Question

If the plane $3x-4y+5z=0$ is parallel to $\frac{2x-1}{2}=\frac{1-y}{3}=\frac{z-2}{a},$ then the value of $a$ is:

• A. $\frac{6}{4}$
• B. $-3$
• C. $\frac{1}{4}$
• D. $\frac{3}{4}$

Answer 1

The correct option is B $-3$

For the plane $3x-4y+5z=0$ to be parallel to $\frac{2x-1}{2}=\frac{1-y}{3}=\frac{z-2}{a}$, the $D.R^{\prime }s$ of the line should be perpendicular to that of the normal of the plane.
The corresponding $D.R^{\prime }s$ of the line are $(1,-3,a)$ and that of normal of plane are $(3,-4,5)$
$\therefore 3\cdot 1-4\cdot (-3)+5a=0$
$\Rightarrow 15+5a=0$
$\Rightarrow a=-\frac{15}{5}=-3$