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

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(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

$YOZ$

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(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:

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