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)

$l_{1}^2+m_{1}^2+n_{1}^2 =1\ and\ l_{2}^2+m_{2}^2+n_{2}^2 =1$ .............(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, $ll_{1}+mm_{1}+nn_{1} = 0 \and\ ll_{2}+mm_{2}+nn_{2} = 0$

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 $\overrightarrow{r}.(\widehat{i}+2\widehat{j}-5\widehat{k})+9=0$

Answer:

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

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

$\vec{l} = \vec{r} + \lambda\vec{n}$ , where $\lambda \epsilon R$

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

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

Answer:

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

So, we have

The position vector of the point $A(a,b,c)$ is, $\vec{r_{A}} = a\widehat{i}+b\widehat{j}+c\widehat{k}$

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

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

Therefore the equation we get,

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

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

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

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

So, this is the required equation of the plane .

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

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

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, $\frac{x-5}{3-5} =\frac{y-1}{4-1} = \frac{z-6}{1-6}$

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

$\implies x=5-2k,\ y=3k+1,\ z=6-5k$

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 $\left ( 0,\frac{17}{2},\frac{-13}{2} \right )$

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, $\frac{x-5}{3-5} =\frac{y-1}{4-1} = \frac{z-6}{1-6}$

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

$\implies x=5-2k,\ y=3k+1,\ z=6-5k$

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

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

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, $(3,-4,-5)\ and\ (2,-3,1)$ .

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

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

$\implies x=3-k,\ y=k-4,\ z=6k-5$

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

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

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

$\implies 5k-3=7$

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 :

$\vec n = \vec n_1\times\vec n_2$

$\vec n = (\hat i+2\hat j +3\hat k )\times (3\hat i + 3\hat j +\hat k)$

$\vec n =\begin{vmatrix} \hat i &\hat j &\hat k \\ 1 &2 &3 \\ 3& 3& 1 \end{vmatrix}=\hat i(2-9)-\hat j(1-9)+\hat k (3-6)$

$\vec n =-7\hat i+8\hat j-3\hat k$

Now, as we know

the equation of a plane in vector form is :

$\vec r\cdot\vec n=d$

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

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

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

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

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

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

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

$\vec{a_{2}} = -4\widehat{i}+\widehat{k}$

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

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

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

$D_{1} = \frac{\left | (\widehat{i}+\widehat{j}+p\widehat{k}).(3\widehat{i}+4\widehat{j}-12\widehat{k})+13 \right |}{3\widehat{i}+4\widehat{j}-12\widehat{k}}$

$D_{1} = \frac{\left | 3+4-12p+13 \right |}{\sqrt{3^2+4^2+(-12)^2}}$

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

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

$D_{2} = \frac{\left | (-3\widehat{i}+\widehat{k}).(3\widehat{i}+4\widehat{j}-12\widehat{k})+13 \right |}{3\widehat{i}+4\widehat{j}-12\widehat{k}}$

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

$D_{2} = \frac{8}{13}$ .........................(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.

$\therefore D_{1} =D_{2}$

$\implies \frac{\left | 20-12p \right |}{13} =\frac{8}{13}$

therefore we have,

$\implies 12p =12$

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 $\overrightarrow{r}.\left ( \widehat{i}+\widehat{j}+\widehat{k} \right )=1$ and $\overrightarrow{r}.\left (2 \widehat{i}+3\widehat{j}-\widehat{k} \right )+4=0$ and parallel to x-axis.

Answer:

So, the given planes are:

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

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

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

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

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

$\implies 2\lambda+1 = 0$

$\implies \lambda = -\frac{1}{2}$

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

$\implies \vec{r}.\left [ -\frac{1}{2}\widehat{j}+\frac{3}{2}\widehat{k} \right ]+(-3)= 0$

$\implies \vec{r}(\widehat{j}-3\widehat{k})+6= 0$

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.

Here, the direction ratios of normal are $1,2,$ and $-3$ and the point P is $(1,2,-3)$ .

