The foot of the perpendicular drawn from the | vedclass

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(A) The normal vector $\vec{n}$ to the plane is the vector from the origin $(0, 0, 0)$ to the foot of the perpendicular $(4, -2, -5)$.Thus,$\vec{n} =

Vedclass · Accepted Answer

$4x - 2y - 5z = 45$

Vedclass · Answer

(A) The normal vector $\vec{n}$ to the plane is the vector from the origin $(0, 0, 0)$ to the foot of the perpendicular $(4, -2, -5)$.Thus,$\vec{n} = \langle 4 - 0, -2 - 0, -5 - 0 \rangle = \langle 4, -2, -5 \rangle$.The equation of a plane with normal $\vec{n} = \langle a, b, c \rangle$ passing through a point $(x_1, y_1, z_1)$ is given by $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$.Substituting the values,we get $4(x - 4) - 2(y + 2) - 5(z + 5) = 0$.Expanding this,$4x - 16 - 2y - 4 - 5z - 25 = 0$.$4x - 2y - 5z - 45 = 0$,which simplifies to $4x - 2y - 5z = 45$.

The foot of the perpendicular drawn from the origin to the plane is $(4, -2, -5)$. Hence,the equation of the plane is

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