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

NCERT Exemplar Class 12 Maths Solutions

Three Dimensional Geometry Exercise 11.3
Page number: 235-240, Total Questions: 49

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.

Answer:

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),
${(\cos {60}^{\circ })}^2+{(m=\cos {45}^{\circ })}^2+{(n=\cos \alpha )}^2=1$
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
$\overrightarrow{OA}=\overrightarrow{OA}(l\hat{i}+m\hat{j}+n\hat{k})$..........(ii)
We have,
$$\begin{gathered} l=\cos {60}^{\circ }=\frac{1}{2} \\ m=cos{45}^{\circ }=\frac{1}{\sqrt{2}} \\ n=\cos \alpha =\pm \frac{1}{2} \end{gathered}$$
Inserting these values of l, m and n in equation (ii),
$\overrightarrow{OA}=\left|\overrightarrow{OA}\right|(\frac{1}{2}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{2}\hat{k})$
Also ,Put $\Rightarrow \overrightarrow{OA}=10(\frac{1}{2}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{2}\hat{k})$
$\Rightarrow \overrightarrow{OA}=10\times \frac{1}{2}\hat{i}+10\times \frac{1}{\sqrt{2}}\hat{j}+10\times \frac{1}{2}\hat{k}$
$\Rightarrow \overrightarrow{OA}=5i+10\times \frac{\sqrt{2}}{\sqrt{2}}\times +10\times \frac{1}{\sqrt{2}}\hat{j}\pm 5\hat{k}$
$\Rightarrow \overrightarrow{OA}=5i+\frac{10\times \sqrt{2}}{2}\hat{j}\pm 5\hat{k}$
$\Rightarrow \overrightarrow{OA}=5i+5\sqrt{2}\hat{j}\pm 5\hat{k}$
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).

Answer:

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, $\lambda ϵ\mathbb{R}$

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

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

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

This can be further rearranged, upon which we get:

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

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

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

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

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

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.

Answer:

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 $\lambda ,\mu ϵ\mathbb{R}$
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
$(2\lambda +1,3\lambda +3,4\lambda +3)\equiv (5\mu +4,2\mu +1,\mu )$
$\Rightarrow 2\lambda +1=5\mu +4......(iii)$
$3\lambda +2=2\mu +1.....(iv)$
$4\lambda +3=\mu .....(iv)$
Substituting the value of μ from equation (v) into equation (iv),
$3\lambda +2=2(4\lambda +3)+1$
$\Rightarrow 3\lambda +2=8\lambda +6+1$
$\Rightarrow 3\lambda +2=8\lambda +7$
$\Rightarrow 8\lambda -3\lambda =2-7$
$\Rightarrow 5\lambda =-5$
$\Rightarrow \lambda =-\frac{5}{5}$
$\Rightarrow \lambda =-1$
Putting this value of $\lambda$ in eq (v),
$4(-1)+3=\mu$
$\Rightarrow \mu =-4+3$
$\Rightarrow \mu =-1$
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),
$z=4\lambda +3$
$\Rightarrow z=4(-1)+3[\because \lambda =-1]$
$\Rightarrow z=-4+3$
$\Rightarrow z=-1$
And the z-coordinate from equation (ii),
$z=\mu$
$z=-1[\because \mu =-1]$
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

Find the angle between the lines $\overrightarrow{r}=3\hat{i}-2\hat{j}+6\hat{k}+\lambda (2\hat{i}+\hat{j}+2\hat{k})$ and

Answer:

Given lines:
$\overrightarrow{r}=3\hat{i}-2\hat{j}+6\hat{k}+\lambda (2\hat{i}+\hat{j}+2\hat{k})$
$\overrightarrow{r}=(2\hat{i}-5\hat{k})+\mu (6\hat{i}+3\hat{j}+2\hat{k})$
We are instructed to find the angle between the lines.
The line $\overrightarrow{r}=3\hat{i}-2\hat{j}+6\hat{k}+\lambda (2\hat{i}+\hat{j}+2\hat{k})$ is parallel to the vector
$2\hat{i}+\hat{j}+2\hat{k}$
Let
$\overrightarrow{b_1}=2\hat{i}+\hat{j}+2\hat{k}$

