1. ## Proving Vectors

A) By considering the angles between the vectors, show that a+b and
a-b are perpendicular when |a|=|b|

Note: all bolded above are vectors (arrows on top)

Below is my attempt on the above questions, please let me know if I have the correct answer (as there are no answers in the back for these questions). Demonstrate any errors that I may have, the correct way. Thanks in advance!

A) well I drew two right angled triangles
the first has a as the base, b as the height and hypotenuse is a+b, the second has a as base, -b as height, and hypotenuse is a-b

2. Originally Posted by skeske1234
A) By considering the angles between the vectors, show that a+b and
a-b are perpendicular when |a|=|b|
$\displaystyle \left( {\overrightarrow a + \overrightarrow b } \right) \cdot \left( {\overrightarrow a - \overrightarrow b } \right) = \overrightarrow a \overrightarrow { \cdot a} - \overrightarrow b \cdot \overrightarrow b$

$\displaystyle \overrightarrow a \overrightarrow { \cdot a} = \left\| {\overrightarrow a } \right\|^2$

3. what happens to the bb?

4. Originally Posted by skeske1234
what happens to the bb?
He thought you could work that out for yourself! If, as Plato says, $\displaystyle \vec{a}\cdot\vec{a}= |\vec{a}|^2$,l what do you think $\displaystyle \vec{b}\cdot\vec{b}$?

Do you see what happened to $\displaystyle \vec{a}\cdot\vec{b}$ and $\displaystyle \vec{b}\cdot\vec{a}$?

And, of course, a condition for this problem is that $\displaystyle |\vec{a}|= |\vec{b}|$.

The problem with your attempted proof of (A) is that you started with a right triangle: assuming that $\displaystyle \vec{a}+ \vec{b}$ and $\displaystyle \vec{a}- \vec{b}$ are perpendicular. You are asked to prove the opposite: that if $\displaystyle |\vec{a}|= |\vec{b}|$ then $\displaystyle \vec{a}- \vec{b}$ and $\displaystyle \vec{a}+ \vec{b}$ are perependicular. You were "going the wrong way".

5. So is this right then:
(a+b) (a-b)= aa - bb
=|a|^2 - |b|^2

and how does this answer prove that a+b and a-b are perpendicular?
if |a|=|b| then wouldnt the answer be 0? hm.. so zero is equal to |a+b| and |a-b| .. how does that determine that they are perpendicular though?

6. Originally Posted by skeske1234
So is this right then:
(a+b) (a-b)= aa - bb
=|a|^2 - |b|^2

and how does this answer prove that a+b and a-b are perpendicular?
if |a|=|b| then wouldnt the answer be 0? hm.. so zero is equal to |a+b| and |a-b| .. how does that determine that they are perpendicular though?
If the dot product of two vectors is 0, then the two vectors are perpendicular.

7. Originally Posted by Plato
$\displaystyle \left( {\overrightarrow a + \overrightarrow b } \right) \cdot \left( {\overrightarrow a - \overrightarrow b } \right) = \overrightarrow a \overrightarrow { \cdot a} - \overrightarrow b \cdot \overrightarrow b$

$\displaystyle \overrightarrow a \overrightarrow { \cdot a} = \left\| {\overrightarrow a } \right\|^2$
I don't know what the question means by "considering the angles between the vectors", though...

The dot product of two vectors x and y is:

|x| |y| cos(theta)

Where theta is the angle between them. When the dot product is zero, and |x| and |y| are different from 0, that means cos(theta) = 0, which for 0<theta<pi means that theta=pi/2 --> perpendicular.

8. Originally Posted by pedrosorio
I don't know what the question means by "considering the angles between the vectors", though...

The dot product of two vectors x and y is:

|x| |y| cos(theta)

Where theta is the angle between them. When the dot product is zero, and |x| and |y| are different from 0, that means cos(theta) = 0, which for 0<theta<pi means that theta=pi/2 --> perpendicular.
Remember that

$\displaystyle \mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos{\theta}$.

If $\displaystyle \mathbf{a}$ and $\displaystyle \mathbf{b}$ are perpendicular, then $\displaystyle \theta = 90^{\circ}$, and so $\displaystyle \cos{\theta} = 0$.

Thus $\displaystyle \mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|(0) = 0$.

So if the dot product of two vectors is 0, then they are perpendicular.