Solving 2 vectors with cross product

[majormaaz]

Homework Statement

Two vectors are given by A = -6 i + 5 j and B = 1 i + 4 j
Find A X B (answer only in terms of i, j, k)
Find the angle between A and B (answer is terms of degrees)

Homework Equations

All I was told was that if I set a 3x3 matrix like this:
i j k
-6 5 0
1 4 0
then AxB is the determinant

The Attempt at a Solution

I made the 3x3 matrix and found the determinant to be only -29 k, which I am told is correct.
I have absolutely no idea on how to approach the angle problem. If I may ask, can someone get me started in the right direction for that problem?

 

[rock.freak667]

Look up the definition of cross product (here).

You will see that you can calculate AxB in terms of a unit vector.

 

[LCKurtz]

Homework Statement

Two vectors are given by A = -6 i + 5 j and B = 1 i + 4 j
Find A X B (answer only in terms of i, j, k)
Find the angle between A and B (answer is terms of degrees)

Homework Equations

All I was told was that if I set a 3x3 matrix like this:
i j k
-6 5 0
1 4 0
then AxB is the determinant

The Attempt at a Solution

I made the 3x3 matrix and found the determinant to be only -29 k, which I am told is correct.
I have absolutely no idea on how to approach the angle problem. If I may ask, can someone get me started in the right direction for that problem?

Have you had the formula $\vec a \cdot \vec b = |\vec a||\vec b|\cos\theta$? That might come in handy.

 

[johnqwertyful]

Have you had the formula $\vec a \cdot \vec b = |\vec a||\vec b|\cos\theta$? That might come in handy.

Also:
$|\vec a \times \vec b| = |\vec a||\vec b|\sin\theta$

 

[majormaaz]

Why thanks, you two! I vaguely remember the dot product equation from precalculus (all those years ago), but the cross product one I've never seen. Thanks!

 

[LCKurtz]

Why thanks, you two! I vaguely remember the dot product equation from precalculus (all those years ago), but the cross product one I've never seen. Thanks!

You're welcome. I might add that since the angle between two vectors is always between 0 and $\pi$, the dot product may be slightly more convenient because that is the principle range of the inverse cosine.