If (2, -1, 3) is the foot of the perpendicula | vedclass

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(B) Let the equation of the plane be $ax + by + cz + d = 0$. Since the line segment joining the origin $(0, 0, 0)$ and the point $(2, -1, 3)$ is perpe

Vedclass · Accepted Answer

$2x - y + 3z - 14 = 0$

Vedclass · Answer

(B) Let the equation of the plane be $ax + by + cz + d = 0$. Since the line segment joining the origin $(0, 0, 0)$ and the point $(2, -1, 3)$ is perpendicular to the plane, the direction ratios of the normal to the plane are given by the coordinates of the foot of the perpendicular: $a = 2 - 0 = 2$ $b = -1 - 0 = -1$ $c = 3 - 0 = 3$ Thus, the equation of the plane is $2x - y + 3z + d = 0$. Since the plane passes through the point $(2, -1, 3)$, we substitute these coordinates into the equation: $2(2) - (-1) + 3(3) + d = 0$ $4 + 1 + 9 + d = 0$ $14 + d = 0$ $d = -14$ Therefore, the equation of the plane is $2x - y + 3z - 14 = 0$.

If $(2, -1, 3)$ is the foot of the perpendicular drawn from the origin to a plane, then the equation of that plane is

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