sum of squares of the direction cosines of a line = 1
=> k² + k² + k² = 1
=> 3k² = 1
$\Rightarrow k=\pm\frac{1}{\sqrt{3}}$ (Option D)

Question:31

Given plane is
$\vec{r}.\left ( \frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k} \right )=1$
Let
$\vec{n}=\frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k}$
$\left |\vec{n} \right |=\sqrt{\frac{2}{7}\hat{i}+\frac{3}{7}\hat{j}-\frac{6}{7}\hat{k} }=1$
=> n is a unit vector
Thus, the equation of the plane is of the form $\vec{r}.\hat{n} =d$, where n is
the unit vector and d is the distance from the origin.
Comparing, we get d =1, hence the distance of the plane from origin is 1
(Option A)

Question:32

The equation of the line is given as
$\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$
The direction vector of this line can be represented as $\vec{b}=3\hat{i}+4 \hat{j}+5\hat{k}$
Also given is the equation of the plane 2x - 2y + z = 5
The normal to this plane is,$\vec{n}=2\hat{i}-2 \hat{j}+\hat{k}$
We also know that the angle $\phi$ between the line with the direction
vector b and the plane with the normal vector n is,
$\sin \phi =\left | \frac{\vec{b}.\vec{n}}{\left | \vec{b} \right |\left | \vec{n} \right |} \right |$
$\Rightarrow \sin \phi =\left |\frac{3(2)+4(-2)+5(1)}{\sqrt{3^{2}+4^{2}+5^{2}}\sqrt{2^{2}+(-2)^{2}+1^{2}}} \right |$
$\Rightarrow \sin \phi =\left |\frac{3}{3\sqrt{50}} \right |$
$=\frac{1}{5\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}$
$=\frac{\sqrt{2}}{10}$
(Option D)

Question:33

The reflection of the point (α, β, γ) in the xy- plane is:
A. (α, β, 0)
B. (0, 0, γ)
C. (-α, -β, -γ)
D. (α, β, -γ)

The equation of the XY plane is Z = 0
Given point (α, β, γ); if we draw a perpendicular from this point in the XY plane, the coordinates of the plane will be
(α, β, 0).
Let the reflection be (x, y, z)
So, $\alpha=\frac{\alpha+x}{2}$
=> x = α
$\beta=\frac{\beta+y}{2}$
=> y = β
$0=\frac{\gamma+z}{2}$
=> z = -γ
The reflection is: (α, β, -γ). (option D)

Question:34

The area of the quadrilateral ABCD, where A (0, 4, 1), B (2, 3, -1), C (4, 5, 0) and D (2, 6, 2) is equal to:
A. 9 square units
B. 18 square units
C. 27 square units
D. 81 square units

Given, A (0, 4, 1), B (2, 3, -1), C (4, 5, 0) and D (2, 6, 2);
$\vec{AB}=\left ( 2-0 \right )\hat{i}+\left ( 3-4 \right )\hat{j}+\left ( -1-1 \right )\hat{k}=2\hat{i}-\hat{j}-2\hat{k}$
$\vec{BC}=\left ( 4-2\right )\hat{i}+\left ( 5-3 \right )\hat{j}+\left ( 0-(-1) \right )\hat{k}=2\hat{i}+2\hat{j}+\hat{k}$
$\vec{CD}=\left ( 2-4\right )\hat{i}+\left ( 6-5 \right )\hat{j}+\left ( 2-0 \right )\hat{k}=-2\hat{i}+\hat{j}+2\hat{k}=-\vec{AB}$
$\vec{DA}=\left ( 0-2\right )\hat{i}+\left ( 4-6 \right )\hat{j}+\left ( 1-2 \right )\hat{k}=-2\hat{i}-2\hat{j}-1\hat{k}=-\vec{BC}$
Since opposite vectors of this parallelogram are equal and opposite, ABCD is a parallelogram and we know the area of a parallelogram is | AB x CD|
$=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & -1 & -2 \\ 2 & 2 & 1 \end{vmatrix}$
$=\left | \hat{i}\left ( -1+4 \right )+\hat{j}\left ( -4-2 \right )+\hat{k}\left ( 4+2 \right ) \right |$
$=\left | 3\hat{i}-6\hat{j}+6\hat{k} \right |$
$=\sqrt{3^{2}+(-6)^{2}+6^{2}}=\sqrt{81}$
= 9 square units (Option A).

