So I understand where the formula for cos between two vectors comes from. However, if i have a paralelogram of basis vec(a) , vec(b) , and i want to find out the angle between its diagonals starting only from the fact that norm(vec(a))=3 and norm(vec(b))=2 and the angle between them pi/3 , how would i go about it?

Aldo Ashley 2022-10-12 Answered
So I understand where the formula for cos between two vectors comes from. However, if i have a paralelogram of basis a , b , and i want to find out the angle between its diagonals starting only from the fact that
a = 3
and
b = 2
and the angle between them pi/3 , how would i go about it?
I know the diagonals are a b and a + b respectively, but im not sure how I shoul apply the formula for sum and difference of vectors.
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Answers (2)

Kenley Rasmussen
Answered 2022-10-13 Author has 13 answers
a = | | a | | d i r ( a ) = ( | | a | | cos θ , | | a | | sin θ ) = ( 3 cos θ , 3 sin θ )
b = | | b | | d i r ( b ) = ( | | b | | cos ( θ + π 3 ) , | | a | | sin ( θ + π 3 ) ) = ( 2 cos ( θ + π 3 ) , 2 sin ( θ + π 3 ) )
Now, you are looking for α = arccos ( a + b ) ( a b ) | | a + b | | | | a b | |
You can easily calculate the parts, indeed
( a + b ) ( a b ) = a 2 b 2 = 9 4 = 5
Besides,
| | a b | | = ( 3 cos θ 2 cos ( θ + π 3 ) ) 2 + ( 3 sin θ 2 sin ( θ + π 3 ) ) 2 = 9 + 4 12 ( cos θ cos ( θ + π 3 ) + sin θ sin ( θ + π 3 ) ) = 7
| | a + b | | = ( 3 cos θ + 2 cos ( θ + π 3 ) ) 2 + ( 3 sin θ + 2 sin ( θ + π 3 ) ) 2 = 9 + 4 + 12 ( cos θ cos ( θ + π 3 ) + sin θ sin ( θ + π 3 ) ) = 19
Here, I've used the fact that
cos θ cos ( θ + π 3 ) + sin θ sin ( θ + π 3 ) = 1 2
Therefore, α = arccos 5 133 .
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Kamila Frye
Answered 2022-10-14 Author has 4 answers
You have
a b 2 = a 2 2 a , b + b 2 = 9 2 × 3 × 2 × cos ( π 3 ) + 4 = 7.
Can you deal with a + b now?
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