Thread: Points, angles and lines of intersection.

a) Determine where the line L : x = (-1, 8, 9)+t(3,-9,-4) and the plane
E : -2x1 + 6x2 + 3x3 + 19 = 0 intersect and calculate the angle between
them.
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.

What have you tried?

Originally Posted by mamt6

a) Determine where the line L : x = (-1, 8, 9)+t(3,-9,-4) and the plane
E : -2x1 + 6x2 + 3x3 + 19 = 0 intersect and calculate the angle between
them.

Note that we can write the line as, .

That is,

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 )

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 normal vector for is given by .

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

i understand how to do a) but im still not sure about b) im pretty sure im meant to convert the parametric equation (E') into a normal equation in the form of E but im not sure how i got about doing it.