If P=(0,1,0) and Q=(0,0,1), then the length o | vedclass

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(B) Let the vector $\vec{v} = \vec{PQ} = (0-0, 0-1, 1-0) = (0, -1, 1)$.The normal vector to the plane $x+y+z=3$ is $\vec{n} = (1, 1, 1)$.The length of

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$\sqrt{2}$

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(B) Let the vector $\vec{v} = \vec{PQ} = (0-0, 0-1, 1-0) = (0, -1, 1)$.The normal vector to the plane $x+y+z=3$ is $\vec{n} = (1, 1, 1)$.The length of the projection of a vector $\vec{v}$ on a plane with normal $\vec{n}$ is given by $|\vec{v}| \sin 	heta$, where $	heta$ is the angle between the vector and the normal.First, calculate the magnitude of $\vec{v}$: $|\vec{v}| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{2}$.Next, calculate $\cos 	heta = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}| |\vec{n}|} = \frac{|(0)(1) + (-1)(1) + (1)(1)|}{\sqrt{2} \cdot \sqrt{1^2+1^2+1^2}} = \frac{|0-1+1|}{\sqrt{2} \cdot \sqrt{3}} = 0$.Since $\cos 	heta = 0$, the vector $\vec{v}$ is parallel to the plane, so $\sin 	heta = 1$.The length of the projection is $|\vec{v}| \sin 	heta = \sqrt{2} \cdot 1 = \sqrt{2}$.

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

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