First slide

Planes in 3D

Question

If the angle between the line $x = \frac{y - 1}{2} = \frac{z - 3}{λ}$ and the plane $x + 2 y + 3 z = 4$ is $\cos^{- 1} (\sqrt{5 / 14})$ then $λ$ =

Moderate

A

$\frac{3}{2}$
B

$\frac{5}{3}$
C

$\frac{2}{3}$
D

$\frac{2}{5}$

Solution

Let $\cos^{- 1} (\sqrt{\frac{5}{14}}) = θ \Rightarrow \cos θ = \frac{\sqrt{5}}{\sqrt{14}}$
$\Rightarrow \sin θ = \frac{3}{\sqrt{14}}$
D.r's of given line $(a_{1}, b_{1}, c_{1}) = (1, 2, λ)$
D.r's of normal to the given plane $(a_{2}, b_{2}, c_{2}) = (1, 2, 3)$
$\sin θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}$
$\frac{3}{\sqrt{14}} = \frac{1 + 4 + 3 λ}{\sqrt{1 + 4 + λ^{2}} \sqrt{\sqrt{1 + 4 + 9}}}$
$λ = \frac{2}{3}$

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