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

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(C) Let the 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

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$\sqrt{2}$

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(C) Let the 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 $P(0, 1, 0)$ and $Q(0, 0, 1)$ onto the plane.The vector $\vec{PQ} = Q - P = (0, -1, 1)$.The length of the projection of a line segment $PQ$ on a plane is given by $L' = \sqrt{|PQ|^2 - |PQ \cdot \hat{n}|^2}$,where $\hat{n}$ is the unit normal vector to the plane.First,calculate the length $|PQ| = \sqrt{(0-0)^2 + (0-1)^2 + (1-0)^2} = \sqrt{0 + 1 + 1} = \sqrt{2}$.The unit normal vector is $\hat{n} = \frac{(1, 1, 1)}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{1}{\sqrt{3}}(1, 1, 1)$.Now,calculate the dot product $\vec{PQ} \cdot \hat{n} = (0, -1, 1) \cdot \frac{1}{\sqrt{3}}(1, 1, 1) = \frac{1}{\sqrt{3}}(0 - 1 + 1) = 0$.Since the dot product is $0$,the line segment $PQ$ is parallel to the plane.Therefore,the length of the projection $L'$ is equal to the length of the original segment $|PQ|$.$L' = \sqrt{|PQ|^2 - 0^2} = |PQ| = \sqrt{2}$.

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

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