Find the vector and Cartesian equations of the plane containing the two lines:

r=2i+j-3k+U(i+2j+5k) and r= 3i +3j+2k+T(3i-2j+5k)

We have the vector equation of two lines as-

r=2i+j-3k+U(i+2j+5k)

r= 3i +3j+2k+T(3i-2j+5k)

We need to find the equation of the plane containing the the given to lines.

So the normal vector to the required plane can be found by the cross product of the vector directions of the two lines as,

So the vector equation of the plane is-

Therefore the Cartesian form of plane is