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 $\frac{x}{4}=\frac{y}{z}=\frac{z}{3}$
$n_1=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 4& 2\\ 4& 2& 3\end{array}\right|=(8\hat{i}-\hat{j}-10\hat{k})$

Also, equation of plane containing $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and DR's of normal to be $n_1=a\hat{i}+b\hat{j}+c\hat{k}$
$\Rightarrow ax+by+cz=\mathrm{0....}$(i)
where, $n_1:n_2=0$
$\Rightarrow 8a-b-10c=\mathrm{0....}$(ii)
and $n_2\perp (2\hat{i}+3\hat{j}+4\hat{k})$
$\Rightarrow 2a+3b+4c=\mathrm{0....}$ (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 $x-2y+z=0$