If P equiv (0, 1, 0) and Q equiv (0, 0, 1),th | vedclass

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(C) Let the given plane be $\pi: x + y + z - 3 = 0$. The normal vector to the plane is $\vec{n} = (1, 1, 1)$.Let $P'$ and $Q'$ be the projections of p

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$\sqrt{2}$

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(C) Let the given plane be $\pi: x + y + z - 3 = 0$. The normal vector to the plane is $\vec{n} = (1, 1, 1)$.Let $P'$ and $Q'$ be the projections of points $P$ and $Q$ onto the plane $\pi$,respectively. The length of the projection of $PQ$ on the plane is the distance $P'Q'$.Let $	heta$ be the angle between the line segment $PQ$ and the plane $\pi$. The length of the projection is given by $L = |PQ| \cos 	heta$.The vector $\vec{PQ} = Q - P = (0 - 0, 0 - 1, 1 - 0) = (0, -1, 1)$.The length $|PQ| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{2}$.The angle $\phi$ between the line $PQ$ and the normal $\vec{n}$ is given by $\cos \phi = \frac{|\vec{PQ} \cdot \vec{n}|}{|\vec{PQ}| |\vec{n}|}$.$\vec{PQ} \cdot \vec{n} = (0)(1) + (-1)(1) + (1)(1) = 0 - 1 + 1 = 0$.Since the dot product is $0$,the line $PQ$ is parallel to the plane $\pi$ (i.e.,$\phi = 90^\circ$,so $	heta = 0^\circ$).Therefore,the length of the projection is $|PQ| \cos(0^\circ) = |PQ| = \sqrt{2}$.

If $P \equiv (0, 1, 0)$ and $Q \equiv (0, 0, 1)$,then the projection of $PQ$ on the plane $x + y + z = 3$ is

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