# Find the angles of the triangle

1. Aug 18, 2013

### utkarshakash

1. The problem statement, all variables and given/known data
Let a be a unit vector and b be a non-zero vector not parallel to a. Find the angles of the triangle, whose two sides are represented by the vectors √3(a x b) and b-(a.b)a

2. Relevant equations

3. The attempt at a solution
The third side will be equal to $\sqrt{3}(a \times b) - \left\{ b-(a.b)a \right\}$

Now taking the dot product of the given two sides will give me zero so the angle between them is 90°. But what about the other two angles? Even if I take the dot product of the remaining sides pairwise it will not give me an expression which can be evaluated as I don't know the lengths of sides of the triangle.

2. Aug 18, 2013

### Staff: Mentor

You know the length of the sides in terms of a and b, that is sufficient to find an expression for the angle.
I did not calculate it, but my intuition is that you will get a result independent of a and b.

3. Aug 18, 2013

### haruspex

I don't think that is borne out, but does it matter? Why cannot the answer be in terms of a and b?

4. Aug 19, 2013

### ehild

It can, but a is unit vector and the result will contain b in the form of b2/b2

ehild

5. Aug 19, 2013

### haruspex

OK, I see it now. Must have made an algebraic error. b - (a.b)a is merely the component of b perpendicular to a. The vector axb only depends on that component of b.

6. Aug 19, 2013

### Staff: Mentor

Exactly, and therefore |b| does not influence the angles. |a| is known, and the other lengths scale properly with the angle between a and b, so the result is independent of the vectors.

7. Aug 19, 2013

### utkarshakash

8. Aug 19, 2013

### Staff: Mentor

Did you calculate that?

9. Aug 20, 2013

### utkarshakash

No. I mean I don't know how to calculate that? My only guess is length of one side of the triangle might be √3. I'm not sure. I have no idea about length of other two sides.

10. Aug 20, 2013

### haruspex

Please post your attempt at calculating the angle between, say, $\sqrt{3}(a \times b) - \left\{ b-(a.b)a \right\}$ and $\left\{ b-(a.b)a \right\}$

11. Aug 20, 2013

### utkarshakash

OK. I took the dot product of the two sides (posted by you) and got something like this

$\left( \vec{a} . \vec{b} \right) ^2 - | \vec{b} |^2$

which after further simplification yields $-b^2 sin^2 \theta$, where θ is the angle between a and b.

Now, to get the angle between these two sides I also need their magnitudes so that I can divide the obtained expression by product of their magnitudes to get the cosine of the angle between them.

Last edited: Aug 20, 2013
12. Aug 20, 2013

### Staff: Mentor

Just continue with the calculations...
You need the magnitudes, how do you calculate them?

13. Aug 21, 2013

### utkarshakash

I took the dot product of the sides with itself and got the magnitudes as $4b^2 sin^2 \theta$ and $b^2 sin^2 \theta$. Now if I will divide the expression obtained earlier with the product of these two I get $\dfrac{-1}{4 b^2 sin^2 \theta}$. Unfortunately, this still contains b.

14. Aug 21, 2013

### Staff: Mentor

I think your values are the squares of the magnitudes and not the actual magnitudes.

15. Aug 21, 2013

### utkarshakash

Oh! That was a silly mistake. Thanks for pointing out.

16. Aug 21, 2013

### Saitama

How do you evaluate $(a \times b).(a \times b)$?

Sorry for the dumb question but how do you conclude that?

17. Aug 21, 2013

### haruspex

Useful rule: $(a \times b).(a \times b) = a^2b^2sin^2(\theta) = a^2b^2(1-cos^2(\theta)) = a^2b^2-(a.b)^2$.
Since a is a unit vector, (a.b)a is the component of b in the direction of a. Its magnitude is |b|cos(θ) and its direction is that of a. Subtracting it from b leaves the component perpendicular to a.

18. Aug 22, 2013

### Saitama

Thanks haruspex. :)