Prove this Vector Operation

[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.

[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.

[Mark44]

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.

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.

[Mark44]

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.

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.

[bigplanet401]

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.


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}

and instead ended up with a dyadic (an outer product).

Notice that the arrows are missing on the RHS of each equation in my previous post.