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

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(A) Let the origin be $O(0,0,0)$ and the foot of the perpendicular be $P(2,-1,3)$.Since $OP$ is the normal to the plane,the direction ratios of the no

Vedclass · Accepted Answer

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

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(A) Let the origin be $O(0,0,0)$ and the foot of the perpendicular be $P(2,-1,3)$.Since $OP$ is the normal to the plane,the direction ratios of the normal are the same as the direction ratios of the line $OP$.The direction ratios of $OP$ are $(2-0, -1-0, 3-0) = (2, -1, 3)$.Thus,the normal vector to the plane is $\vec{n} = 2\hat{i} - \hat{j} + 3\hat{k}$.The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal vector $(a, b, c)$ is given by $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$.Substituting the point $P(2, -1, 3)$ and the normal vector $(2, -1, 3)$ into the equation:$2(x-2) - 1(y-(-1)) + 3(z-3) = 0$$2(x-2) - 1(y+1) + 3(z-3) = 0$$2x - 4 - y - 1 + 3z - 9 = 0$$2x - y + 3z - 14 = 0$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 the plane,then the equation of the plane is

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