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

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(A) Let the plane be $ax + by + cz + d = 0$. Since the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane is $(2, -1, 3)$,the normal v

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

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(A) Let the plane be $ax + by + cz + d = 0$. Since the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane is $(2, -1, 3)$,the normal vector $\vec{n}$ to the plane is the vector from the origin to the foot of the perpendicular,which is $\vec{n} = (2 - 0)\hat{i} + (-1 - 0)\hat{j} + (3 - 0)\hat{k} = 2\hat{i} - \hat{j} + 3\hat{k}$. Thus,the equation of the plane is $2x - y + 3z = D$. Since the point $(2, -1, 3)$ lies on the plane,we substitute these coordinates into the equation: $2(2) - (-1) + 3(3) = D$ $4 + 1 + 9 = D$ $D = 14$. 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 $(0, 0, 0)$ to a plane,then the equation of that plane is:

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