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

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

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$\sqrt{2}$

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(B) The vector representing the line segment $PQ$ is $\vec{PQ} = (0-0, 0-1, 1-0) = (0, -1, 1)$.The normal vector to the plane $x+y+z=3$ is $\vec{n} = (1, 1, 1)$.The length of the projection of a vector $\vec{v}$ onto a plane with normal $\vec{n}$ is given by the formula $L = |\vec{v}| \sin(	heta)$,where $	heta$ is the angle between the vector $\vec{v}$ and the normal $\vec{n}$.First,calculate the magnitude of $\vec{PQ}$:$|\vec{PQ}| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{2}$.Next,calculate the cosine of the angle $	heta$ between $\vec{PQ}$ and $\vec{n}$:$\cos(	heta) = \frac{|\vec{PQ} \cdot \vec{n}|}{|\vec{PQ}| |\vec{n}|} = \frac{|(0)(1) + (-1)(1) + (1)(1)|}{\sqrt{2} \cdot \sqrt{1^2+1^2+1^2}} = \frac{|0 - 1 + 1|}{\sqrt{2} \cdot \sqrt{3}} = 0$.Since $\cos(	heta) = 0$,the angle $	heta = 90^\circ$,which means $\sin(	heta) = 1$.Therefore,the length of the projection is $L = |\vec{PQ}| \sin(90^\circ) = \sqrt{2} \cdot 1 = \sqrt{2}$.

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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