Question

If the length of the perpendicular from the point $(\beta ,0,\beta )(\beta \ne 0)$ to the line, $\frac{x}{1}=\frac{y-1}{0}=\frac{z+1}{-1}$ is $\sqrt{\frac{3}{2}}$, then $\beta$ is equal to :

• A. $-2$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

The correct option is B $-1$

$L:\frac{x}{1}=\frac{y-1}{0}=\frac{z+1}{-1}=\alpha$
Let $C(\alpha ,1,-\alpha -1)$ be the any point on the line $L$.

Let $A$ be $(\beta ,0,\beta )$ and $B(1,0,-1)$ be the point on the line $L$, such that $AC$ is perpendicular to the line $L$.

$\Rightarrow \left(1\right)(\beta -\alpha )+0(0-1)+(-1)(\beta -(-\alpha -1\left)\right)=0\Rightarrow \alpha =-\frac{1}{2}\Rightarrow C(-\frac{1}{2},1,-\frac{3}{2})\because AC=\sqrt{\frac{3}{2}}$
${(\beta +\frac{1}{2})}^2+1+{(\beta +\frac{3}{2})}^2=\frac{3}{2}\Rightarrow {\beta }^2+\beta +\frac{1}{4}+1+{\beta }^2+3\beta +\frac{9}{4}=\frac{3}{2}\Rightarrow 2{\beta }^2+4\beta +2=0\Rightarrow \beta =-1$