Question:35

The locus represented by xy + yz = 0 is:
A. A pair of perpendicular lines
B. A pair of parallel lines
C. A pair of parallel planes
D. A pair of perpendicular planes

Given, xy + yz = 0
=> x (y + z) = 0
=> x = 0 and y + z = 0
Clearly, the above equations are the equations of planes [of the form ax + by + cz + d = 0]
Also, x = 0 has the normal vector $\hat{i}$
And y + z = 0 has the normal vector $\hat{j}+\hat{k}$
And the dot product of these two is
$\hat{i}\left (\hat{j}+\hat{k} \right )=\hat{i}.\hat{j}+\hat{i}.\hat{k}$
= 0
Hence, the planes are perpendicular (Option D).

Question:36

The plane 2x - 3y + 6z - 11 = 0 makes an angle $\sin^{-1}(\alpha)$ with the x-axis. The value of α is:
A. $\frac{\sqrt{3}}{2}$
B. $\frac{\sqrt{2}}{3}$
C. $\frac{2}{7}$
D. $\frac{3}{7}$

Given, the equation of the plane is 2x - 3y + 6z - 11 = 0.
The normal to this plane is,
$\vec{n}=2\hat{i}-3\hat{j}+6\hat{k}$
Also, the x-axis has the direction vector $\vec{b}=\hat{i}$
Also, we know that the angle $\varphi$ between the line with direction vector b and the plane having the normal vector n is:
$\\ \sin \varphi =\left | \frac{\vec{b}.\vec{n}}{\left |\vec{b} \right |.\left |\vec{n} \right |} \right |\\ \Rightarrow \sin \varphi=\left | \frac{1(2)+0(-3)+0(6)}{\sqrt{2^{2}+3^{2}+6^{2}}\sqrt{1^{2}+0^{2}+0^{2}}} \right |\\ \Rightarrow \sin \varphi = \left | \frac{2}{\sqrt{49}} \right |=\frac{2}{7}\\\Rightarrow \varphi =\sin^{-1}\left ( \frac{2}{7} \right )$
On comparing, we find$\alpha =\frac{2}{7}$ (Option C)

Question:37

Fill in the blanks:
A plane passes through the points (2, 0, 0), (0, 3, 0) and (0, 0, 4). The equation of the plane is ______.

We know, the equation of a plane cutting the coordinate axes at (a, 0, 0), (0, b, 0) and (0, 0, c) is given as
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
In this case, a = 2, b = 3, and c = 4.
Therefore, putting these values in the equation of the plane, we get
$\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$

Question:38

Fill in the blanks: The direction cosines of the vector $\left ( 2\hat{i}+2\hat{j}-\hat{k} \right )$ are _______.

If l, m, and n are the direction cosines and the direction ratios of a line are a, b, and c, then we know:
$\\l=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}} \\m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}\\l=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}$
According to the question,
a = 2, b = 2, c = -1
Then
$\sqrt{a^{2}+b^{2}+c^{2}}=\sqrt{2^{2}+2^{2}+(-1)^{2}}=3$
Thus, the direction cosines are
l = 2/3, b = 2/3, c = -1/3

Question:39

Fill in the blanks: The vector equation of the line $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$ is __________.

