Q.
The equation of line parallel to the plane x+y+z=2 and intersecting the line x+y−z=0=x+2y−3z+5 at the point -4,1,3 is
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a
x+4−3=y−1−2=z−31
b
x+41=y−12=z−3−3
c
x+4−3=y−12=z−31
d
x+4−1=y−12=z−3−3
answer is C.
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Detailed Solution
The equation of plane is x+y−z+λ(x+2y−3z+5)=0It is passing through the point -4,1,3It implies that−4+1−3+λ(−4+2−9+5)=0⇒λ=−1Hence, the equation of plane is y−2z+5=0The required line is parallel to both planes y−2z+5=0 and x+y+z=2Normal vectors to above planes aren1=i+j+k,n2=j−2k The vector along the line is perpendicular to both the vectors, so it is along n1×n2Hence,n1×n2=ijk11101−2=−3i+2j+kHence, the direction ratios of the required line are ⟨−3,2,1⟩ Therefore, the equation of the required line is x+4−3=y−12=z−31
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