If the length of the perpendicular from

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 :

  1. $-1$

  2. 2

  3. $-2$

  4. 1


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 $\beta \neq 0)$

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