First slide

Planes in 3D

Question

What is the equation of the plane which passes through the e-axis and is perpendicular to the line $\frac{x - a}{\cos θ} = \frac{y + 2}{\sin θ} = \frac{z - 3}{0} ?$

Moderate

A

$x + y \tan θ = 0$
B

$y + x \tan θ = 0$
C

$x \cos θ - y \sin θ = 0$
D

$x \sin θ - y \cos θ = 0$

Solution

The plane is perpendicular to the line $\frac{x - a}{\cos θ} = \frac{y + 2}{\sin θ} = \frac{z - 3}{0}$
Hence , the direction ratios of the normal of the plane are $\cos θ, \sin θ$ and 0.
Now, the required plane passes through the e-axis. Hence, the point (0, 0, 0) lies on the plane.
Thus, we get equation of the plane as
$\begin{array}{l} \cos θ (x - 0) + \sin θ (y - 0) + 0 (z - 0) = 0 \\ \cos θx + \sin θy = 0 \\ x + ytan θ = 0 \end{array}$

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