Hello,

plz try to solve that question.Thanks

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- Apr 4th 2007, 02:42 AMm777angle between the vectors ..........
Hello,

plz try to solve that question.Thanks - Apr 4th 2007, 09:15 AMAryth
a) Ok, so we have two vectors

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. - Apr 4th 2007, 09:22 AMJhevon
Here's (a)

- Apr 4th 2007, 09:43 AMJhevon
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 - Apr 5th 2007, 12:01 AMearboth
Hello,

to b)

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.

EB - Apr 5th 2007, 11:52 AMJhevon
- Apr 6th 2007, 08:11 AMearboth
Hello, Jhevon,

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?)

EB - Apr 6th 2007, 10:04 AMJhevon
i understand perfectly earboth.

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