NCERT Solutions for Miscellaneous Exercise Chapter 11 Class 12 - Three Dimensional Geometry

Question 1: Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, – 1), (4, 3, – 1).

Answer:

We can assume the line joining the origin, be OA where $O(0,0,0)$ and the point $A(2,1,1)$ and PQ be the line joining the points $P(3,5,-1)$ and $Q(4,3,-1)$ .

Then the direction ratios of the line OA will be $(2-0),(1-0),and(1-0)=2,1,1$ and that of line PQ will be

$(4-3),(3-5),and(-1+1)=1,-2,0$

So to check whether line OA is perpendicular to line PQ then,

Applying the relation we know,

$a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

$\Rightarrow 2(1)+1(-2)+1(0)=2-2+0=0$

Therefore OA is perpendicular to line PQ.

Question 2: If l ₁ , m ₁ , n ₁ and l ₂ , m ₂ , n ₂ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are $m_1{n}_2-m_2{n}_1,n_1{l}_2-n_2{l}_1,l_1{m}_2-l_2{m}_1$ .

Answer:

Given that $l_1,m_1,n_{1}and{l}_2,m_2,n_2$ are the direction cosines of two mutually perpendicular lines.

Therefore, we have the relation:

$l_1{l}_2+m_1{m}_2+n_1{n}_2=0$ .........................(1)

$l_1^2+m_1^2+n_1^2=1and{l}_2^2+m_2^2+n_2^2=1$ .............(2)

Now, let us assume $l,m,n$ be the new direction cosines of the lines which are perpendicular to the line with direction cosines. $l_1,m_1,n_{1}and{l}_2,m_2,n_2$

Therefore we have, $l{l}_1+m{m}_1+n{n}_1=0\text{and}l{l}_2+m{m}_2+n{n}_2=0$

Or, $\frac{l}{m_1{n}_2-m_2{n}_1}=\frac{m}{n_1{l}_2-n_2{l}_1}=\frac{n}{l_1{m}_2-l_2{m}_1}$

$\Rightarrow \frac{l^2}{(m_1{n}_2-m_2{n}_1{)}^2}=\frac{m^2}{(n_1{l}_2-n_2{l}_1{)}^2}=\frac{n^2}{(l_1{m}_2-l_2{m}_1{)}^2}$

$\Rightarrow \frac{l^2+m^2+n^2}{(m_1{n}_2-m_2{n}_1{)}^2+(n_1{l}_2-n_2{l}_1{)}^2+(l_1{m}_2-l_2{m}_1{)}^2}$ ......(3)

So, l,m,n are the direction cosines of the line.

where, $l^2+m^2+n^2=1$ ........................(4)

Then we know that,

$\Rightarrow (l_1^2+m_1^2+n_1^2)(l_2^2+m_2^2+n_2^2)-(l_1{l}_2+m_1{m}_2+n_1{n}_2{)}^2$

$=(m_1{n}_2-m_2{n}_1{)}^2+(n_1{l}_2-n_2{l}_1{)}^2+(l_1{m}_2-l_2{m}_1{)}^2$

So, from the equation (1) and (2) we have,

$(1)(1)-(0)=(m_1{n}_2-m_2{n}_1{)}^2+(n_1{l}_2-n_2{l}_1{)}^2+(l_1{m}_2-l_2{m}_1{)}^2$

Therefore, $(m_1{n}_2-m_2{n}_1{)}^2+(n_1{l}_2-n_2{l}_1{)}^2+(l_1{m}_2-l_2{m}_1{)}^2=1$ ..(5)

Now, we will substitute the values from the equation (4) and (5) in equation (3), to get

$\frac{l^2}{(m_1{n}_2-m_2{n}_1{)}^2}=\frac{m^2}{(n_1{l}_2-n_2{l}_1{)}^2}=\frac{n^2}{(l_1{m}_2-l_2{m}_1{)}^2}=1$

Therefore we have the direction cosines of the required line as;

$l=m_1{n}_2-m_2{n}_1$

$m=n_1{l}_2-n_2{l}_1$

$n=l_1{m}_2-l_2{m}_1$

Question 3: If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

Answer:

Direction ratios of AB are $(4-1),(5-2),(7-3)=3,3,4$

and Direction ratios of CD are $(2-(-4)),(9-3),(2-(-6))=6,6,8$

So, it can be noticed that, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=\frac{1}{2}$

Therefore, AB is parallel to CD.

Thus, we can easily say the angle between AB and CD which is either $0^{\circ }or{180}^{\circ }$ .

Question 4: Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane $\overrightarrow{r}.(\hat{i}+2\hat{j}-5\hat{k})+9=0$

Answer:

Given that the plane is passing through the point $A(1,2,3)$ so, the position vector of the point A is $\overrightarrow{r_A}=\hat{i}+2\hat{j}+3\hat{k}$ and perpendicular to the plane $\overrightarrow{r}.(\hat{i}+2\hat{j}-5\hat{k})+9=0$ whose direction ratios are $1,2,and-5$ and the normal vector is $\overrightarrow{n}=\hat{i}+2\hat{j}-5\hat{k}$

So, the equation of a line passing through a point and perpendicular to the given plane is given by,

$\overrightarrow{l}=\overrightarrow{r}+\lambda \overrightarrow{n}$ , where $\lambda ϵR$

$\Rightarrow \overrightarrow{l}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda (\hat{i}+2\hat{j}-5\hat{k})$

Question 5: Find the equation of the plane passing through (a, b, c) and parallel to the plane $\overrightarrow{r}.(\hat{i}+\hat{j}+\hat{k})=2$ .

