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
.. not sure about this
Do you see what happened to and ?
And, of course, a condition for this problem is that .
The problem with your attempted proof of (A) is that you started with a right triangle: assuming that and are perpendicular. You are asked to prove the opposite: that if then and are perependicular. You were "going the wrong way".
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?
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.