Thus, the equation of the required plane is

$1(x-1)+2(y-2)-3(z+3) = 0$

$\implies x+2y -3z-14 = 0$

Question 13: Find the equation of the plane which contains the line of intersection of the planes $\overrightarrow{r}.(\widehat{i}+2\widehat{j}+3\widehat{k})-4=0,\overrightarrow{r}.\left ( 2\widehat{i}+\widehat{j}-\widehat{k} \right )+5=0$ and which is perpendicular to the plane $\overrightarrow{r}.(5\widehat{i}+3\widehat{j}-6\widehat{k})+8=0$

Answer:

The equation of the plane passing through the line of intersection of the given plane in $\overrightarrow{r}.(\widehat{i}+2\widehat{j}+3\widehat{k})-4=0,\overrightarrow{r}.\left ( 2\widehat{i}+\widehat{j}-\widehat{k} \right )+5=0$

$\left [ \vec{r}.(\widehat{i}+2\widehat{j}+3\widehat{k})-4 \right ]+\lambda\left [ \vec{r}.\left ( 2\widehat{i}+\widehat{j}-\widehat{k} \right )+5 \right ] = 0$

$\vec{r}.[(2\lambda+1)\widehat{i}+(\lambda+2)\widehat{j}+(3-\lambda)\widehat{k}]+(5\lambda-4)= 0$ ,,,,,,,,,,,,,(1)

The plane in equation (1) is perpendicular to the plane, Therefore $5(2\lambda+1)+3(\lambda+2) -6(3-\lambda) = 0$

$\implies 19\lambda -7 = 0$

$\implies \lambda = \frac{7}{19}$

Substituting $\lambda = \frac{7}{19}$ in equation (1), we obtain

$\implies \vec{r}.\left [ \frac{33}{19}\widehat{i}+\frac{45}{19}\widehat{j}+\frac{50}{19}\widehat{k} \right ] -\frac{41}{19} = 0$

$\implies \vec{r}.(33\widehat{i}+45\widehat{j}+50\widehat{k}) -41 = 0$ .......................(4)

So, this is the vector equation of the required plane.

The Cartesian equation of this plane can be obtained by substituting $\implies \vec{r}= (x\widehat{i}+y\widehat{j}+z\widehat{k})$ in equation (1).

$(x\widehat{i}+y\widehat{j}+z\widehat{k}).(33\widehat{i}+45\widehat{j}+50\widehat{k}) -41 = 0$

Therefore we get the answer $33x+45y+50z -41 = 0$

Question 14: Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the line $\overrightarrow{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+ \lambda\left ( 3\widehat{i}+4\widehat{j}+2\widehat{k} \right )$ and the plane $\overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+\widehat{k} \right )=5$ .

Answer:

Given,

Equation of a line :

$\overrightarrow{r}=2\widehat{i}-\widehat{j}+2\widehat{k}+ \lambda\left ( 3\widehat{i}+4\widehat{j}+2\widehat{k} \right )$

Equation of the plane

$\overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+\widehat{k} \right )=5$

Let's first find out the point of intersection of line and plane.

putting the value of $\vec r$ into the equation of a plane from the equation from line

$\left (2\widehat{i}-\widehat{j}+2\widehat{k}+ \lambda\left ( 3\widehat{i}+4\widehat{j}+2\widehat{k} \right ) \right )\cdot (\hat i-\hat j+\hat k)=5$

$(2+3\lambda)-(4\lambda -1)+(2+2\lambda)=5$

$\lambda+5=5$

$\lambda=0$

Now, from the equation, any point p in line is

$P=(2+3\lambda,4\lambda-1,2+2\lambda)$

So the point of intersection is

$P=(2+3*0,4*0-1,2+2*0)=(2,-1,2)$

SO, Now,

The distance between the points (-1,-5,-10) and (2,-1,2) is

$d=\sqrt{(2-(-1))^2+(-1-(-5))^2+(2-(-10))^2}=\sqrt{9+16+144}$

$d=\sqrt{169}=13$

Hence the required distance is 13.

