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 $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is

• A. x +2y - 2z = 0
• B. 3x + 2y - 2z = 0
• C. x - 2y + z =0
• D. 5x + 2y - 4z = 0

Answer 1

The correct option is C

x - 2y + z =0

The DR's of normal to the plane containing $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}and\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$
$n_1=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 4& 2\\ 4& 2& 3\end{array}\right|=(8\hat{i}-\hat{j}-10\hat{k})$
n(a,b,c)


Also, equation of plane containing$\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and DR's of
normal to be $n_1=a\hat{i}+b\hat{j}+c\hat{k}\Rightarrow ax+by+cz=0\left(i\right)$
Where, $n_1.n_2=0\Rightarrow 8a-b-10c=0\left(ii\right)$
and $n_2\perp (2\hat{i}+3\hat{j}+4\hat{k})\Rightarrow 2a+3b+4c=0\left(iii\right)$
From Eqs (ii) and (iii),
$\frac{a}{-4+30}=\frac{b}{-20-32}=\frac{c}{24+2}\Rightarrow \frac{a}{-4+30}=\frac{b}{-20-32}=\frac{c}{24+2}\Rightarrow \frac{a}{26}=\frac{b}{-52}=\frac{c}{26}\Rightarrow \frac{a}{1}=\frac{b}{-2}=\frac{c}{1}\left(iv\right)$

From eqs (i) and (iv), we get $x-2y+z=0$