Vectors question

[Panphobia]

So I have a vectors test tomorrow, and I understand all the questions for review but could not get the final answer for one thinking question here is the question,

code:

A and B are vectors Two vectors A + 3B and 4A - 3B make an angle of 90 degrees when tail-to-tail Find the angle between A and B if |A| = 3|B|. (Hint: u dot u = |u|^2)

I started out by taking the dot product of A+3B and 4A-3B which gave me (period is dot in dot product) 4A . A + 4A . 3B - 3B . A - 3B . 3B = 0 , I am pretty sure you can simplify this more until you get to |a||b|cos(theta) but I am not sure how, can anyone steer me in the right direction?

[michaelp]

Your first step is right. Try and use the hint to simplify what you have more.

[Panphobia]

|4A|^2 + 4A . 3B - 3B . A - |9B|^2=0
lost now

[Dreadnought]

Hopefully, you know that the dot product is linear and commutative, that is

code:

For any vectors A, B and scalars s, t (real numbers) we have (sA).(tB) = (st)(A.B) and A.B = B.A

As an exercise, use the definition of the dot product to prove these.

[Panphobia]

that was actually one of my projects, um could |3B| be turned to 3|B| not sure

[michaelp]

Note that |4A|^ 2 != 4|A|^2, and |9B|^2 != 9|B|^2.

[Dreadnought]

Panphobia wrote:

that was actually one of my projects, um could |3B| be turned to 3|B| not sure

Just go by definition
We define |A|^2 := (A.A)
then apply linearity to |sA|

EDIT: Also, because of how |A| is defined, it is always nicer to work with |A|^2, in this case you can do this (if you are careful, though not required).

[Panphobia]

No wonder most of the class didnt even get one mark on this question last year, psh

[Panphobia]

ok so i did this so far, am I right?
4a.a - 3b.a + 12a.b - 9b.b=0
4|a|^2 - 15b.a - 9|b|^2=0
4|3b||3b| - 15b.a - 9|b|^2 = 0
36|b|^2 - 15b.a - 9|b|^2 = 0

[Dreadnought]

Careful
- 3(B.A) + 12(A.B) = 15(A.B) ?

[Panphobia]

oh yea

[Dreadnought]

You have (-3)(A.B) + (12)(A.B) = (-3 + 12)(A.B) = ???

[Panphobia]

ok i think I am completely done, point out errors please
(4a-3b).(a+3b)=0
4a.a - 3b.a + 12a.b - 9b.b=0
4|a|^2 + 9b.a - 9|b|^2=0
4|3b||3b| + 9b.a - 9|b|^2 = 0
36|b|^2 + 9b.a - 9|b|^2 = 0
27|b|^2 = -9b.a
-1/3 = b.a/b.b
-1/3 = |b||a|cosQ/|b|^2
-1/3 = |a|cosQ/|b|
-1/3 = 3|b|cosQ/|b|
-1/3 = 3cosQ
cosQ = -1/9
Q = 96.4

[Dreadnought]

Almost, check this step carefully

27|b|^2 = -9b.a
-1/3 = b.a/b.b

[Panphobia]

27|b|^2 = -9b.a
-1/3 = b.a/b.b

ok look either way you put it
|b|^2 = -9b.a/27
and then wouldnt b.a go b.a/|b|^2 = -9/27? or am I confused lol