Question

The value of m for which straight line $2x-3y+4=0,x-4y+4z+5$ is parallel to the plane $2x-y+mz-2=0$

• A. $3$
• B. $-2$
• C. $\frac{-16}{5}$
• D. $-4$

Answer 1

The correct option is B $-2$

Direction ratios of normal to planes are $(3,-2,1)$ and $(4,-3,4)$ respectively.

So, the direction ratio of required line is cross product of these normal.

i.e., $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -2& 1\\ 4& -3& 4\end{array}\right|=-5\hat{i}-8\hat{j}-\hat{k}$

Therefore, direction ratio of line is $(5,8,1)$

Now line is parallel to $2x-y+mz-2=0$

So, dot product of line and normal to plane$=0$

$\therefore 10-8+m=0$

$m=-2$.