Question

The equation of the plane containing the straight $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=$ line 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. $x-2y+z=0$
• C. $5x+2y-4z=0$
• D. $3x+2y-3z=0$

Answer 1

Answer: (B)

$x-2y+z=0$

Plane 1: $ax+by+cz=0$ contains line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$

$\therefore 2a+3b+4c=0\cdots \left(i\right)$

Plane 2: $a^{\prime }x+b^{\prime }y+c^{\prime }z=0$ is perpendicular to plane containing lines.

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

$\therefore 3a^{\prime }+4b^{\prime }+2c^{\prime }=0$ and $4a^{\prime }+2b^{\prime }+3c^{\prime }=0$

$\Rightarrow \frac{a^{\prime }}{12-4}=\frac{b^{\prime }}{8-9}=\frac{c^{\prime }}{6-16}$

$\Rightarrow 8a-b-10c=0\cdots$ (ii)

From (i) and (ii), we get

$\frac{a}{-30+4}=\frac{b}{32+20}=\frac{c}{-2-24}$

$\Rightarrow$ Equation of plane $x-2y+z=0$