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

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

Question

(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

Vedclass · Accepted Answer

pair of perpendicular planes

Vedclass · Answer

(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 -

Similar Questions

Distance between the two planes: $2x + 3y + 4z = 4$ and $4x + 6y + 8z = 12$ is

If the foot of the perpendicular drawn from the point $P(2,0,-3)$ to the plane $\pi$ is $F(1,-2,0)$ and the equation of the plane $\pi$ is $ax+by-3z+d=0$,then $a+b+d=$

The equation of the planes parallel to the plane $x - 2y + 2z - 3 = 0$ which are at a unit distance from the point $(1, 2, 3)$ is $ax + by + cz + d = 0$. If $(b - d) = K(c - a)$,then the positive value of $K$ is

The length of the perpendicular from the origin to the plane passing through the point $a$ and containing the line $r = b + \lambda c$ is

The mirror image of the point $A(1, 2, 3)$ in a plane is $B\left(-\frac{7}{3}, -\frac{4}{3}, -\frac{1}{3}\right)$. Which of the following points lies on this plane?

Vedclass Test Series

Exam Paper Generator

Online Exam Module