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The foot of the perpendicular drawn from the point (4,2,3) to the line joining the points (1,-2,3) and (1,1,0) lies on the plane:  
Option: 1 2x +y-z =1
Option: 2 x-2y +z=1
Option: 3 x-2y+z =1
Option: 4 x+2y -z =1

Answers (1)

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x=1, y=-3\lambda -2, z=3+3\lambda

L = (0,3,-3)

P=(1,-3\lambda -3,3+3\lambda)

A = (4,2,3)

AP=(3,4-3\lambda ,-3\lambda)

\vec{AP}.\vec{L}=0

3.0+(4+3\lambda).(-3)+(3\lambda).(-3)=0

0-12+9\lambda +9\lambda =0

\lambda =\frac{2}{3}

P=(1,0,1)

Posted by

Suraj Bhandari

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