First slide

Planes in 3D

Question

If angle $θ$ between the line $\frac{x + 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{2}$ and the plane $2 x - y + \sqrt{λ} z + 4 = 0$ is such that $\sin θ = \frac{1}{3},$ the value of $λ$ is

Moderate

A

$\frac{- 3}{5}$
B

$\frac{5}{3}$
C

$\frac{- 4}{3}$
D

$\frac{3}{4}$

Solution

The line is $\frac{x + 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{2}$ and the plane is $2 x - y + \sqrt{λ} z + 4 = 0$
If $θ$ is the angle between the line and the plane, then 90^o - $θ$ is the angle between the line and normal to the plane.
Thus,
$\begin{array}{l} \cos (90^{\circ} - θ) = \frac{(1) (2) + (2) (- 1) + (2) (\sqrt{λ})}{\sqrt{1 + 4 + 4} \sqrt{4 + 1 + λ}} \\ or \sin θ = \frac{2 - 2 + 2 \sqrt{λ}}{3 \sqrt{5 + λ}} or \frac{1}{3} = \frac{2 \sqrt{λ}}{3 \sqrt{5 + λ}} \\ or \sqrt{5 + λ} = 2 \sqrt{λ} \\ or 5 + λ = 4 λ \\ or 3 λ = 5 \\ or λ = \frac{5}{3} \end{array}$

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