Answer:

Given that the plane is passing through $(a,b,c)$ and is parallel to the plane $\overrightarrow{r}.(\hat{i}+\hat{j}+\hat{k})=2$

So, we have

The position vector of the point $A(a,b,c)$ is, $\overrightarrow{r_A}=a\hat{i}+b\hat{j}+c\hat{k}$

and any plane which is parallel to the plane, $\overrightarrow{r}.(\hat{i}+\hat{j}+\hat{k})=2$ is of the form,

$\overrightarrow{r}.(\hat{i}+\hat{j}+\hat{k})=\lambda$ . .......................(1)

Therefore the equation we get,

$(a\hat{i}+b\hat{j}+c\hat{k}).(\hat{i}+\hat{j}+\hat{k})=\lambda$

Or, $a+b+c=\lambda$

So, now substituting the value of $\lambda =a+b+c$ in equation (1), we get

$\overrightarrow{r}.(\hat{i}+\hat{j}+\hat{k})=a+b+c$ .................(2)

So, this is the required equation of the plane .

Now, substituting $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$ in equation (2), we get

$(x\hat{i}+y\hat{j}+z\hat{k}).(\hat{i}+\hat{j}+\hat{k})=a+b+c$

Or, $x+y+z=a+b+c$

Question 6: Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,1) crosses the YZ-plane.

Answer:

We know that the equation of the line that passes through the points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by the relation;

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

and the line passing through the points, $\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}$

$⟹\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=k(say)$

$⟹x=5-2k,y=3k+1,z=6-5k$

And any point on the line is of the form $(5-2k,3k+1,6-5k)$ .

So, the equation of the YZ plane is $x=0$

Since the line passes through YZ- plane,

we have then,

$5-2k=0$

$\Rightarrow k=\frac{5}{2}$

or $3k+1=3(\frac{5}{2})+1=\frac{17}{2}$ and $6-5k=6-5(\frac{5}{2})=\frac{-13}{2}$

So, therefore the required point is $(0,\frac{17}{2},\frac{-13}{2})$

Question 7: Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.

Answer:

We know that the equation of the line that passes through the points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by the relation;

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

and the line passing through the points, $\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}$

$⟹\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=k(say)$

$⟹x=5-2k,y=3k+1,z=6-5k$

And any point on the line is of the form $(5-2k,3k+1,6-5k)$ .

So, the equation of ZX plane is $y=0$

Since the line passes through YZ- plane,

we have then,

$3k+1=0$

$\Rightarrow k=-\frac{1}{3}$

or $5-2k=5-2(-\frac{1}{3})=\frac{17}{3}$ and $6-5k=6-5(\frac{-1}{3})=\frac{23}{3}$

So, therefore the required point is $(\frac{17}{3},0,\frac{23}{3})$

Question 8: Find the coordinates of the point where the line through (3, – 4, – 5) and (2, – 3, 1) crosses the plane 2x + y + z = 7.

Answer:

We know that the equation of the line that passes through the points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by the relation;

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

and the line passing through the points, $(3,-4,-5)and(2,-3,1)$ .

$⟹\frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}=k(say)$

$⟹\frac{x-3}{-1}=\frac{y+4}{-1}=\frac{z+5}{6}=k(say)$

$⟹x=3-k,y=k-4,z=6k-5$

And any point on the line is of the form. $(3-k,k-4,6k-5)$

This point lies on the plane, $2x+y+z=7$

$\therefore 2(3-k)+(k-4)+(6k-5)=7$

$⟹5k-3=7$

or $k=2$ .

Hence, the coordinates of the required point are $(3-2,2-4,6(2)-5)$ or $(1,-2,7)$ .

Question 9: Find the equation of the plane passing through the point (– 1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

Answer:

Given

two planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

the normal vectors of these plane are

$n_1=\hat{i}+2\hat{j}+3\hat{k}$

$n_2=3\hat{i}+3\hat{j}+\hat{k}$

Since the normal vector of the required plane is perpendicular to the normal vector of given planes, the required plane's normal vector will be :

$\overrightarrow{n}={\overrightarrow{n}}_1\times {\overrightarrow{n}}_2$

$\overrightarrow{n}=(\hat{i}+2\hat{j}+3\hat{k})\times (3\hat{i}+3\hat{j}+\hat{k})$

$\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ 3& 3& 1\end{array}\right|=\hat{i}(2-9)-\hat{j}(1-9)+\hat{k}(3-6)$

$\overrightarrow{n}=-7\hat{i}+8\hat{j}-3\hat{k}$

Now, as we know

the equation of a plane in vector form is :

$\overrightarrow{r}\cdot \overrightarrow{n}=d$

$\overrightarrow{r}\cdot (-7\hat{i}+8\hat{j}-3\hat{k})=d$

Now Since this plane passes through the point (-1,3,2)

$(-\hat{i}+3\hat{j}+2\hat{k})\cdot (-7\hat{i}+8\hat{j}-3\hat{k})=d$

$7+24-6=d$

$d=25$

Hence the equation of the plane is

$\overrightarrow{r}\cdot (-7\hat{i}+8\hat{j}-3\hat{k})=25$

Question 10: If the points (1, 1, p) and (– 3, 0, 1) be equidistant from the plane $\overrightarrow{r}.(3\hat{i}+4\hat{j}-12\hat{k})+13=0$ then find the value of p.

