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

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(A) The given equation of the plane is $by + cz + d = 0$. Since the variable $x$ is missing from the equation,the normal vector to the plane is $\vec{

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$YOZ$-plane

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(A) The given equation of the plane is $by + cz + d = 0$. Since the variable $x$ is missing from the equation,the normal vector to the plane is $\vec{n} = 0\hat{i} + b\hat{j} + c\hat{k}$. The normal vector lies in the $YOZ$-plane. $A$ plane is perpendicular to another plane if their normal vectors are perpendicular. The normal to the $YOZ$-plane is the $x$-axis,which is $\hat{i}$. Since the dot product of the normal vector $\vec{n} = b\hat{j} + c\hat{k}$ and the $x$-axis vector $\hat{i}$ 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

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