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Prove this Vector Operation

  • Thread starter BioBabe91
  • Start date
  • #1
13
0

Homework Statement


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


Homework Equations


quadratic equation, definition of absolute value


The Attempt at a Solution


[tex]|\vec{a}-\vec{b}| = \sqrt{\vec{a}^{2}-2\vec{a}\bullet\vec{b}+\vec{b}^{2}}[/tex]
and then I got
[tex]|\vec{a}-\vec{b}| = \sqrt{\vec{a}^{2}-2|\vec{a}||\vec{b}|cos\Phi+\vec{b}^{2}}[/tex]
So then cosine of the angle is equal to -1, and I don't know how to go from there.
 
Last edited:

Answers and Replies

  • #2
104
0
Looks like a notation issue.

[tex]
\begin{align}
\vec{a}^2 &= a^2\notag\\

|\vec{a}|&=a\notag
\end{align}
[/tex]

Everything under the radical is a scalar.
 
  • #3
33,181
4,860
Looks like a notation issue.

[tex]
\begin{align}
\vec{a}^2 &= a^2\notag\\

|\vec{a}|&=a\notag
\end{align}
[/tex]

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.
 
  • #4
33,181
4,860

Homework Statement


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


Homework Equations


quadratic equation, definition of absolute value


The Attempt at a Solution


[tex]|\vec{a}-\vec{b}| = \sqrt{\vec{a}^{2}-2\vec{a}\bullet\vec{b}+\vec{b}^{2}}[/tex]
and then I got
[tex]|\vec{a}-\vec{b}| = \sqrt{\vec{a}^{2}-2|\vec{a}||\vec{b}|cos\Phi+\vec{b}^{2}}[/tex]
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.
 
  • #5
104
0
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:
[tex]
\begin{align}
\mathbf{a}^2 &= \mathbf{a} \cdot \mathbf{a} = a^2 = |\mathbf{a}|^2
\intertext{and}
|\mathbf{a}| &= a.
\end{align}
[/tex]

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

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