# Prove this Vector Operation

• BioBabe91
This is because vectors in R^3 are defined by their components in the x, y, and z directions, and the cross product of two vectors is just their vector product in these directions. However, in order for a vector to have a direction in space, it needs to have a magnitude (or length). This means that the vectors in your problem have a negative magnitude, which means that the cross product of them won't have a direction.In summary, you're trying to solve for the magnitude of a vector which has the opposite direction of its original vector. You're missing some crucial information, such as the fact that vectors in R^3 need to have a magnitude in order to have a direction in space. Once you havef

## Homework Statement

if vectors $$\vec{a}$$ and $$\vec{b}$$ have opposite directions, how to show that |$$\vec{a}$$| + |$$\vec{b}$$| = |$$\vec{a}$$ - $$\vec{b}$$|?

## Homework Equations

quadratic equation, definition of absolute value

## 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.

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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.

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.

## Homework Statement

if vectors $$\vec{a}$$ and $$\vec{b}$$ have opposite directions, how to show that |$$\vec{a}$$| + |$$\vec{b}$$| = |$$\vec{a}$$ - $$\vec{b}$$|?

## Homework Equations

quadratic equation, definition of absolute value

## 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.

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
$$\mathbf{a}^2 = \mathbf{aa} = \mathbf{a} \otimes \mathbf{a}$$