Answer:

Given that the points $A(1,1,p)$ and $B(-3,0,1)$ are equidistant from the plane

$\overrightarrow{r}.(3\hat{i}+4\hat{j}-12\hat{k})+13=0$

So we can write the position vector through the point $(1,1,p)$ is $\overrightarrow{a_1}=\hat{i}+\hat{j}+p\hat{k}$

Similarly, the position vector through the point $B(-3,0,1)$ is

$\overrightarrow{a_2}=-4\hat{i}+\hat{k}$

The equation of the given plane is $\overrightarrow{r}.(3\hat{i}+4\hat{j}-12\hat{k})+13=0$

and We know that the perpendicular distance between a point whose position vector is $\overrightarrow{a}$ and the plane, $\overrightarrow{n}=3\hat{i}+4\hat{j}-12\hat{k}$ and $d=-13$

Therefore, the distance between the point $A(1,1,p)$ and the given plane is

$D_1=\frac{|(\hat{i}+\hat{j}+p\hat{k}).(3\hat{i}+4\hat{j}-12\hat{k})+13|}{3\hat{i}+4\hat{j}-12\hat{k}}$

$D_1=\frac{|3+4-12p+13|}{\sqrt{3^2+4^2+(-12{)}^2}}$

$D_1=\frac{|20-12p|}{13}$ nbsp; .........................(1)

Similarly, the distance between the point $B(-1,0,1)$ , and the given plane is

$D_2=\frac{|(-3\hat{i}+\hat{k}).(3\hat{i}+4\hat{j}-12\hat{k})+13|}{3\hat{i}+4\hat{j}-12\hat{k}}$

$D_2=\frac{|-9-12+13|}{\sqrt{3^2+4^2+(-12{)}^2}}$

$D_2=\frac{8}{13}$ .........................(2)

And it is given that the distance between the required plane and the points, $A(1,1,p)$ and $B(-3,0,1)$ is equal.

$\therefore D_1=D_2$

$⟹\frac{|20-12p|}{13}=\frac{8}{13}$

therefore we have,

$⟹12p=12$

or $p=1$ or $p=\frac{7}{3}$

Question 11: Find the equation of the plane passing through the line of intersection of the planes $\overrightarrow{r}.(\hat{i}+\hat{j}+\hat{k})=1$ and $\overrightarrow{r}.(2\hat{i}+3\hat{j}-\hat{k})+4=0$ and parallel to x-axis.

Answer:

So, the given planes are:

$\overrightarrow{r}.(\hat{i}+\hat{j}+\hat{k})=1$ and $\overrightarrow{r}.(2\hat{i}+3\hat{j}-\hat{k})+4=0$

The equation of any plane passing through the line of intersection of these planes is

$[\overrightarrow{r}.(\hat{i}+\hat{j}+\hat{k})-1]+\lambda [\overrightarrow{r}.(2\hat{i}+3\hat{j}-\hat{k})+4]=0$

$\overrightarrow{r}.[(2\lambda +1)\hat{i}+(3\lambda +1)\hat{j}+(1-\lambda )\hat{k}]+(4\lambda +1)=0$ ..............(1)

Its direction ratios are $(2\lambda +1),(3\lambda +1),$ and $(1-\lambda )$ = 0

The required plane is parallel to the x-axis.

Therefore, its normal is perpendicular to the x-axis.

The direction ratios of the x-axis are 1,0, and 0.

$\therefore 1.(2\lambda +1)+0(3\lambda +1)+0(1-\lambda )=0$

$⟹2\lambda +1=0$

$⟹\lambda =-\frac{1}{2}$

Substituting $\lambda =-\frac{1}{2}$ in equation (1), we obtain

$⟹\overrightarrow{r}.[-\frac{1}{2}\hat{j}+\frac{3}{2}\hat{k}]+(-3)=0$

$⟹\overrightarrow{r}(\hat{j}-3\hat{k})+6=0$

So, the Cartesian equation is $y-3z+6=0$

Question 12: If O be the origin and the coordinates of P be (1, 2, – 3), then find the equation of the plane passing through P and perpendicular to OP.

Answer:

We have the coordinates of the points $O(0,0,0)$ and $P(1,2,-3)$ respectively.

Therefore, the direction ratios of OP are $(1-0)=1,(2-0)=2,and(-3-0)=-3$

And we know that the equation of the plane passing through the point $(x_1,y_1,z_1)$ is

$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$ where a,b,c are the direction ratios of normal.