The equation of the line is given as
$\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$
Clearly, the line passes through A (5, -4, 6) and has the direction ratios 3, 7, and 2.
Also, the position vector of A is $\vec{a}=5\hat{i}-4\hat{j}+6\hat{k}$
The direction vector of the given line will be:
$\vec{b}=3\hat{i}+7\hat{j}+2\hat{k}$
Also, the vector equation of a line passing through the given point whose position vector is a and b is:
$\vec{r}=\vec{a}+\lambda \vec{b}$
Hence, the required equation of the line will be:
$\vec{r}=\left ( 5\hat{i}-4\hat{j}+6\hat{k} \right )+\lambda\left ( 3\hat{i}+7\hat{j}+2\hat{k}\right )$

Question:40

Fill in the blanks: The Cartesian equation $\vec{r}.\left ( \hat{i}+\hat{j}-\hat{k} \right )=2$ of the plane is ______.

By expanding the dot product given in the question, we can get the Cartesian equation of the plane.
Given: $\vec{r}.\left ( \hat{i}+\hat{j}-\hat{k} \right )=2$
$\vec{r}=x\hat{i}+y\hat{j}-z\hat{k}$
Putting
$\Rightarrow \left ( x\hat{i}+y\hat{j}+z\hat{k} \right )\left ( \hat{i}+\hat{j}-\hat{k} \right )=2$
$\Rightarrow x + y - z = 2$
Thus, the required Cartesian equation is x + y - z = 2.

Question:41

State True or False for the given statement:
The unit vector normal to the plane $x + 2y +3z - 6 = 0$ is $\frac{1}{\sqrt{14}}\hat{i}+\frac{2}{\sqrt{14}}\hat{j}+\frac{3}{\sqrt{14}}\hat{k}$

Given, the equation of the plane is x + 2y + 3z - 6 = 0
The normal to this plane will be The unit vector of this normal is:
$\\ \vec{n}=\frac{\vec{n}}{\left |\vec{n} \right |}\\ \vec{n}=\frac{\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{1^{2}+2^{2}+3^{2}}}=\frac{1}{\sqrt{14}}\hat{i}+\frac{2}{\sqrt{14}}\hat{j}+\frac{3}{\sqrt{14}}\hat{k}$
Therefore, the statement is True

Question:42

Fill in the blanks:
The vector equation of the line that passes through the points (3, 4, -7) and (1, -1, 6) is ______.

The position vector of the first point (3, 4, -7) is $\vec{a}=3\hat{i}+4\hat{j}-7\hat{k}$
And the position vector of the second point(1, -1, 6): $\vec{b}=\hat{i}-\hat{j}+6\hat{k}$

Also, the vector equation of a line passing through two points with position vectors a and b is given by:
$\vec{r}=\vec{a}+\lambda\left ( \vec{b}-\vec{a} \right )$
Thus, the required line equation is
$\vec{r}=3\hat{i}+4\hat{j}-7\hat{k}+\lambda\left ( \hat{i}-\hat{j}+6\hat{k}-\left ( 3\hat{i}+4\hat{j}-7\hat{k} \right ) \right )$
$\vec{r}=3\hat{i}+4\hat{j}-7\hat{k}+\lambda\left ( -2\hat{i}-5\hat{j}+13\hat{k} \right )$

Question:43

State True or False for the given statement:
The intercepts made by the plane $2x - 3y + 5z + 4 = 0$ on the coordinate axes are $-2, \frac{4}{3},\frac{-4}{5}$

To begin with, we convert the given plane equation to intercept form:
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$where a, b, and c are the intercepts on x, y, and z, axes respectively.
Given, $2x - 3y + 5z + 4 = 0$
$\Rightarrow -2x + 3y - 5z = 4$
Dividing this equation on both sides by 4,
$\Rightarrow \frac{-1}{2}x + \frac{3}{4}y - \frac{5}{4}z = 1$
On comparison, we get the intercepts -2, 4/3, and -4/5 respectively.
Therefore, the statement is True.

