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(B) Let the line be $\frac{x}{1} = \frac{y - 1}{0} = \frac{z + 1}{-1} = \lambda$. Any point $C$ on the line is given by $(\lambda, 1, -\lambda - 1)$.

Vedclass · Accepted Answer

$-1$

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(B) Let the line be $\frac{x}{1} = \frac{y - 1}{0} = \frac{z + 1}{-1} = \lambda$. Any point $C$ on the line is given by $(\lambda, 1, -\lambda - 1)$. The given point is $P(\beta, 0, \beta)$. The direction vector of the line is $\vec{v} = (1, 0, -1)$. The vector $\vec{PC} = (\lambda - \beta, 1 - 0, -\lambda - 1 - \beta) = (\lambda - \beta, 1, -\lambda - \beta - 1)$. Since $PC$ is perpendicular to the line,$\vec{PC} \cdot \vec{v} = 0$: $(\lambda - \beta)(1) + (1)(0) + (-\lambda - \beta - 1)(-1) = 0$ $\lambda - \beta + \lambda + \beta + 1 = 0$ $2\lambda + 1 = 0 \implies \lambda = -\frac{1}{2}$. Thus,the point $C$ is $(-\frac{1}{2}, 1, -\frac{1}{2})$. The length of the perpendicular $PC$ is $\sqrt{(\beta - (-\frac{1}{2}))^2 + (0 - 1)^2 + (\beta - (-\frac{1}{2}))^2} = \sqrt{\frac{3}{2}}$. Squaring both sides: $(\beta + \frac{1}{2})^2 + 1 + (\beta + \frac{1}{2})^2 = \frac{3}{2}$. $2(\beta + \frac{1}{2})^2 = \frac{3}{2} - 1 = \frac{1}{2}$. $(\beta + \frac{1}{2})^2 = \frac{1}{4}$. $\beta + \frac{1}{2} = \pm \frac{1}{2}$. Case $1$: $\beta + \frac{1}{2} = \frac{1}{2} \implies \beta = 0$ (Rejected as $\beta 
eq 0$). Case $2$: $\beta + \frac{1}{2} = -\frac{1}{2} \implies \beta = -1$. Therefore,$\beta = -1$.

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

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