Question

If the plane $2x-y+z=0$ is parallel to the line $\frac{[2x-1]}{2}=\frac{[2-y]}{2}=\frac{[z+1]}{a}$, then the value of $a$ is

• A. $4$
• B. $-4$
• C. $2$
• D. $-2$
• E. $0$

Answer 1

The correct option is B

$-4$

Explanation for the correct answer:

Writing the equation of the given line in standard form we get

$\frac{x-{\displaystyle \frac{1}{2}}}{1}=\frac{y-2}{-2}=\frac{z+1}{a}$

We can observe that the direction ratios of the given line are $\left(1,-2,a\right)$.

Hence the vector along the line is written as $\hat{i}-2\hat{j}+a\hat{k}$

$2x-y+z=0$ is the equation of the given plane.

The direction ratios of the normal to the given plane are $\left(2,-1,1\right)$.

Hence, the normal vector is written as $2\hat{i}-\hat{j}+\hat{k}$

As the plane is parallel to the given line, the normal to the plane must be perpendicular to the given line.

Hence, the dot product of the normal vector and the vector along the given line must be zero.

$\therefore$$\left(2\hat{i}-\hat{j}+\hat{k}\right)·\left(\hat{i}-2\hat{j}+a\hat{k}\right)=0$

$\Rightarrow$ $2+2+a=0$

$\Rightarrow$ $a=-4$

So, the value of $a$ is $-4$ for the given plane to be parallel to the given line.

Hence, option(B) is the correct answer.