First slide

Planes in 3D

Question

If the angles made by the normal of the plane $2 x + 3 y - 4 z - 16 = 0$ with the coordinates axes X, Y, Z are , then $C o s^{- 1} k_{1}, C o s^{- 1} k_{2}, C o s^{- 1} k_{3}$ respectively are

Moderate

A

$\frac{2}{\sqrt{29}}, \frac{3}{\sqrt{29}}, \frac{4}{\sqrt{29}}$
B

$\frac{2}{\sqrt{29}}, - \frac{3}{\sqrt{29}}, \frac{4}{\sqrt{29}}$
C

$\frac{2}{\sqrt{29}}, \frac{3}{\sqrt{29}}, - \frac{4}{\sqrt{29}}$
D

$\frac{1}{2}, \frac{1}{3}, - \frac{1}{4}$

Solution

The d.r’s of normal of the plane $2 x + 3 y - 4 z - 16 = 0$ are $(2, 3, - 4)$ and the d.c’s are $(\frac{2}{\sqrt{29}}, \frac{3}{\sqrt{29}}, \frac{- 4}{\sqrt{29}})$
$∴ \cos α = \frac{2}{\sqrt{29}}, \cos β = \frac{3}{\sqrt{29}}, \cos γ = \frac{- 4}{\sqrt{29}}$
Where $α, β, γ$ are the angle made by the normal with X, Y, Z axes respectiely.
$∴ α = C o s^{- 1} (\frac{2}{\sqrt{29}}), β = C o s^{- 1} (\frac{3}{\sqrt{29}})$
$γ = C o s^{- 1} (\frac{- 4}{\sqrt{29}})$
$∴ k_{1}, k_{2}, k_{3}$ values are $\frac{2}{\sqrt{29}}, \frac{3}{\sqrt{29}}, \frac{- 4}{\sqrt{29}}$ respectively

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