Find the distance of the point (0,-3,2) from the plane x+2y-z=1, measured parallel to the line (x+1)/2=(y+1)/2=z/3 The answer given in the book is: 10 units But my answer comes to:?153 units.

Asked by Manoj | 1st Jun, 2013, 06:11: PM

Expert Answer:

The equation of line passing through P(0,-3,2) and parallel to   (x+1)/2=(y+1)/2=z/3 is
 L2: (x)/2=(y+3)/2=(z-2)/3 = t
Hence, x = 2t, y = -3+2t, z = 2+3t
 
Now, finding intersection point between the plane x+2y-z=1 and L2
2t+2(-3+2t) - (2+3t) = 1
-8-3t = 1
-3t = 9 
t = -3
Hence, the intersection point is (-6, -9,-7)
Hence the distance = sqrt((-6-0)^2 + (-9+3)^2 + (-7-2)^2 )  = sqrt(36+36+81)  = sqrt(153)
 
Yes your answer is correct, the answer in the book is wrong. 

Answered by  | 2nd Jun, 2013, 05:09: AM

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