# Finding Vector Angle using cosine law

Tags:
1. Sep 16, 2018

### alexi_b

1. The problem statement, all variables and given/known data
Vectors A and B have equal magnitudes of 4.93. If the sum of A and B is the vector 6.79j, determine the angle between A and B

2. Relevant equations
c^2 = a^2 + b^2 -2abcos(theta)

3. The attempt at a solution
I just rearranged the formula above so that I could solve for the angle. I keep getting the same answer (87.0 or 87.1 degrees) but its still wrong. Is it the fact that the sum is 6.79j with j meaning something? I know its a unit vector that indicates direction but I wasn't too sure. Please help!!

2. Sep 16, 2018

### kuruman

Draw a picture of the two vectors and their sum. Remember that the angle between the two vectors is obtained by putting the vectors tail to tail, not tip to tail as you would to draw the sum.

3. Sep 16, 2018

### alexi_b

Yes I have done that, I also tried subrtacted from 180 which i probably should have mentioned earlier and get 93.0 when rounded to 3 sig figs and its still not right

4. Sep 16, 2018

### kuruman

The answer should be less that 90o. The sum is the base of an isosceles triangle of equal sides 4.93. If it were a right isosceles triangle, the base would be $4.93\sqrt{2}=6.95.$ It is given as 6.79, so the angle you want is (slightly) less than 90o. I agree with your first answer of 87o. How do you know it is not the right answer?

On (belated) edit:
The angle should be (slightly) greater than 90o. I got confused about the same point that I cautioned OP not to be confused about.

Last edited: Sep 17, 2018
5. Sep 16, 2018

### alexi_b

It's an online assignment where I enter my answer and it will either tell me if its right or wrong. I need 3 significant digits so my answers were 87.0 and 87.1 which were both marked wrong

6. Sep 16, 2018

### kuruman

Check the input numbers again. If they agree, talk to your instructor. Sometimes there are are errors in the algorithms that generate the online answers. Better yet, send your solution in detail to your instructor before the assignment is due and ask why it was marked wrong.

7. Sep 16, 2018

### alexi_b

okay thanks so much, I appreciate your help!!

8. Sep 17, 2018

### SammyS

Staff Emeritus
I get a somewhat different answer.

In the law of cosines as you have written it, θ (theta) is not the angle between (the directions of) vectors A and B. The θ in the standard math form you used is the supplement of the angle between vectors A and B.

If ∅ is the angle between the vectors, then use:
$c^2 = a^2 + b^2 +2ab \cos(\phi)$​

9. Sep 17, 2018

### robphy

Note:
$\vec C=(\vec A+\vec B)$
Compute the dot-product
$\vec C\cdot \vec C=(\vec A+\vec B)\cdot (\vec A+\vec B)$
and expand it out. What is $\vec A\cdot\vec B$? (Can you solve for $\vec A\cdot\vec B$? [Polarization identity.])

10. Sep 17, 2018

### kuruman

Sorry, isn't that what I said in post #2?

11. Sep 17, 2018

### SammyS

Staff Emeritus

Yes that is what you said.

I have withdrawn that post.

However, the angle of 87° is the supplement of the correct angle.

12. Sep 17, 2018

### kuruman

It would be that. I edited post #4 to echo my agreement.

13. Sep 17, 2018

### robphy

Note that, from the polarization identity [expand $(\vec A+\vec B)\cdot(\vec A+\vec B)$ and solve for $\cos\theta$ ]
we have
$\cos\theta=\frac{1}{2AB}\left( (\vec A+\vec B)\cdot(\vec A+\vec B)-\vec A\cdot\vec A -\vec B\cdot\vec B \right)$
And the OP was given $(\vec A+\vec B)\cdot(\vec A+\vec B) =6.79^2$ and $\vec A\cdot\vec A = \vec B\cdot\vec B = 4.93^2$.
Note that the term in parenthesis is negative since $6.79^2 - 2(4.93)^2 = -2.5057$.
Thus $\cos\theta<0$, so $90^\circ <|\theta|<180^\circ$.

http://m.wolframalpha.com/input/?i=arccos(.5*((6.79^2-2*(4.93)^2)/(4.93)^2))+in+degrees
yields 92.95 degrees.

14. Sep 18, 2018

### SammyS

Staff Emeritus
Somehow, I missed this post. As @robphy has posted above, the answer to 4 digits is: 92.95°, which agrees with your answer to 3 sig. figs.

Using a TI-94 calculator and not rounding any intermediate steps I get 92.954747591° (Actually used a simpler method to get θ/2 using adjacent/hypotenuse for cos(θ/2) . )