Prove this Vector Operation

1. Apr 28, 2009

BioBabe91

1. The problem statement, all variables and given/known data
if vectors $$\vec{a}$$ and $$\vec{b}$$ have opposite directions, how to show that |$$\vec{a}$$| + |$$\vec{b}$$| = |$$\vec{a}$$ - $$\vec{b}$$|?

2. Relevant equations
quadratic equation, definition of absolute value

3. The attempt at a solution
$$|\vec{a}-\vec{b}| = \sqrt{\vec{a}^{2}-2\vec{a}\bullet\vec{b}+\vec{b}^{2}}$$
and then I got
$$|\vec{a}-\vec{b}| = \sqrt{\vec{a}^{2}-2|\vec{a}||\vec{b}|cos\Phi+\vec{b}^{2}}$$
So then cosine of the angle is equal to -1, and I don't know how to go from there.

Last edited: Apr 28, 2009
2. Apr 28, 2009

bigplanet401

Looks like a notation issue.

\begin{align} \vec{a}^2 &= a^2\notag\\ |\vec{a}|&=a\notag \end{align}

Everything under the radical is a scalar.

3. Apr 29, 2009

Staff: Mentor

This makes no sense. In the first equation you have the square of a vector, which implies multiplication of some kind. The only kinds of multiplication that are available are 1) multiplication by a scalar, 2) the dot product, and 3) (for vectors in R^3) the cross product.

In the second equation, you claim that the magnitude of a vector is equal to the vector itself.

4. Apr 29, 2009

Staff: Mentor

You're given that the vectors have the opposite directions, which means that a = -kb for some positive scalar k. Also, since a.b = |a||b|cos(theta), and theta = pi, you have a.b = -|a||b|.

Put both of these ideas together, and the result you want should fall out pretty readily.

5. Apr 29, 2009

bigplanet401

Again, I think it's a notation issue. You'll often see shorthands like this in the literature. Since you're using boldface:
\begin{align} \mathbf{a}^2 &= \mathbf{a} \cdot \mathbf{a} = a^2 = |\mathbf{a}|^2 \intertext{and} |\mathbf{a}| &= a. \end{align}

I can see where there might be problems, if instead we did something like
$$\mathbf{a}^2 = \mathbf{aa} = \mathbf{a} \otimes \mathbf{a}$$