The foot of the perpendicular drawn from the | vedclass

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Question

(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,$\vec{n} = (4 - 0)\hat{i} + (-2 - 0)\hat{j} + (5 - 0)\hat{k} = 4\hat{i} - 2\hat{j} + 5\hat{k}$.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 $(4, -2, 5)$ and the normal vector $(4, -2, 5)$:$4(x - 4) - 2(y - (-2)) + 5(z - 5) = 0$$4(x - 4) - 2(y + 2) + 5(z - 5) = 0$$4x - 16 - 2y - 4 + 5z - 25 = 0$$4x - 2y + 5z - 45 = 0$$4x - 2y + 5z = 45$.Therefore,the correct option is $A$.

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

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