Then, we can say the line $\overrightarrow{r}=3\hat{i}-2\hat{j}+6\hat{k}+\lambda (2\hat{i}+\hat{j}+2\hat{k})$ is parallel to vector $\overrightarrow{b_1}=2\hat{i}+\hat{j}+2\hat{k}$
Similarly, let $\overrightarrow{b_2}=6\hat{i}+3\hat{j}+2\hat{k}$
Then, we can say is $\overrightarrow{r}=2\hat{j}-5\hat{k}+\mu (6\hat{i}+3\hat{j}+2\hat{k})$ parallel to the vector $\overrightarrow{b_2}=6\hat{i}+3\hat{j}+2\hat{k}$
If we take θ as the angle between the lines, then cosine θ is:
$\cos \theta =\frac{\overrightarrow{b_1}\overrightarrow{b_2}}{\left|\overrightarrow{b_1}\right|\left|\overrightarrow{b_2}\right|}$
Substituting the values of $\overrightarrow{b_1}=2\hat{i}+\hat{j}+2\hat{k}$ and $\overrightarrow{b_2}=6\hat{i}+3\hat{j}+2\hat{k}$ in the above equation,
We get
$\cos \theta =\frac{(2\hat{i}+\hat{j}+2\hat{k})(6\hat{i}+3\hat{j}+2\hat{k})}{|2\hat{i}+\hat{j}+2\hat{k}||6\hat{i}+3\hat{j}+2\hat{k}|}$
Here,
$(2\hat{i}+\hat{j}+2\hat{k})(6\hat{i}+3\hat{j}+2\hat{k})=(2\times 6)+(1\times 3)+(2\times 2)$
$(2\hat{i}+\hat{j}+2\hat{k})(6\hat{i}+3\hat{j}+2\hat{k})=12+3+4$
$(2\hat{i}+\hat{j}+2\hat{k})(6\hat{i}+3\hat{j}+2\hat{k})=19...........(i)$
Also,
$|2\hat{i}+\hat{j}+2\hat{k}||6\hat{i}+3\hat{j}+2\hat{k}|=\sqrt{2^2+1^2+2^2}\sqrt{6^2+3^2+2^2}$
$|2\hat{i}+\hat{j}+2\hat{k}||6\hat{i}+3\hat{j}+2\hat{k}|=\sqrt{4+1+4}\sqrt{36+9+4}$
$|2\hat{i}+\hat{j}+2\hat{k}||6\hat{i}+3\hat{j}+2\hat{k}|=\sqrt{9}\sqrt{49}$
$|2\hat{i}+\hat{j}+2\hat{k}||6\hat{i}+3\hat{j}+2\hat{k}|=3\times 7$
$|2\hat{i}+\hat{j}+2\hat{k}||6\hat{i}+3\hat{j}+2\hat{k}|=21............(ii)$
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).

Answer:

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
$$\begin{gathered} L_1:\frac{x}{4}=\frac{y+1}{6}=\frac{z+1}{2}=\lambda \\ \frac{x}{4}=\lambda ,\frac{y+1}{6}=\lambda ,\frac{z+1}{2}=\lambda \end{gathered}$$
We must find the values of x, y, and z. Therefore,
$$\begin{gathered} Take\frac{x}{4}=\lambda \\ \Rightarrow x=4\lambda \\ Take\frac{y+1}{6}=\lambda \end{gathered}$$
$$\begin{gathered} \Rightarrow y+1=6\lambda \\ \Rightarrow y=6\lambda -1 \\ Take\frac{z+1}{2}=\lambda \end{gathered}$$
$$\begin{gathered} \Rightarrow z+1=2\lambda \\ \Rightarrow z=2\lambda -1 \end{gathered}$$
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
$L_2:\frac{x-3}{-7}=\frac{(y-9)}{-5}=\frac{z-4}{0}=\mu$
$\Rightarrow \frac{x-3}{-7}=\mu ,\frac{(y-9)}{-5}=\mu ,\frac{z-4}{0}=\mu$
We must find the values of x, y, and z. Therefore,
$$\begin{gathered} Take\frac{x-3}{-7}=\mu \\ \Rightarrow x-3=-7\mu \end{gathered}$$
$$\begin{gathered} \Rightarrow x=-7\mu +3 \\ Take\frac{(y-9)}{-5}=\mu \end{gathered}$$
$$\begin{gathered} \Rightarrow y-9=-5\mu \\ \Rightarrow y=-5\mu +9 \\ Take\frac{z-4}{0}=\mu \end{gathered}$$
$$\begin{gathered} \Rightarrow z-4=0 \\ \Rightarrow z=4 \end{gathered}$$
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
$(4\lambda ,6\lambda -1,2\lambda -1)\equiv (-7\mu +3,-5\mu +9,4)$
$$\begin{gathered} \Rightarrow 4\lambda =-7\mu +3...(i) \\ 6\lambda -1=-5\mu +9..(ii) \\ 2\lambda -1=4(iii) \end{gathered}$$
From equation (iii), we get
$$\begin{gathered} 2\lambda -1=4 \\ \Rightarrow 2\lambda =4+1 \\ \Rightarrow 2\lambda =5 \end{gathered}$$
$\Rightarrow \lambda =\frac{5}{2}$
Substituting the value of λ in equation (i),
$$\begin{gathered} 4\left(\frac{5}{2}\right)=-7\mu +3 \\ \Rightarrow 2\times 5=-7\mu +3 \end{gathered}$$
$$\begin{gathered} \Rightarrow 10=-7\mu +3 \\ \Rightarrow 7\mu =3-10 \\ \Rightarrow 7\mu =-7 \end{gathered}$$
$$\begin{gathered} \Rightarrow -\frac{7}{7} \\ \Rightarrow \mu =-1 \end{gathered}$$
Substituting these values of λ and μ in equation (ii),
$6\left(\frac{5}{2}\right)-1=-5(-1)+9$
$\Rightarrow 3\times 5-1=5+9$
$$\begin{gathered} \Rightarrow 15-1=14 \\ \Rightarrow 14=14 \end{gathered}$$
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.