Question:44

State True or False for the given statement: The angle between the line $r=\left ( 5 \hat{i}-\hat{j}-4\hat{k} \right )+\lambda\left ( 2 \hat{i}-\hat{j}+k \right )$ and the plane $r.\left ( 3 \hat{i}-4\hat{j}-\hat{k} \right )+5=0\, \, \, \, \, \, \sin^{-1}\frac{5}{2\sqrt{91}}$ is

We know, the angle $\phi$ between the plane with normal vector n and the line with direction vector b is denoted by:
$\sin\varphi\frac{\vec{b}.\vec{n}}{\left |\vec{b} \right |.\left |\vec{n} \right |}$
Given equation of the line is $r=\left ( 5 \hat{i}-\hat{j}-4\hat{k} \right )+\lambda\left ( 2 \hat{i}-\hat{j}+k \right )$
Hence, its direction vector will be:
$\vec{b}=2\hat{i}-\hat{j}+\hat{k}$
Given equation of the plane is $\vec{r}.\left (3\hat{i}-4\hat{j}-\hat{k} \right )+5=0$
Hence, its normal vector will be:
$\vec{n}=3\hat{i}-4\hat{j}-\hat{k}$
Thus, we have:
$\sin\varphi =\left | \frac{\left ( 2\hat{i}-\hat{j}+\hat{k} \right )\left ( 3\hat{i}-4\hat{j}-\hat{k} \right )}{\sqrt{2^{2}+(-1)^{2}+1^{2}}\sqrt{3^{2}+(-4)^{2}+(-1)^{2}}} \right |\\ \Rightarrow \sin \varphi=\frac{2(3)-1(-4)+1(-1)}{\sqrt{6}\sqrt{26}}=\frac{9}{\sqrt{156}}=\frac{9}{2\sqrt{39}}$
$\varphi =\sin^{-1}\frac{9}{2\sqrt{39}}$
Therefore, the given statement is False.

Question:45

State True or False for the given statement:
The angle between the planes $r.\left ( 2\hat{i}-3\hat{j}+\hat{k} \right )=1$ and $\bar{r}.\left ( \hat{i}-\hat{j} \right )=4$ is $\cos^{-1}\frac{-5}{\sqrt{58}}$

In vector form, if we take θ as the angle between the two planes
$\vec{r}.\vec{n_{1}}=\vec{d_{1}}$ and $\vec{r}.\vec{n_{2}}=\vec{d_{2}}$
Then
$\theta=\frac{\left | \vec{n_{1}}.\vec{n_{2}} \right |}{\left | \vec{n_{1}} \right |\left | \vec{n_{2}} \right |}$
Now, the given planes are $\vec{r}.\left ( 2\hat{i}-3\hat{j}+\hat{k} \right )=1$ and $\vec{r}.\left ( \hat{i}-\hat{j}\right )=4$

Here, $\vec{n_{1}}=2 \hat{i}-3\hat{j}+\hat{k}$ and $\vec{n_{2}}=\hat{i}-\hat{j}$
Therefore,
$\theta =\cos^{-1}\frac{2(1)+3(1)+1(0)}{\sqrt{2^{2}+(-3)^{2}+1^{2}}\sqrt{1^{2}+(-1)^{2}+0^{2}}}$
$=\cos^{-1}\frac{5}{\sqrt{2}\sqrt{14}}\\=\cos^{-1}\frac{5}{2\sqrt{7}}$
The statement is False.

Question:46

State True or False for the given statement:
The line $r=2\hat{i}-3\hat{j}-\hat{k}+\lambda\left ( \hat{i}-\hat{j}+2\hat{k} \right )$ lies in the plane $r.\left (3\hat{i}+\hat{j}-\hat{k} \right )+2=0$

