What have you tried?
Note that we can write the line as, .
You can use that to find the point of intersection. As for the angle of inetersection, use the formula
Here and are vectors and is the angle between them. (Hint: one of your vectors should be lying in the plane )
The normal vector for is given by .
b) Given the two planes
E : . = 2.x + 3x2 + 7x3 + 2 = 0, E’ : x=(3,0,0)+ s(-2,0,0)+t(3,0,1)
find the line of intersection and the angle between them.
The angle between the planes is the same as the angle between their normal vectors. Use the formula I gave you just above.
As for the line of intersection, we can find it if we find a point on the line and a vector in the direction of the line. The latter can be found by taking the cross product of the two normal vectors (Why?). For the former, set in both planes. This will tell you on what lines the planes intersect with the xy-plane. Then just find the intersection point of those two lines (remember, the z-coordinate of those points is 0), that will give you a point on the line of intersection.
Now see if you can get anywhere with those hints