The point on the plane 2x - 2y + 4z + 5 = 0 t | vedclass

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Question

(A) Let $P$ be the point nearest to $A = \left(1, \frac{3}{2}, 2\right)$ on the plane $2x - 2y + 4z + 5 = 0$. The normal vector to the plane is $\vec{

Vedclass · Accepted Answer

$\left(0, \frac{5}{2}, 0\right)$

Vedclass · Answer

(A) Let $P$ be the point nearest to $A = \left(1, \frac{3}{2}, 2\right)$ on the plane $2x - 2y + 4z + 5 = 0$. The normal vector to the plane is $\vec{n} = (2, -2, 4)$. The line passing through $A$ and perpendicular to the plane is given by $P = A + c\vec{n} = \left(1 + 2c, \frac{3}{2} - 2c, 2 + 4c\right)$. Since $P$ lies on the plane,substitute the coordinates into the plane equation: $2(1 + 2c) - 2\left(\frac{3}{2} - 2c\right) + 4(2 + 4c) + 5 = 0$. $2 + 4c - 3 + 4c + 8 + 16c + 5 = 0$. $24c + 12 = 0 \Rightarrow c = -\frac{1}{2}$. Substituting $c = -\frac{1}{2}$ back into the expression for $P$: $P = \left(1 + 2(-\frac{1}{2}), \frac{3}{2} - 2(-\frac{1}{2}), 2 + 4(-\frac{1}{2})\right) = \left(0, \frac{5}{2}, 0\right)$.

The point on the plane $2x - 2y + 4z + 5 = 0$ that is nearest to $\left(1, \frac{3}{2}, 2\right)$ is

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