The equation of the line is given as
$\Rightarrow \vec{r}=2\hat{i}-3\hat{j}-\hat{k}+\lambda \left ( \hat{i}-\hat{j}+2\hat{k} \right )\\ \Rightarrow \vec{r}=\left ( 2+\lambda \right )\hat{i}+\left ( -3-\lambda \right )\hat{j}+\left ( -1+2\lambda \right )\hat{k}$
Any point lying on this line will satisfy the plane equation if the line itself lies in the plane. Also, any point on this line will have a position vector:
$\vec{a}=\left ( 2+\lambda \right )\hat{i}+\left ( -3-\lambda \right )\hat{j}+\left ( -1+2\lambda \right )\hat{k}$
Given equation of the plane $\vec{r}.\left ( 3\hat{i}+\hat{j}-\hat{k} \right )+2=0$
If we put a in the above equation,
$\left (\left (2 + \lambda \right )\hat{i}+ \left(-3 - \lambda \right) \hat{j} + \left (-1 + 2\lambda \right)\hat{k} \right ).\left ( 3\hat{i}+\hat{j}-\hat{k} \right ) + 2 \\ = \left (2 + \lambda \right ) \left(3\right) + \left(-3 - \lambda \right) \left(1\right) + \left (-1 + 2\lambda \right)\left(-1 \right) + 2 \\ = 6 - 3\lambda - 3 - \lambda + 1 - 2\lambda + 2 \\ = 5 - 6\lambda \neq R.H.S$
Thus, the line does not lie in the given plane.
Therefore, the given statement is False.

Question:47

State True or False for the given statement.
The vector equation of the line $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$ is $r=5\hat{i}-4\hat{j}+6\hat{k}+\lambda\left ( 3\hat{i}+7\hat{j}+2\hat{k} \right )$

The given equation of the line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$
It is clear from the equation that this line passes through A (5, -4, 6) and has the direction ratios 3, 7 and 2.
The position vector of A is $\vec{a}=5\hat{i}-4\hat{j}+6\hat{k}$
And the direction vector of the line will be
We know, the vector equation of a line that passes through a given point with position vector a and b is given as
$\vec{r}=3\hat{i}+7\hat{j}+2\hat{k}$
Hence, the required line equation will be:
$\hat{r}=\left ( 5\hat{i}-4\hat{j}+6\hat{k}\right )\lambda\left ( 3\hat{i}+7\hat{j}+2\hat{k} \right )$
Thus, the statement is True.

Question:48

State True and False for the given statement:
The equation of a line, which is parallel to $2 \hat{i}+\hat{j}+3\hat{k}$ and which passes through (5, -2, 4) is $\frac{x-5}{2}=\frac{y+2}{-1}=\frac{z-4}{3}$

We know, the equation of a line in Cartesian form is
$\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$
, where a, b and c are the direction ratios and (x1, y1, z1) is a particular point on the line.
The given line is parallel to therefore it has 2, 1, 3 as direction ratios.
(a = 2, b = 1, c = 3)
The line passes through (5, -2, 4)
Substituting these values, we get the equation of line:
$\frac{x-5}{2}=\frac{y+2}{1}=\frac{z-4}{3}$
Thus, the given statement is False.

Question:49

State True or False for the given statement:
If the foot of the perpendicular drawn from the origin to a plane is (5, -3, -2) then the equation of the plane is $r.\left ( 5\hat{i}-3\hat{j}-2\hat{k} \right )=38$

Let us take O as the origin, P as the foot of the perpendicular drawn from origin to the plane.
Then the position vector OP is:
$\vec{n}=\vec{OP}=5\hat{i}-3\hat{j}-2\hat{k}$
The unit vector of n is:
$\vec{n}=\frac{\vec{n}}{\left | \vec{n} \right |}\\ \hat{n}=\frac{5\hat{i}-3\hat{j}-2\hat{k}}{\sqrt{5^{2}+(-3)^{2}+(-2)^{2}}}=\frac{5}{\sqrt{38}}\hat{i}-\frac{3}{\sqrt{38}}\hat{j}-\frac{2}{\sqrt{4}}\hat{k}$
$OP =\sqrt{\left ( 5-0 \right )^{2}+\left ( -3-0 \right )^{2}+\left (-2-0 \right )^{2}} \\ =\sqrt{25+9+4}\\ =\sqrt{38}$
Now, the equation of the plane with unit normal vector n and having a perpendicular drawn from the origin d is:
$\vec{r}.\hat{n}=d$
Therefore,
Equation of the given plane will be,
$\vec{r}.\left ( \frac{5}{\sqrt{38}}\hat{i}-\frac{3}{\sqrt{38}}\hat{j}-\frac{2}{\sqrt{4}}\hat{k} \right )=\sqrt{38}\\ \Rightarrow \vec{r}.\left ( 5\hat{i}-3\hat{j}-2\hat{k} \right )=38$
=> The given statement is True.