Answer:

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’
$\Rightarrow y=\frac{x-q^{\prime }}{p^{\prime }}$
From z = r’y + s’;
r’y = z - s’
$\Rightarrow y=\frac{z-s^{\prime }}{r^{\prime }}$
So,
$\frac{x-q^{\prime }}{p^{\prime }}=y=\frac{z-s^{\prime }}{r^{\prime }}$
Or,
$L_2:\frac{x-q^{\prime }}{p^{\prime }}=\frac{y}{1}=\frac{z-s^{\prime }}{r^{\prime }}.......(ii)$
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 $p^{\prime }\hat{i}+\hat{j}+r^{\prime }\hat{k}$ (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,
$$\begin{gathered} (p\hat{i}+\hat{j}+r\hat{k}).(p^{\prime }\hat{i}+\hat{j}+r^{\prime }\hat{k}) \\ \Rightarrow p{p}^{\prime }+1+r{r}^{\prime }=0 \end{gathered}$$
(since, in vector dot product, $(x\hat{i}+y\hat{j}+z\hat{k})(x^{\prime }\hat{i}+y^{\prime }\hat{j}+z^{\prime }\hat{k})=x{x}^{\prime }+y{y}^{\prime }+z{z}^{\prime }=0$
Or,
$p{p}^{\prime }+r{r}^{\prime }+1=0$
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.

Answer:

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,
$MidpointofAB=(\frac{2+4}{2},\frac{3+5}{2},\frac{4+8}{2})$
$MidpointofAB=(\frac{6}{2},\frac{8}{2},\frac{12}{2})$
= (3, 4, 6).
We can represent this as a position vector, $\overrightarrow{a}=3\hat{i}+4\hat{j}+6\hat{k}$
Next, we must find the normal of the plane, $\overrightarrow{n}$
$$\begin{gathered} \overrightarrow{n}=(4-2)\hat{i}+(5-3)\hat{j}+(8-4)\hat{k} \\ \Rightarrow \overrightarrow{n}=2\hat{i}+2\hat{j}+4\hat{k} \end{gathered}$$
We know, the equation of the plane which perpendicularly bisects the line joining two given points is
$(\overrightarrow{r}-\overrightarrow{a})\overrightarrow{n}=0$
Where,
$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$
Substituting the values in the above equation,
$((x\hat{i}+y\hat{j}+z\hat{k})-(3\hat{i}+4\hat{j}+6\hat{k})).(2\hat{i}+2\hat{j}+4\hat{k})=0$
$\Rightarrow (x\hat{i}+y\hat{j}+z\hat{k}-3\hat{i}-4\hat{j}-6\hat{k}).(2\hat{i}+2\hat{j}+4\hat{k})=0$
$\Rightarrow (x\hat{i}-3\hat{i}+y\hat{j}-4\hat{j}+z\hat{k}-6\hat{k}).(2\hat{i}+2\hat{j}+4\hat{k})=0$
$\Rightarrow ((x-3)\hat{i}+(y-4)\hat{j}+(z-6)\hat{k}).(2\hat{i}+2\hat{j}+4\hat{k})=0$
$\Rightarrow 2(x-3)+2(y-4)+4(z-6)=0$
Upon further simplification,
$\Rightarrow 2x-6+2y-8+4z-24=0$
$\Rightarrow 2x+2y+4z-6-8-24=0$
$\Rightarrow 2x+2y+4z-38=0$
$\Rightarrow 2(x+y+2z-19)=0$
$$\begin{gathered} \Rightarrow x+y+2z-19=0 \\ \Rightarrow x+y+2z=19 \end{gathered}$$
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.

