Given equation of line is
Let
x=y=z−1=t
we get
x=ty=tz=1+t
Given Equation of cone is
x2+y2+z2+2xy+2yz+2xz=0→t2+t2+(1+t)2+2t2+2t(1+t)+2t(1+t)=0→9t2+6t+1=0
→(3t+1)2=0→t=−31
so we get,
x=−31y=−31z=1−31=32
So given line intersect with cone at (3−1,3−1,32) .
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