## Main Subtopics of NCERT Exemplar Class 12 Maths Solutions Chapter 11

The main topics and sub topics covered in this chapter of are as follows:

• Direction aspects of a line
• Direction Cosines
• Direction Ratios
• Relation between the direction cosines of a line
• Direction Cosines of a line passing through two points
• Equation of a line in space
• Equation of a line through a given point and parallel to a given vector
• Equation of a line passing through two given points
• Derivation of Cartesian form from Vector form.
• Angle between two lines
• Shortest distance between two lines
• Distance between two skew lines
• Distance between parallel lines
• Planes
• Equation of a plane in Normal form
• Equation of a plane perpendicular to a given vector through a given point
• Equation of a plane passing through three non-collinear points
• Planes passing through the intersection of two given planes
• Coplanarity of two lines
• Angle between two planes
• Distance of a point from a plane
• Angle between a line and a plane

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## What will the students learn in NCERT Exemplar Class 12 Maths Solutions Chapter 11?

• NCERT exemplar Class 12 Maths solutions chapter 11 would not only help you find volumes of objects such as cubes, cylinders, pyramids, or location of coordinates but also to expand your learning and potential to achieve academic excellence.
• The design of assembly systems in manufacturing automobiles and nanotechnology is done with 3D geometry. Various graphic designers, visual artists, and game developers use the concepts of three dimensions to create computer graphics, visual graphs, virtual reality programs, and video games.
• Geometry acts as a fundamental tool to calculate the location of interplanetary and galactic objects in space and to draft a possible trajectory and find out the entry points to a planet’s atmosphere for a space vehicle's journey.
• Thus, NCERT exemplar solutions for Class 12 Maths chapter 11 exposes one to a wide array of fields like art, technology, architecture, astronomy and physics, and aspects of geographic information systems.

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## NCERT Exemplar Class 12 Maths Solutions

 Chapter 1 Relations and Functions Chapter 2 Inverse Trigonometric Functions Chapter 3 Matrices Chapter 4 Determinants Chapter 5 Continuity and Differentiability Chapter 6 Applications of Derivatives Chapter 7 Integrals Chapter 8 Applications of Integrals Chapter 9 Differential Equations Chapter 10 Vector Algebra Chapter 12 Linear Programming Chapter 13 Probability

## Importance of NCERT Exemplar Class 12 Maths Solutions Chapter 11

· NCERT Exemplar Class 12 Maths chapter 11 solutions define Dimension as the standard measure of an object's size and shape.

· We would use the concept of Vector algebra to make three-dimensional geometry organised and straightforward.

· In Class 12 Maths NCERT exemplar solutions chapter 11, we will study the direction aspects of a line joining two points, including the direction cosines and the ratios, discuss the equations of lines and planes in space, measure the shortest distance between two lines and learn more about the Cartesian form analytical and geometric representation.

### NCERT Exemplar Class 12 Solutions

 NCERT Exemplar Class 12 Chemistry Solutions NCERT Exemplar Class 12 Physics Solutions NCERT Exemplar Class 12 Biology Solutions

### Also, check NCERT Solutions for questions given in the book:

 Chapter 1 Relations and Functions Chapter 2 Inverse Trigonometric Functions Chapter 3 Matrices Chapter 4 Determinants Chapter 5 Continuity and Differentiability Chapter 6 Application of Derivatives Chapter 7 Integrals Chapter 8 Application of Integrals Chapter 9 Differential Equations Chapter 10 Vector Algebra Chapter 11 Three Dimensional Geometry Chapter 12 Linear Programming Chapter 13 Probability

#### Must Read NCERT Solution subject wise

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#### Also Check NCERT Books and NCERT Syllabus here

1. What are the topics covered in this chapter?

This entire chapter talks about the dimensional geometry which covered vector usage to measure and determine line, planes and angles.

