Planes in 3D

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Question

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

Moderate

Solution

Let $A = (2 λ + 1, 4 λ + 3, 3 λ + 2)$
Let this point lie on the line such that AP is parallel to the plane $\bar{A P}$ is perpendicular to normal to the plane.
$\Rightarrow \vec{A P} ⊥ (3 \hat{i} + 2 \hat{j} - 2 \hat{k})$
$\Rightarrow 3. (2 λ - 2) + 2 (4 λ - 5) - 2 (3 λ) = 0$
$\Rightarrow λ = 2$
therefore, A is (5,11,8), P = (3, 8, 2) , AP = 7

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Similar Questions

The coordinates of the point where the line through $(3, - 4, - 5)$ and $(2, - 3, 1)$ crosses the plane $2 x + y + z = 7$ is

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