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

Mylo O'Moore

Mylo O'Moore

Answered question

2021-09-23

Find an equation for the plane that contains the two lines
l1:x=t+2,y=3t5,z=5t+1
and
l2:x=5t,y=3t10,z=92t

Answer & Explanation

casincal

casincal

Skilled2021-09-24Added 82 answers

The vector i+3j+5k is parallel to the first line and the vector i+3j2k 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+3j2k)
=21i3j+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(x2)3(y+5)+6(z1)=0
Or simplifying the coefficients a bit, we could write this as
7(x2)+(y+5)2(z1)=0
Result: 7(x2)+(y+5)2(z1)=0

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