Find an equation for the plane that contains the two lines l1:x=t+2,y=3t−5,z=5t+1 and l2:x=5−t,y=3t−10,z=9−2t

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Find an equation for the plane that contains the two lines  l1:x=t+2,y=3t−5,z=5t+1  and  l2:x=5−t,y=3t−10,z=9−2t

casincal · Accepted Answer

The vector i+3j+5k is parallel to the first line and the vector −i+3j−2k is parallel to the second line. So the cross product of these two is perpendicular to the containing these two lines:  n=(i+3j+5k)×(−i+3j−2k)  =−21i−3j+6j  As the lines are contained in the plane, if we find a single point on either line, that point will also be in the plane. Notice that (2,-5,1) is a point in the first line. So the equation of the plane is then  −21(x−2)−3(y+5)+6(z−1)=0  Or simplifying the coefficients a bit, we could write this as  7(x−2)+(y+5)−2(z−1)=0  Result: 7(x−2)+(y+5)−2(z−1)=0

Find an equation for the plane that contains the two lines l_1:x=t+2,y=3t

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