• # question_answer 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 : [JEE Main 10-4-2019 Morning] A) -1B) 2C) -2                                D) 1

Solution :

One of the point on line is $P\left( 0,1,1 \right)$ and given point is $Q\left( \beta ,0,\beta \right).$ So, $\overrightarrow{PQ}=\beta \hat{i}-\hat{j}+(\beta +1)\hat{k}$ Hence, ${{\beta }^{2}}+1+{{(\beta +1)}^{2}}-\frac{{{\left( \beta -\beta -1 \right)}^{2}}}{2}=\frac{3}{2}$ $\Rightarrow 2{{\beta }^{2}}+2\beta =0$$\Rightarrow \beta =0,-1$ $\Rightarrow \beta =-1$(as$\beta \ne 0$)

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