# Dot/Cross product vector problem.

1. Sep 7, 2011

### QuarkCharmer

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
Well, I have the whole thing drawn out, and calculated some information that will be needed.

$|A|=7.92$
$|B|=8.28$

I am assuming that the question means that vector C is perpendicular with vector A, meaning, there is 90 degrees between them.

a.) Find the x component of C

It says that the scalar product of C and B is 13, so I am guessing that means I have to know the angle between those two vectors. Which would be the angle between B and A minus 90 degrees right?

How do I find the angle between B and A ?

Alright, so:
$$A\bullet B = (4.8)(-3.9)+(-6.3)(7.3)$$
$$A\bullet B = -64.71$$
and since $A\bullet B = |A||B|cos(\theta)$
$$|A||B|cos(\theta) = -64.71$$

$$cos(\theta) = \frac{-64.71}{(7.92)(8.28)}$$
$$\theta = 170.67 degrees$$

Then the angle between B and C is 80.67 degrees, which looks about right.

So then, the scalar/dot product of C and B is:
$$C \bullet B = |8.28||C|cos(80.67) = 13$$
$$|C| = \frac{13}{cos(80.67)|8.28|}$$
$$|C| = 9.68$$

Now what?
I don't see how the components of C are coming out of this.

Last edited: Sep 7, 2011
2. Sep 7, 2011

### cepheid

Staff Emeritus
The fact that the scalar product of C and B is 13 also means that CxBx + CyBy = 13 (by the definition of scalar product). Does that help at all?

3. Sep 7, 2011

### QuarkCharmer

I was thinking of that but then I would still be stuck with two unknowns.
$$C_{x}(-3.9)+C_{y}(7.3)=13$$

but you probably said that for a reason (thanks) so I will look for a second system perhaps.

4. Sep 7, 2011

### cepheid

Staff Emeritus
Right, but you also know that C is perpendicular to A, which should give you a second equation involving Cx and Cy. Two equations and two unknowns ==> an exact solution can be found.

5. Sep 7, 2011

### QuarkCharmer

Hmm,

I found the angle from the positive x-axis to A to be -52.67 degrees (used arctan on vector A's components).

If the angle between C and A is 90, then the angle between C and the positive x-axis must be 90-|-52.67| = 37.33

Since I know |C| = 9.68 I could just get the x/y components of vector C right there?

I get:
$$C_{x} = 7.69$$
$$C_{y} = 5.87$$
Does that seem right? (Edit: No, it's not right at all, ugh. Looking in to cepheid's solution)

Last edited: Sep 7, 2011
6. Sep 7, 2011

### cepheid

Staff Emeritus
If two vectors are perpendicular, then their dot product is zero. You can see this just by noting that cos(90°) = 0, but it's also helpful to understand the geometric argument. Recall that when you're taking the dot product of A and B, you sort of "project" A onto B by drawing a line starting from the tip of A that is perpendicular to A and extend it until it lands on B. This marks out the component of A that is in the same direction as B. So the dot product can be interpreted as the magnitude of B multiplied by the component of A that is in the direction of B.

However, if the two vectors are perpendicular, then there is NO component of A that is in the direction of B. When you draw a line from the tip of A perpendicular to A, it never intersects B.

In any case, the dot product being zero gives you your second equation involving the x and y components.

7. Sep 7, 2011

### QuarkCharmer

Ah that's right. Sorry, I'm kind of teaching myself ahead of the course some).

My two systems of equations are:

-3.9x + 7.3y = 13
4.8x - 6.3y = 0

x=-1.3739
y= 1.047

Which both to two significant digits, is still incorrect?
x=-1.4
y= 1.1

8. Sep 7, 2011

### cepheid

Staff Emeritus
Yeah, I get different answers. My strategy was to use the lower equation to solve for x in terms of y, and plug that expression for x into the upper equation in order to solve for y. Once you have y, you have x.

9. Sep 7, 2011

### QuarkCharmer

Yes, definitely made a mistake on the system of equations.

The proper answer was x=7.8 and y=6.0 (to 2 s.f.)

I sure made that one more complicated than it needed to be. Thanks for the help, I appreciate it.

10. Sep 7, 2011

### cepheid

Staff Emeritus
You're welcome!