The image of the point (1,6,3) in the line | vedclass

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Question

(A) Let the given line be $ \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} = \lambda $. Any point $ R $ on the line is given by $ R(\lambda, 1+2\lambda, 2+3\

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$ (1,0,7) $

Vedclass · Answer

(A) Let the given line be $ \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} = \lambda $. Any point $ R $ on the line is given by $ R(\lambda, 1+2\lambda, 2+3\lambda) $. Let $ P $ be the point $ (1,6,3) $. The vector $ \vec{PR} $ is $ (\lambda-1, 2\lambda-5, 3\lambda-1) $. Since $ PR $ is perpendicular to the line with direction ratios $ (1,2,3) $,their dot product is zero: $ 1(\lambda-1) + 2(2\lambda-5) + 3(3\lambda-1) = 0 $ $ \lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 0 $ $ 14\lambda - 14 = 0 \Rightarrow \lambda = 1 $. Substituting $ \lambda = 1 $ in $ R $,we get $ R(1, 3, 5) $. Since $ R $ is the midpoint of $ PQ $,where $ Q(x_1, y_1, z_1) $ is the image of $ P $,$ \frac{x_1+1}{2} = 1 \Rightarrow x_1 = 1 $ $ \frac{y_1+6}{2} = 3 \Rightarrow y_1 = 0 $ $ \frac{z_1+3}{2} = 5 \Rightarrow z_1 = 7 $. Thus,the image $ Q $ is $ (1,0,7) $.

The image of the point $ (1,6,3) $ in the line $ \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} $ is

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