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

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(A) The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal vector $\vec{n} = (a, b, c)$ is given by $a(x-x_1) + b(y-y_1) + c(z-

Vedclass · Accepted Answer

$x+2y+3z=14$

Vedclass · Answer

(A) The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal vector $\vec{n} = (a, b, c)$ is given by $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$. Here, the foot of the perpendicular from the origin $(0,0,0)$ to the plane is $(1,2,3)$. This point $(1,2,3)$ lies on the plane, so it serves as $(x_1, y_1, z_1)$. The vector from the origin to the foot of the perpendicular acts as the normal vector $\vec{n}$ to the plane. Thus, $\vec{n} = (1-0, 2-0, 3-0) = (1, 2, 3)$. Substituting these values into the plane equation: $1(x-1) + 2(y-2) + 3(z-3) = 0$ $x - 1 + 2y - 4 + 3z - 9 = 0$ $x + 2y + 3z - 14 = 0$ $x + 2y + 3z = 14$.

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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