In three-dimensional space,if f(x, y, z) = xy | vedclass

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(C) Given the equation $f(x, y, z) = xy + xz = 0$. Factoring the expression,we get $x(y + z) = 0$. This equation is satisfied if $x = 0$ or $y + z = 0

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pair of perpendicular planes

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(C) Given the equation $f(x, y, z) = xy + xz = 0$. Factoring the expression,we get $x(y + z) = 0$. This equation is satisfied if $x = 0$ or $y + z = 0$. In three-dimensional space,$x = 0$ represents the $yz$-plane. The equation $y + z = 0$ represents a plane passing through the $x$-axis. The normal vector to the plane $x = 0$ is $\vec{n_1} = (1, 0, 0)$. The normal vector to the plane $y + z = 0$ is $\vec{n_2} = (0, 1, 1)$. The dot product of the normal vectors is $\vec{n_1} \cdot \vec{n_2} = (1)(0) + (0)(1) + (0)(1) = 0$. Since the dot product of the normal vectors is $0$,the two planes are perpendicular to each other.

In three-dimensional space,if $f(x, y, z) = xy + xz$,then the locus of all points which satisfy the equation $f(x, y, z) = 0$ is -

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