- #1
finkeroid99
Hello I was doing some study for a maths test involving vectors when I came across this question:
For non zero vectors show that: |a-b|=|a+b| if and only if a and b are perpendicular.
Deduce that a parallelogram is a rectangle if and only if its diagonals are equal in length.
I did the first question using dot products:
|a-b|=|a+b|
|a-b|^2=|a+b|^2
(a-b).(a.b)=(a+b).(a+b)
|a|^2+|b|^2-2|a||b|cosX=|a|^2+|b|^2+2|a||b|cosX
-2|a||b|cosX=+2|a||b|cosX
-cosX=cosX
Therefore cosX=0
x=90, 270, 480...etc.
Thus vector a and vector b must be perpendicular for|a-b|=|a+b| to be valid.
I'm not sure if this way is correct though...
Could someone please check if the stuff that I've done above is right? Also, could I get some help doing the second part of teh question? Thank you it will be much apprechiated.
For non zero vectors show that: |a-b|=|a+b| if and only if a and b are perpendicular.
Deduce that a parallelogram is a rectangle if and only if its diagonals are equal in length.
I did the first question using dot products:
|a-b|=|a+b|
|a-b|^2=|a+b|^2
(a-b).(a.b)=(a+b).(a+b)
|a|^2+|b|^2-2|a||b|cosX=|a|^2+|b|^2+2|a||b|cosX
-2|a||b|cosX=+2|a||b|cosX
-cosX=cosX
Therefore cosX=0
x=90, 270, 480...etc.
Thus vector a and vector b must be perpendicular for|a-b|=|a+b| to be valid.
I'm not sure if this way is correct though...
Could someone please check if the stuff that I've done above is right? Also, could I get some help doing the second part of teh question? Thank you it will be much apprechiated.