The projection of the line L:x+1−1=y2=z−13 on the plane π:x−2y+z=6 is the line of intersection of this plane π with the plane (containing the line L)
A
2x+y+2=0
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B
3x+y−z=2
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C
2x−3y+8z=3
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D
none of these
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Solution
The correct option is A2x+y+2=0 Equation of the plane through (−1,0,1) is a(x+1)+b(y−0)+c(z−1)=0…(1)
which is parallel to the given line and perpendicular to the given plane. −a+2b+3c=0…(2) and a−2b+c=0…(3)
From Eqs. (2) and (3), we get c=0,a=2b