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 $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$
$6x+4y+3z=12...(i)$
Given, points are $A=(-1,6,2)$
$B=(1,2,3)$
$C=(-2,3,4)$
DR's of $BC=(-2,-1,3-2,4-3)$
$=(-3,1,1)$
$\therefore$ Equation of plane passing through $A(-1,6,2)$ and having DR's $(-3,1,1)$ is given by
$a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0$
$-3(x+1)+1(y-6)+1(z-2)=0$
$-3x-3+y-6+z-2=0$
$-3x+y+z-11=0$
$-3x+y+z=11..(ii)$
Angle between the planes (i) and (ii) is
$\cos \theta =\frac{|a_1{a}_2+b_1{b}_2+c_1{c}_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$=\frac{|(6)(-3)+(4)(1)+(3)(1)|}{\sqrt{36+16+9}\sqrt{9+1+1}}$
$=\frac{|-18+4+3|}{\sqrt{61}\sqrt{11}}$
$=\frac{(\sqrt{11}{)}^2}{\sqrt{61}\sqrt{11}}$
$=\frac{\sqrt{11}}{\sqrt{61}}$
$\cos \theta =\sqrt{\frac{11}{61}}$
$\therefore \theta ={\cos }^{-1}\sqrt{\frac{11}{61}}$