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

Answer: (C)

$x-2y+z=0$

vector normal to plane $P_1=n_1=\left[\begin{array}{ccc}i& j& k\\ 3& 4& 2\\ 4& 2& 3\end{array}\right]=8i-j-10k$

vector normal to plane $P_2=n_2=\left[\begin{array}{ccc}i& j& k\\ 8& -1& -10\\ 2& 3& 4\end{array}\right]=26i-52j+26k$

$P_2$ is perpendicular to plane $P_1$ and is leaking the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$

The plane $P_2$ having normal $n_2$ passing through $(0,0,0)$ is

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

$x-2y+z=0$