First slide

Planes in 3D

Question

The length of the projection of the line segment joining the points $(- 1, - 2, 2) and (1, 0, 2)$ on the plane $x + 3 y - 5 z - 2 = 0$ is equal to

Moderate

A

$\frac{6 \sqrt{6}}{\sqrt{35}}$
B

$\frac{216}{35}$
C

$\frac{\sqrt{8}}{\sqrt{35}}$
D

None

Solution

Find the foot of the perpendiculars   $P, Q$   of the points $A (- 1, - 2, 2) and B (1, 0, 2)$ on the plane $x + 3 y - 5 z - 2 = 0$ and then find the distance between those two points.
The length $P Q$ is projection of    $A B$ on the plane
It implies $P Q = A B \sin θ$ , where $θ$ is angle between the line $\bar{A B}$ and the normal vector to the plane.
Hence,
$\begin{matrix} P Q = \sqrt{2^{2} + 2^{2}} \sqrt{(1 - \cos^{2} θ)} \\ = \sqrt{8} \sqrt{1 - \frac{2 (1) + 2 (3) + 0 (- 5)}{\sqrt{4 + 4 + 0} \sqrt{1 + 9 + 25}}} \\ = \sqrt{8} \sqrt{1 - \frac{8}{\sqrt{8} \sqrt{35}}} \\ = \sqrt{8 \times (1 - \frac{8}{35})} \\ = \sqrt{\frac{8 \times 27}{35}} \\ = \frac{6 \sqrt{6}}{\sqrt{35}} \end{matrix}$
Therefore, the projection is $\frac{6 \sqrt{6}}{\sqrt{35}}$

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