x+2y-z=4
2x+y+z=8
calculate the angle between the planes
i got 47 degree correct
obtain the unit vector in the direction of the line of intersection of the two planes? I'm not too sure about this problem
Okay, so you have your two planes
1: x+2y-z=4
2: 2x+y+z=8
I assume you've figured out by now that the normal to plane 1 is going to be in the direction (1, 2, -1) and the normal to plane 2 is going to be in the direction (2, 1, 1). I guess you then used the scalar (dot) product between them to work out what the angle was.
The next bit: Unit vector in the line of intersection.
Well, the unit bit just means it has a total length of 1 so let's worry about that later. First of all, work out what direction it's in.
How do we do this? Think about your two planes. If you're having trouble visualising it, hold up some sheets of A4 to help you. Each plane has its own normal coming out of it at 90 degrees. The line of intersection is in BOTH planes, right? (fairly obviously, since it has to be in both for it to be the line of intersection)
If a line is in a plane, then it's perpendicular to the normal of that plane. If it's in both planes, then it must be perpendicular to both normals. Again, that has to be true, by definition pretty much. Think about that for a minute and convince yourself that it's right.
If you want to generate a vector perpendicular to two other vectors, how do you do it? That's right. The vector (or "cross") product.
So it'll be in direction (1, 2, -1) CROSS (2, 1, 1).
Once you know the direction, you just have to divide by the total magnitude to make sure it has length 1.
Hope this helps.
A vector in the direction of the line of intersection of two planes can be obtained by taking the cross-product of the normal vectors to the planes; so compute $\displaystyle [1,2,-1]$x$\displaystyle [2,1,1]$ and then find the magnitude of this vector and divide by the magnitude to obtain the unit vector.
Edit: Fedex beat me to it, and he has a nice explanation.
I think Mathstud was reminding you how to calculate the angle between the two planes.
If you call the normal to plane 1 n1 and the normal to plane 2 n2, then
n1 DOT n2 = |n1||n2|cos(theta)
Rearranging that will give you the angle between the normals as Mathstud showed. The angle between the normals must be the angle between the planes too (if you think about that by holding up pieces of paper and stuff, you'll see that that's true).
I don't know if your answer is right, by the way. I can't work out inverse cosines in my head :-D I'll let you check it.