Vector Question -- |A+B| is 78 times greater than |A-B|

[smakhtar]

Homework Statement

Vectors A and B have the same magnitude. |A+B| is 78 times greater than |A-B|, for this to happen, what must be the angle between vectors A and B?

Homework Equations

78((A^2) + (B^2)-2ABcos(theta))=((A^2)+(B^2)+2ABcos(theta))
Made it by using dot product.

The Attempt at a Solution

What I did was used dot product and got 78((A^2) + (B^2)-2ABcos(theta))=((A^2)+(B^2)+2ABcos(theta)). And I let B=A because they have the same magnitude. I simplify this equation further and cancel things out and I got the wrong answer. The answer I got is 12.9 deg. What did I do wrong?

 

[Simon Bridge]

Sketch out the addition and subtraction by head-to-tail.
You can use geometry (i.e. the cosine rule) on the two triangles to find the lengths of A+B and A-B.

The requirement that |A+B|/|A-B|=78 gives the third equation you need to solve the system.

 

[smakhtar]

So I got (x^2)=(A^2)+(B^2)-2AB(cos180 - theta)
and 78(x^2)=(A^2)+(B^2)-2ABcos(theta)
I rearranged for x in the first equation, and subbed that into the second, made B=A because they have the same magnitude. I canceled out the A's, and I got to 78(2+theta)=1-cos(theta). How do I solve for theta when there is theta alone and theta with cos?

 

[collinsmark]

So I got (x^2)=(A^2)+(B^2)-2AB(cos180 - theta)
and 78(x^2)=(A^2)+(B^2)-2ABcos(theta)

By the way, before going much further, shouldn't we be squaring the 78 value somewhere along the way?

[e.g., (78x)² = 78²x²]

 

[Simon Bridge]

So I got (x^2)=(A^2)+(B^2)-2AB(cos180 - theta)
and 78(x^2)=(A^2)+(B^2)-2ABcos(theta)
I rearranged for x in the first equation, and subbed that into the second, made B=A because they have the same magnitude. I canceled out the A's, and I got to 78(2+theta)=1-cos(theta). How do I solve for theta when there is theta alone and theta with cos?

... well you could have divided the second equation by the first one rather than substitute - it's algebraically easier.
... but if x=|A+B|, then shouldn't the second equation start out (x/78)^2 ... ?

You also simplify by putting |A|=|B|=a

You still get something that looks a bit like what you got - just a tad tidier.
To simplify - follow your nose: solve for cosθ and take the inverse.

 

[DEvens]

How does |A+B| relate to (A+B) dot (A+B) ? Are they equal?

 

[Fredrik]

How does |A+B| relate to (A+B) dot (A+B) ? Are they equal?

You should find it pretty easy to verify that C·C=|C|² for all vectors C. In particular, this means that |A+B| is the square root of (A+B)·(A+B).