Question

Equation of the plane containing the straight line $\frac{x}{2}=\frac{y}{3}= \frac{z}{4}$ and perpendicular to the plane containing the staight lines $\frac{x}{2}=\frac{y}{4}= \frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}= \frac{z}{3}$ is

• A. $x + 2y - 2z = 0$
• B. $3x + 2y - 2z = 0$
• C. $x - 2y + 2 = 0$
• D. $5x + 2y - 42 = 0$

Answer 1

Answer: (C)

The DR's of normal to the plane containing $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{z}=\frac{z}{3}$
$n _{1}= \begin{vmatrix} \hat{ i } & \hat{ j } & \hat{ k } \\3 & 4 & 2 \\4 & 2 & 3\end{vmatrix}=(8 \hat{ i }-\hat{ j }-10 \hat{ k })$

Also, equation of plane containing $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and DR's of normal to be $n _{1}=a \hat{ i }+b \hat{ j }+c \hat{ k }$
$\Rightarrow a x+b y+c z=0 ....$(i)
where, $n _{1}: n _{2}=0$
$\Rightarrow 8 a-b-10 c=0 ....$(ii)
and $ n _{2} \perp(2 \hat{ i }+3 \hat{ j }+4 \hat{ k }) $
$ \Rightarrow 2 a+3 b+4 c=0 ....$ (iii)
From Eqs (ii) and (iii)
$\frac{a}{-4+30}=\frac{b}{-20-32}=\frac{c}{24+2}$
$\Rightarrow \frac{a}{26}=\frac{b}{-52}=\frac{c}{26}$
$\Rightarrow \frac{a}{1}=\frac{b}{-2}=\frac{c}{1} ....$(iv)
Form Eqs. (i) and (iv), required equation of plane is $x-2 y+z=0$