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(C) Let the foot of the perpendicular drawn from the point $P(3, -1, 11)$ to the line be $L$. Since $L$ lies on the line $\frac{x}{2} = \frac{y-2}{3}

Vedclass · Accepted Answer

$\sqrt{53}$

Vedclass · Answer

(C) Let the foot of the perpendicular drawn from the point $P(3, -1, 11)$ to the line be $L$. Since $L$ lies on the line $\frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4} = t$,the coordinates of $L$ are $(2t, 3t+2, 4t+3)$. The direction ratios of the line $PL$ are $(2t-3, 3t+2-(-1), 4t+3-11)$,which simplifies to $(2t-3, 3t+3, 4t-8)$. Since $PL$ is perpendicular to the line with direction ratios $(2, 3, 4)$,their dot product must be zero: $2(2t-3) + 3(3t+3) + 4(4t-8) = 0$. $4t - 6 + 9t + 9 + 16t - 32 = 0$. $29t - 29 = 0 \implies t = 1$. Substituting $t=1$ into the coordinates of $L$,we get $L(2, 5, 7)$. The length of the perpendicular $PL$ is the distance between $P(3, -1, 11)$ and $L(2, 5, 7)$: $PL = \sqrt{(2-3)^2 + (5-(-1))^2 + (7-11)^2} = \sqrt{(-1)^2 + 6^2 + (-4)^2} = \sqrt{1 + 36 + 16} = \sqrt{53}$.

The length of the perpendicular drawn from the point $(3, -1, 11)$ to the line $\frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ is

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