Question

I need to find the equation for the plane, and it contains the lines r1(t)= <1+2t,-2+3t,4> and r2(t)=<3-5t,-1+17t,-28t>. Then I have to write the equation in the form ax+by+cz=d.

Answer 1

take the cross product of `direction vectors of lines` to get the normal to the plane

Answer 2

r1(t)= <1+2t,-2+3t,4> and r2(t)=<3-5t,-1+17t,-28t>.

direction vector of r1 = (2, 3, 0)
direction vector of r2 = (-5, 17, -28)

take the cross product, what do u get ?