plz try to solve that question.Thanks
a = <3, 2, sqrt(3)>
b = <3, 1, sqrt(6)>
We should begin by determining some kind of formula for this:
||a||*||b||cos(x) = a*b
So first we find the magnitude of each of these vectors, which is:
||a|| = (9 + 4 + 3)^.5 = 4
||b|| = (9 + 1 + 6)^.5 = 4
We should also find the dot product of these two vectors:
3*3 + 2*1 + sqrt(9) = 9 + 2 + 3 = 14
So now we plug it all in
4*4 cos(x) = 14
cos(x) = 14/16
cos(x) = .875
arccos(.875) = 28.95 degrees or .505 radians
As for b I cannot help you... Unfortunately.
Here's part (b), see Vectors - Cross Product for more info, just do a search for "same plane" by typing Ctrl-F and entering "same plane" in the dialogue box that appears
You may want to check that I found the determinant correctly, I always confuse myself with all the numbers, lol
3 vectors lie in a plane if you can find values for r and s so that the following equation is true;
r*u + s*v = w
r*<4, -6, 1> + s*<3, 1, -2> = <3, -2, 12>
Transform into a system of linear equations:
4r + 3s = 3
-6r + s = -2
r - 2s = 12
Use the first 2 equations to calculate r and s: r = 9/22 and s = 5/11
Now check if these values satisfy the 3rd equation. Obviously they don't do that. Therefore the 3 vectors are not complanar.
you guessed right.
If a poster asks for help here he obviously has some difficulties to understand the math concerning his problem.
So I try to use as much basic math to explain because I believe (and I hope) that at least the basics are understood.
(Somehow I have the impression that this text is far away from being English. Sorry! Do you understand it nevertheless?)
I always try to do a problem the simplest way as well, unless they ask for a particular method. It's just that i didn't remember you could do it your way. i tend to forget a lot of the math i've gone through, lol. that's one of my reasons for joining this site