Dot Product Distributive?

[mikexz]

a,b are vectors ---- ". " is the dot product function

Under what conditions is
(a +b) . (a-b) = 0

I have absolutely no idea how to start this, does FOIL apply to this type of situation b/c distributive property apply to dot products?

I tried googling, and I got this
http://mathproofs.blogspot.com/
-on the dot product for the cosine law proof, I got really confused at what happened on the 2 and 3rd step in the 3rd box. the one with A . (A-B) - B. (A-B)

Is that some sort of property I don't know about? What's going on? Help please

thanks

 

[skeeter]

a,b are vectors ---- ". " is the dot product function

Under what conditions is
(a +b) . (a-b) = 0

Vector (a+b) would have to be perpendicular to vector (a-b) for their dot product to be 0.
Using a geometric approach (i.e. make a sketch), place vector a and b tail to tail. If vector a is one side of a parallelogram and vector b is the side of the parallelogram adjacent to a, then (a+b) and (a-b) would be represented by the two diagonals of the parallelogram. Now, what condition is needed for those two diagonals to be perpendicular?

 

[soroban]

Hello, mikexz!

$\displaystyle \text{Under what conditions is: }\;\left(\vec a +\vec b\right) \cdot \left(\vec a-\vec b\right) \;= \;0$

$\displaystyle \text{Let: }\;\begin{array}{ccc} \vec a &=&\langle p,q\rangle \\ \vec b &=& \langle r,s\rangle \end{array} \quad\Rightarrow\quad\begin{array}{ccc}\vec a + \vec b &=& \langle p+r,q+s\rangle \\ \vec a-\vec b &=& \langle p-r,q-s\rangle \end{array}$

$\displaystyle \text{Then: }\;\left(\vec a +\vec b\right)\cdot\left(\vec a - \vec b\right) \:=\:0 \quad\Rightarrow\quad \langle p+r,q+s\rangle \cdot \langle p-r, q-s\rangle \:=\:0$

. . $\displaystyle (p+r)(p-r) + (q+s)(q-s) \:=\:0 \quad\Rightarrow\quad p^2-r^2 + q^2-s^2\;=\;0$

$\displaystyle \text{Hence: }\;p^2+q^2\;=\;r^2+s^2$


$\displaystyle \text{Therefore: }\vec a\text{ and }\vec b \text{ must have equal magnitudes: }\;\left|\vec{a}\right| \:=\:\left|\vec{b}\right|$