Find the distance of the point P(3, 4, 4) from the point, where the line joining the points A(3, - 4, - 5) and B(2,-3,-1) intersects the plane 2x + y + z = 7.

 

 

Answers (1)
S safeer

\\ $The line passing through the points, (3,-4,-5) and (2,-3,1) is given by: $ \\ \frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5} \\\\ \frac{\mathrm{x}-3}{-1}=\frac{\mathrm{y}+4}{1}=\frac{\mathrm{z}+5}{6}=\lambda $ $ \\\\ \mathrm{x}=-\lambda +3 \\$ $ \mathrm{y}=\lambda -4 \\ z = 6\lambda -5

\\ $ For the the point of intersection of the given line and plane $ \\ $ 2(-\lambda+3)+\lambda-4+6 \lambda-5=7$ $ \\ 5 \lambda=10$ $ \\ \lambda=2 $ $ \\ (-\lambda+3, \lambda-4,6 \lambda-5) \equiv (1,-2,7)$

\\ $ Distance between points (3,4,4) and (1-2,7)$ =\sqrt{(3-1)^{2}+(4+2)^{2}+(4-7)^{2}} \\ =\sqrt{4+36+9} \\ =\sqrt{49} \\ =7 \mathrm{units}

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