# Finding Vector Angle using cosine law

• alexi_b
In summary, two vectors A and B with equal magnitudes of 4.93 have a sum of 6.79j. To determine the angle between A and B, the law of cosines can be used, but the angle obtained is the supplement of the actual angle between the vectors. The correct angle can be found by using the dot product and expanding it out. Additionally, when finding the angle between two vectors, they should be placed tail to tail, not tip to tail as when drawing the sum. f

## Homework Statement

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

## Homework Equations

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

## 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![/B]

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.

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.
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

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:
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?
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

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.

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.
okay thanks so much, I appreciate your help!

## Homework Statement

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

## Homework Equations

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

## 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![/B]
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)##​

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.])

What is generally referred to as the angle between two vectors is the angle formed when they are tail to tail.
Sorry, isn't that what I said in post #2?
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.

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

Yes that is what you said.

I have withdrawn that post.

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

However, the angle of 87° is the supplement of the correct angle.
It would be that. I edited post #4 to echo my agreement.

• SammyS
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.

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
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) . )