# Dot product

1. Oct 4, 2003

### PiRsq

The angle between vectors a and b is cos^-1(4/21). Find p if a = [6,3,-2] and b = [-2, p, -4]

I did:

cos x= 4/21 = a.b/|a||b|

The result comes out to be p=8/3 but it only satisfies that a dot b is 4 and not |a||b| = 21...What am I doing wrong?

2. Oct 4, 2003

### Hurkyl

Staff Emeritus
Two fractions being equal does not mean their numerators are equal (e.g. 1/2 = 2/4); you can't simply discard the denominator when solving an equation.

3. Oct 4, 2003

### PiRsq

Oh so I cant simply say a.b is equal to 4 right? But if I do not do that then I get for cos theta:

cos theta = 3p-4/(7)root of p^2+20

I have no clue now of how to do it

4. Oct 4, 2003

### Hurkyl

Staff Emeritus
Well, you can cross multiply to get rid of the fractions.

(In case you don't remember, that is to multiply both sides by both denominators, thus going from a/b=c/d to a*d=b*c)

Then if only you knew an operation you can do to an equation to undo a square root...

5. Oct 4, 2003

### PiRsq

Yes so:

3p-4/root of p^2+20 = 4/21

Then I square the left side and cross multiply to get:

21(9p^2-24p+16)=4[7(p^2+20)]

am I right so far?

6. Oct 4, 2003

### Hurkyl

Staff Emeritus
Almost; when you square an equation, you have to square both sides... and you seem to have misplaced a 7 in the first equation in your post.

Incidentally, you should group your terms with parenthesis to make them more clear (and accurate): the LHS should be written something like

(3p-4)/( 7*sqrt(p^2+20) )

(where 'sqrt' stands for square root)

7. Oct 4, 2003

### PiRsq

Precisely, I did:

(3p-4)/( 7*sqrt(p^2+20) )=4^2/21^2 then I end up getting numbers in the millions

Eventually I end up with an answer of 1.63 for p by using the quadratic formula, and my book says the answer is simply 4

8. Oct 4, 2003

### Hurkyl

Staff Emeritus
Lemme see your work so I can see what went wrong.

Incidentally, one tip is to look for simplifications you can make at every step of the problem; for example, before doing anything, notice that the denominator of both sides is divisible by 7; you could multiply the equation through by 7 to cancel that out and reduce the size of the numbers with which you have to work.

9. Oct 4, 2003

### PiRsq

Ah yes, now I get the answer 4 and another negative number. But what is that negative number?

10. Oct 4, 2003

### Hurkyl

Staff Emeritus
Squaring an equation is not an invertable operation; e.g.

(2)^2 = 4 and (-2)^2 = 4

When you perform a noninvertable operation to an equation, it says "The solution(s) to the original equation is (are) among the solutions to this new equation". Generally it's good practice to check your solutions when you get them, but it becomes a necessity when you use noninvertable operations like squaring.

In many types of problems where there squaring introduces "false" solutions, the false solutions correspond to some sort of reversal of sign or direction. In this particular case, it corresponds to the case when cos &theta; = -4/21

11. Oct 4, 2003

### PiRsq

Ok great, thanks Hurkyl!