The foot of the perpendicular drawn from the | vedclass

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Question

(C) Let the point be $P(1, 3, 4)$ and the plane be $2x - y + z + 3 = 0$. Let $A(x_1, y_1, z_1)$ be the foot of the perpendicular from $P$ to the plane

Vedclass · Accepted Answer

$(-1, 4, 3)$

Vedclass · Answer

(C) Let the point be $P(1, 3, 4)$ and the plane be $2x - y + z + 3 = 0$. Let $A(x_1, y_1, z_1)$ be the foot of the perpendicular from $P$ to the plane.The direction ratios of the line $PA$ are $(x_1 - 1, y_1 - 3, z_1 - 4)$.The normal vector to the plane is $\vec{n} = (2, -1, 1)$.Since $PA$ is perpendicular to the plane,the direction ratios of $PA$ are proportional to the normal vector:$\frac{x_1 - 1}{2} = \frac{y_1 - 3}{-1} = \frac{z_1 - 4}{1} = \lambda$This gives $x_1 = 2\lambda + 1$,$y_1 = -\lambda + 3$,and $z_1 = \lambda + 4$.Since $A$ lies on the plane $2x - y + z + 3 = 0$,we substitute these coordinates into the plane equation:$2(2\lambda + 1) - (-\lambda + 3) + (\lambda + 4) + 3 = 0$$4\lambda + 2 + \lambda - 3 + \lambda + 4 + 3 = 0$$6\lambda + 6 = 0$$6\lambda = -6 \implies \lambda = -1$.Substituting $\lambda = -1$ back into the expressions for $x_1, y_1, z_1$:$x_1 = 2(-1) + 1 = -1$$y_1 = -(-1) + 3 = 4$$z_1 = (-1) + 4 = 3$Thus,the foot of the perpendicular is $(-1, 4, 3)$.

The foot of the perpendicular drawn from the point $(1, 3, 4)$ to the plane $2x - y + z + 3 = 0$ is:

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