Let A(2,5,7) be the image of the point B(1,-2 | vedclass

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(B) Since $A(2,5,7)$ is the image of $B(1,-2,3)$ with respect to the plane $\pi$, the point $C$ is the midpoint of $AB$. $C = \left( \frac{2+1}{2}, \f

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$\frac{1}{\sqrt{6}}, \frac{-1}{\sqrt{6}}, \frac{2}{\sqrt{6}}$

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(B) Since $A(2,5,7)$ is the image of $B(1,-2,3)$ with respect to the plane $\pi$, the point $C$ is the midpoint of $AB$. $C = \left( \frac{2+1}{2}, \frac{5-2}{2}, \frac{7+3}{2} \right) = \left( \frac{3}{2}, \frac{3}{2}, 5 \right)$. Given $D = (2,1,6)$, the direction ratios of the line segment $CD$ are $(2 - \frac{3}{2}, 1 - \frac{3}{2}, 6 - 5) = (\frac{1}{2}, -\frac{1}{2}, 1)$. To find the direction cosines, we divide by the magnitude $\sqrt{(\frac{1}{2})^2 + (-\frac{1}{2})^2 + 1^2} = \sqrt{\frac{1}{4} + \frac{1}{4} + 1} = \sqrt{\frac{6}{4}} = \frac{\sqrt{6}}{2}$. The direction cosines are $\left( \frac{1/2}{\sqrt{6}/2}, \frac{-1/2}{\sqrt{6}/2}, \frac{1}{\sqrt{6}/2} \right) = \left( \frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}} \right)$.

Let $A(2,5,7)$ be the image of the point $B(1,-2,3)$ with respect to a plane $\pi$. Let $C$ be the point where $AB$ meets the plane $\pi$. Let $D=(2,1,6)$. Then the direction cosines of $CD$ are

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