Question 15: Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes $\overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+2\widehat{k} \right )=5$ and $\overrightarrow{r}.\left ( 3\widehat{i}+\widehat{j}+\widehat{k} \right )=6$ .

Answer:

Given

A point through which line passes

$\vec a=\hat i+2\hat j+3\hat k$

two plane

$\overrightarrow{r}.\left ( \widehat{i}-\widehat{j}+2\widehat{k} \right )=5$ And

$\overrightarrow{r}.\left ( 3\widehat{i}+\widehat{j}+\widehat{k} \right )=6$

it can be seen that normals of the planes are

$\vec n_1=\hat i-\hat j+2\hat k$

$\vec n_2=3\hat i+\hat j+\hat k$
since the line is parallel to both planes, its parallel vector will be perpendicular to normals of both planes.

So, a vector perpendicular to both these normal vector is

$\vec d=\vec n_1\times\vec n_2$

$\vec d=\begin{vmatrix} \hat i &\hat j & \hat k\\ 1 &-1 &2 \\ 3& 1 & 1 \end{vmatrix}=\hat i(-1-2)-\hat j(1-6)+\hat k(1+3)$

$\vec d=-3\hat i+5\hat j+4\hat k$

Now a line which passes through $\vec a$ and parallels to $\vec d$ is

$L=\vec a+\lambda\vec d$

So the required line is

$L=\vec a+\lambda\vec d$

$L=\hat i+2\hat j+3\hat k+\lambda(-3\hat i+5\hat j+4\hat k)$

$L=(1-3\lambda)\hat i+(2+5\lambda)\hat j+(3+4\lambda)\hat k$

Question 16: Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{p^{2}}$ .

Answer:

The equation of plane having a, b and c intercepts with x, y and z-axis respectively is given by
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}= 1$
The distance p of the plane from the origin is given by
$\\p = \left | \frac{\frac{0}{a}+\frac{0}{b}+\frac{0}{c}-1}{\sqrt{(\frac{1}{a})^2+(\frac{1}{b})^2(\frac{1}{c})^2}} \right |\\ \\ p = \left | \frac{-1}{\sqrt{(\frac{1}{a})^2+(\frac{1}{b})^2(\frac{1}{c})^2}} \right |\\ \\ \frac{1}{p^2}= \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$
Hence proved

Question 17: Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is

(A) 2 units (B) 4 units (C) 8 units (D) $\frac{2}{\sqrt{29}}unit$

Answer:

Given equations are
$2x+3y+4z= 4 \ \ \ \ \ \ \ \ \ -(i)$
and
$4x+6y+8z= 12\\ 2(2x+3y+4z)= 12\\ 2x+3y+4z = 6 \ \ \ \ \ \ \ \ \ \ -(ii)$
Now, it is clear from equation (i) and (ii) that given planes are parallel
We know that the distance between two parallel planes $ax+by +cz = d_1 \ and \ ax+by +cz = d_2$ is given by
$D= \left | \frac{d_2-d_1}{\sqrt{a^2+b^2+c^2}} \right |$
Put the values in this equation
we will get,
$D= \left | \frac{6-4}{\sqrt{2^2+3^2+4^2}} \right |$
$D= \left | \frac{2}{\sqrt{4+9+16}} \right |= \left | \frac{2}{\sqrt{29}} \right |$
Therefore, the correct answer is (D)

Question 18: The planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are

(A) Perpendicular (B) Parallel (C) intersect y-axis (D) passes through $\left ( 0,0,\frac{5}{4} \right )$

Answer:

Given equations of planes are
$2x-y+4z=5 \ \ \ \ \ \ \ \ \ -(i)$
and
$5x-2.5y+10z=6\\ 2.5(2x-y+4z)=6\\ 2x-y+4z= 2.4 \ \ \ \ \ \ \ \ \ -(ii)$
Now, from equation (i) and (ii) it is clear that given planes are parallel to each other
$\because \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\Rightarrow \frac{2}{2}= \frac{-1}{-1}=\frac{4}{4}\Rightarrow 1=1=1$
Therefore, the correct answer is (B)