In space,the equation by + cz + d = 0 represe | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given equation of the plane is $by + cz + d = 0$. This can be written as $0 \cdot x + b \cdot y + c \cdot z + d = 0$. The normal vector to thi

Vedclass · Accepted Answer

$YOZ$

Vedclass · Answer

(A) The given equation of the plane is $by + cz + d = 0$. This can be written as $0 \cdot x + b \cdot y + c \cdot z + d = 0$. The normal vector to this plane is $\vec{n_1} = (0, b, c)$. The $YOZ$ plane (or $yz$-plane) has the equation $x = 0$,which can be written as $1 \cdot x + 0 \cdot y + 0 \cdot z = 0$. The normal vector to the $YOZ$ plane is $\vec{n_2} = (1, 0, 0)$. Two planes are perpendicular if the dot product of their normal vectors is zero. $\vec{n_1} \cdot \vec{n_2} = (0 	imes 1) + (b 	imes 0) + (c 	imes 0) = 0 + 0 + 0 = 0$. Since the dot product is $0$,the plane $by + cz + d = 0$ is perpendicular to the $YOZ$ plane.

In space,the equation $by + cz + d = 0$ represents a plane perpendicular to the plane:

Similar Questions

Find the equation of the plane passing through the origin and parallel to the plane $3x - 4y + 5z - 6 = 0$.

The distance of a point $(1, 2, -1)$ from the plane $x - 2y + 4z + 10 = 0$ is

Prove that if a plane has the intercepts $a, b, c$ and is at a distance of $p$ units from the origin,then $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{p^{2}}$.

Find the equation of the plane that passes through the three points $(1, 1, -1)$,$(6, 4, -5)$,and $(-4, -2, 3)$.

If $A$ and $B$ are the feet of the perpendiculars drawn from point $Q(a, b, c)$ to the planes $yz$ and $zx$ respectively,then the equation of the plane passing through the points $A, B$ and the origin $O(0, 0, 0)$ is $.........$

Vedclass Test Series

Exam Paper Generator

Online Exam Module