Find the distance of the point P(3, 8, 2) fro | vedclass

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(D) Let the point $Q$ on the line be $(2\lambda + 1, 4\lambda + 3, 3\lambda + 2)$.Since the line $PQ$ is parallel to the plane $3x + 2y - 2z + 15 = 0$

Vedclass · Accepted Answer

$7$

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(D) Let the point $Q$ on the line be $(2\lambda + 1, 4\lambda + 3, 3\lambda + 2)$.Since the line $PQ$ is parallel to the plane $3x + 2y - 2z + 15 = 0$,the vector $\vec{PQ}$ must be perpendicular to the normal vector of the plane $\vec{n} = (3, 2, -2)$.$\vec{PQ} = (2\lambda + 1 - 3, 4\lambda + 3 - 8, 3\lambda + 2 - 2) = (2\lambda - 2, 4\lambda - 5, 3\lambda)$.Since $\vec{PQ} \cdot \vec{n} = 0$,we have:$3(2\lambda - 2) + 2(4\lambda - 5) - 2(3\lambda) = 0$$6\lambda - 6 + 8\lambda - 10 - 6\lambda = 0$$8\lambda - 16 = 0 \implies \lambda = 2$.Substituting $\lambda = 2$ in the coordinates of $Q$,we get $Q = (2(2) + 1, 4(2) + 3, 3(2) + 2) = (5, 11, 8)$.The distance $PQ = \sqrt{(5-3)^2 + (11-8)^2 + (8-2)^2} = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.

Find the distance of the point $P(3, 8, 2)$ from the line $\frac{x-1}{2} = \frac{y-3}{4} = \frac{z-2}{3}$ measured parallel to the plane $3x + 2y - 2z + 15 = 0$.

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