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(A) Let $P = (1, 2, 3)$ be the given point and $L$ be the foot of the perpendicular from $P$ to the line.Any point on the line $\frac{x - 6}{3} = \fra

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) Let $P = (1, 2, 3)$ be the given point and $L$ be the foot of the perpendicular from $P$ to the line.Any point on the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2} = \lambda$ is given by $L = (3\lambda + 6, 2\lambda + 7, -2\lambda + 7)$.The direction ratios of the line $PL$ are $(3\lambda + 6 - 1, 2\lambda + 7 - 2, -2\lambda + 7 - 3) = (3\lambda + 5, 2\lambda + 5, -2\lambda + 4)$.Since $PL$ is perpendicular to the given line with direction ratios $(3, 2, -2)$,the dot product of their direction ratios must be zero:$3(3\lambda + 5) + 2(2\lambda + 5) + (-2)(-2\lambda + 4) = 0$$9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0$$17\lambda + 17 = 0 \implies \lambda = -1$.Substituting $\lambda = -1$ into the coordinates of $L$,we get $L = (3(-1) + 6, 2(-1) + 7, -2(-1) + 7) = (3, 5, 9)$.The length of the perpendicular $PL$ is the distance between $P(1, 2, 3)$ and $L(3, 5, 9)$:$PL = \sqrt{(3 - 1)^2 + (5 - 2)^2 + (9 - 3)^2} = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ units.

Find the length of the perpendicular drawn from the point $(1, 2, 3)$ to the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2}$.

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