Answer:

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:
$$\begin{gathered} \overrightarrow{r}.\hat{n}=d \\ \Rightarrow (x\hat{i}+y\hat{j}+z\hat{k}).(l\hat{i}+m\hat{j}+n\hat{k})=d \end{gathered}$$
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,
$$\begin{gathered} l=m=n \\ \cos \alpha =\cos \beta =\cos \gamma (ii) \end{gathered}$$
Also, we know,
${\cos }^2\alpha ={\cos }^2\beta ={\cos }^2\gamma =1$
$\Rightarrow {\cos }^2\alpha ={\cos }^2\alpha ={\cos }^2\alpha =1(from(ii))$
$\Rightarrow 3{\cos }^2\alpha =1$
$\Rightarrow {\cos }^2\alpha =\frac{1}{3}$
$\Rightarrow \cos \alpha =\frac{1}{\sqrt{3}}$
This means, $l=m=n=\frac{1}{\sqrt{3}}$
if we substitute the values of l, m and n in equation (i),
$\left(\frac{1}{\sqrt{3}}\right)x+\left(\frac{1}{\sqrt{3}}\right)y+\left(\frac{1}{\sqrt{3}}\right)z=d[whered=3\sqrt{3}]$
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.

Answer:

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 = $\overrightarrow{PQ}$
$\overrightarrow{PQ}=3\hat{i}-2\hat{j}+6\hat{k}$
The vector equation of a plane is represented by $(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{n}=0$
Putting the obtained values in this equation,
$$\begin{gathered} \overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k} \\ \overrightarrow{a}=\hat{i}-3\hat{j}+3\hat{k} \\ \overrightarrow{n}=3\hat{i}-2\hat{j}+6\hat{k} \end{gathered}$$
We get,
$\Rightarrow ((x\hat{i}+y\hat{j}+z\hat{k})-(\hat{i}-3\hat{j}+3\hat{k})).(3\hat{i}-2\hat{j}+6\hat{k})=0$
$\Rightarrow (x\hat{i}-y\hat{j}+z\hat{k}-\hat{i}+3\hat{j}-3\hat{k}).(3\hat{i}-2\hat{j}+6\hat{k})=0$
$\Rightarrow (x\hat{i}-\hat{i}+y\hat{j}+3\hat{j}+z\hat{k}-3\hat{k}).(3\hat{i}-2\hat{j}+6\hat{k})=0$
$\Rightarrow ((x-1)\hat{i}+(y+3)\hat{j}+(z-3)\hat{k}).(3\hat{i}-2\hat{j}+6\hat{k})=0$
$\Rightarrow 3(x-1)+(-2)(y+3)+6(z-3)=0$
$$\begin{gathered} \Rightarrow 3(x-1)-2(y+3)+6(z-3)=0 \\ \Rightarrow 3x-3-2y-6+6z-18=0 \end{gathered}$$
$\Rightarrow 3x-2y+6z-3-6-18=0$
$$\begin{gathered} \Rightarrow 3x-2y+6z-9-18=0 \\ \Rightarrow 3x-2y+6z-27=0 \end{gathered}$$
$\Rightarrow 3x-2y+6z=27$
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).

Answer:

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:
$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$
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,
$$\begin{gathered} \left|\begin{array}{ccc}x-2& y-1& z-0\\ 3-2& -2-1& -2-0\\ 3-2& 1-1& 7-0\end{array}\right|=0 \\ \left|\begin{array}{ccc}x-2& y-1& z-0\\ 1& -3& -2\\ 1& 0& 7\end{array}\right|=0 \end{gathered}$$
$\left|\begin{array}{ccc}x-2& y-1& z-0\\ 1& -3& -2\\ 1& 0& 7\end{array}\right|=(x-2)((-3\times 7)-(-2\times 0))$

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