If the foot of the perpendicular from (0,0,0) | vedclass

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Question

(A) The normal vector $\vec{n}$ to the plane is the vector joining the origin $(0,0,0)$ to the foot of the perpendicular $(1,2,3)$.$\vec{n} = (1-0)\ha

Vedclass · Accepted Answer

$x+2y+3z=14$

Vedclass · Answer

(A) The normal vector $\vec{n}$ to the plane is the vector joining the origin $(0,0,0)$ to the foot of the perpendicular $(1,2,3)$.$\vec{n} = (1-0)\hat{i} + (2-0)\hat{j} + (3-0)\hat{k} = \hat{i} + 2\hat{j} + 3\hat{k}$.The equation of a plane passing through a point $(x_0, y_0, z_0)$ with normal vector $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ is given by $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.Substituting the point $(1,2,3)$ and the normal vector $(1,2,3)$,we get:$1(x-1) + 2(y-2) + 3(z-3) = 0$$x - 1 + 2y - 4 + 3z - 9 = 0$$x + 2y + 3z - 14 = 0$$x + 2y + 3z = 14$.

If the foot of the perpendicular from $(0,0,0)$ to a plane is $(1,2,3)$,then the equation of the plane is

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