Find the angles of the triangle

[utkarshakash]

Homework Statement

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

Homework Equations

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.

 

[mfb]

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.

 

[haruspex]

my intuition is that you will get a result independent of a and b.

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

 

[ehild]

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

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

ehild

 

[haruspex]

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

ehild

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.

 

[mfb]

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.

 

[utkarshakash]

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.

Can you please help me out? I'm still not getting it.

 

[mfb]

You know the length of the sides in terms of a and b

Did you calculate that?

 

[utkarshakash]

Did you calculate that?

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.

 

[haruspex]

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.

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\}

 

[utkarshakash]

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\}

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.

 

[mfb]

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

 

[utkarshakash]

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

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.

 

[mfb]

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

 

[utkarshakash]

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

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

 

[Saitama]

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.

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

b - (a.b)a is merely the component of b perpendicular to a.

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

 

[haruspex]

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

Useful rule: $(a \times b).(a \times b) = a^2b^2sin^2(\theta) = a^2b^2(1-cos^2(\theta)) = a^2b^2-(a.b)^2$.

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

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.

 

[Saitama]

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.

Thanks haruspex. :)