2. Are these solutions helpful in board exams?

Yes, for those who want a clear picture of how to solve questions in three-dimensional geometry, our NCERT exemplar Class 12 Maths solutions chapter 11 can be highly supportive.

3. How to take help from these solutions?

The best way is to use these solutions as reference, while one is solving the questions for practicing.

Yes, these questions and NCERT exemplar Class 12 Maths solutions chapter 11 can be downloaded by using the webpage to PDF tool available online.

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Here are some options you can explore to get admission in a good school even though admissions might be closed for many:

• Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

• Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush.  Reach out to the schools directly to see if they have any open seats in 10th grade.

• Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.

• Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.

In India, the design and coding fields offer exciting career options that can leverage your interest in both. Here's how you can navigate this path:

• Graphic Design Focus: Consider a Bachelor's degree in Graphic Design or a design diploma. Build a strong portfolio showcasing your creative skills. Learn the basics of HTML, CSS, and JavaScript to understand web development better. Many online resources and bootcamps offer these introductory courses.

• Coding Focus: Pursue a Computer Science degree or a coding bootcamp in India. These programs are intensive but can equip you with strong coding skills quickly. While building your coding prowess, take online courses in graphic design principles and UI/UX design.

Engineering Subjects (for a Degree):

• Information Technology (IT): This offers a good mix of web development, networking, and database management, all valuable for web design/development roles.

• Human-Computer Interaction (HCI): This is a specialized field that bridges the gap between design and computer science, focusing on how users interact with technology. It's a perfect choice if you're interested in both aspects.

• Passing NIOS in October 2024 will make you eligible for NIT admissions in 2025 . NIT admissions are based on your performance in entrance exams like JEE Main, which typically happen in January and April. These exams consider the previous year's Class 12th board results (or equivalent exams like NIOS).

Here's why 2025 is more likely:

• JEE Main 2024 Admissions: The application process for NITs through JEE Main 2024 is likely complete by now (May 2024). They consider your 2023 Class 12th marks (CBSE in this case).
• NIOS Results: Since NIOS results typically come out after the NIT admission process, your October 2024 NIOS marks wouldn't be available for JEE Main 2024.

• Focus on JEE Main: Since you have a computer science background, focus on preparing for JEE Main 2025. This exam tests your knowledge in Physics, Chemistry, and Mathematics, crucial for engineering programs at NITs.
• NIOS Preparation: Utilize the time between now and October 2024 to prepare for your NIOS exams.
• Eligibility Criteria: Remember, NITs typically require a minimum of 75% marks in Class 12th (or equivalent) for general category students (65% for SC/ST). Ensure you meet this criteria in your NIOS exams.

Yes, scoring above 99.9 percentile in CAT significantly increases your chances of getting a call from IIM Bangalore,  with your academic background. Here's why:

• High CAT Score: A score exceeding  99.9 percentile is exceptional and puts you amongst the top candidates vying for admission. IIM Bangalore prioritizes  CAT scores heavily in the shortlisting process.

• Strong Academics: Your 96% in CBSE 12th and a B.Tech degree demonstrate a solid academic foundation, which IIM Bangalore also considers during shortlisting.

However, the shortlisting process is multifaceted:

• Other Factors: IIM Bangalore considers other factors beyond CAT scores, such as your work experience (if any), XAT score (if you appear for it), academic diversity, gender diversity, and performance in the interview and Written Ability Test (WAT) stages (if shortlisted).

Here's what you can do to strengthen your application:

• Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.

• Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.

Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

Good Luck

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9