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 straight 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 $x-2y+z=0$

perpendicular to $(3,4,2)$ and $(4,2,3)$

$\Rightarrow$ $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 4& 2\\ 4& 2& 3\end{array}\right|=\hat{i}(12-4)-\hat{j}(9-8)+\hat{k}(6-16)=8\hat{i}-\hat{j}-10\hat{k}$

The point is $(8,-1,-10)$

normal of plane $=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 8& -1& -10\\ 2& 3& 4\end{array}\right|=\hat{i}(-4+30)-\hat{j}(32+20)+\hat{j}(24+2)=26\hat{i}-52\hat{j}+26\hat{k}$

equation of plane $26\hat{i}-52\hat{j}+26\hat{k}.x\hat{i}-y\hat{j}+z\hat{k}=0$

$26x-52y+26z=0$

$x-2y+z=0$