Question

A plane is making intercepts $2,3,4$ on $X, Y$ and $Z$ -axes respectively. Another plane is passing through the point $(-1,6,2)$ and is perpendicular to the line joining the points $(1,2,3)$ and $(-2,3,4)$. Then angle between the two planes is

• A. $90^{\circ}$
• B. $\cos ^{-1} \sqrt{\frac{12}{61}}$
• C. $\cos ^{-1} \sqrt{\frac{11}{61}}$
• D. $\cos ^{-1} \sqrt{\frac{5}{6}}$

Answer 1

Answer: (C)

Given, $X$ - intercept $(a)=2$
$Y$ - intercept $(b)=3$
$Z$ - intercept $(c)=4$
$\therefore $ Equation of the plane is $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$
$6 x+4 y+3 z=12\,\,\,...(i)$
Given, points are $A=(-1,6,2)$
$B=(1,2,3)$
$C=(-2,3,4)$
DR's of $B C =(-2,-1,3-2,4-3)$
$=(-3,1,1)$
$\therefore $ Equation of plane passing through $A(-1,6,2)$ and having DR's $(-3,1,1)$ is given by
$a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0$
$-3(x+1)+1(y-6)+1(z-2)=0$
$-3 x-3+y-6+z-2=0$
$-3 x+y+z-11=0$
$-3 x+y+z=11\,\,\,..(ii)$
Angle between the planes (i) and (ii) is
$\cos \theta =\frac{\left|a_{1} \,a_{2}+b_{1}\, b_{2}+c_{1} \,c_{2}\right|}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} $
$=\frac{|(6)(-3)+(4)(1)+(3)(1)|}{\sqrt{36+16+9} \sqrt{9+1+1}} $
$=\frac{|-18+4+3|}{\sqrt{61} \sqrt{11}} $
$=\frac{(\sqrt{11})^{2}}{\sqrt{61} \sqrt{11}} $
$=\frac{\sqrt{11}}{\sqrt{61}}$
$ \cos \theta =\sqrt{\frac{11}{61}} $
$\therefore \theta= \cos ^{-1} \sqrt{\frac{11}{61}} $