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(D) 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

(D) 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 given 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 $L(2(1), 3(1) + 2, 4(1) + 3) = L(2, 5, 7)$. The length of the perpendicular $PL$ is the distance between $P(3, -1, 11)$ and $L(2, 5, 7)$: $PL = \sqrt{(2 - 3)^2 + (5 - (-1))^2 + (7 - 11)^2} = \sqrt{(-1)^2 + 6^2 + (-4)^2} = \sqrt{1 + 36 + 16} = \sqrt{53}$.

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

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