Question:
If the length of the perpendicular from the point $(\beta, 0, \beta)(\beta \neq 0)$ to the line, $\frac{x}{1}=\frac{y-1}{0}=\frac{z+1}{-1}$ is
$\sqrt{\frac{3}{2}}$, then $\beta$ is equal to :
Correct Option: 1
Solution:
One of the point on line is $\mathrm{P}(0,1,-1)$ and given point is $Q(\beta, 0, \beta)$.
So, $\overrightarrow{\mathrm{PQ}}=\beta \hat{\mathrm{i}}-\hat{\mathrm{j}}+(\beta+1) \hat{\mathrm{k}}$
Hence, $\beta^{2}+1+(\beta+1)^{2}-\frac{(\beta-\beta-1)^{2}}{2}=\frac{3}{2}$
$\Rightarrow 2 \beta^{2}+2 \beta=0$
$\Rightarrow \beta=0,-1$
$\Rightarrow \beta=-1 \quad$ (as $\beta \neq 0)$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.