Three vectors, V_1,V_2,V_3 are in the RR^2 plane where V_1+V_2+V_3=vec(0) and the magnitudes of these vectors are the same. Show that the angle between any two of these vectors is 120 degrees.

empalhaviyt

empalhaviyt

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2022-08-18

Three vectors, V 1 , V 2 , V 3 are in the R 2 plane where V 1 + V 2 + V 3 = 0 and the magnitudes of these vectors are the same. Show that the angle between any two of these vectors is 120 degrees.
My try:
V 1 + V 2 + V 3 = 0 , so V 1 + V 2 = V 3 . This would mean that ( V 1 + V 2 ) 2 = V 3 2 which is | V 1 | + | V 2 | + 2 V 1 V 2 = | V 3 | . But since the magnitude is the same, we get 2 | V 1 | + 2 V 1 V 2 = | V 1 | which is 2 V 1 V 2 = | V 1 |
The formula for the angle is cos ( a ) = V 1 V 2 | V 1 | | V 2 | = 0.5 | V 1 | | V 1 | 2 = 0.5 | V 1 |
I know that the solution would be a triangle and that cos ( a ) = 0.5
But that would mean that my angle is dependent on V 1 which isn't the case. So what am I doing wrong?

Answer & Explanation

Jake Landry

Jake Landry

Beginner2022-08-19Added 22 answers

V 3 = V 1 V 2 implies that V 3 2 = V 1 + V 2 , V 1 + V 2 = V 1 2 + V 2 2 + 2 V 1 , V 2 = V 2 where V 1 = V 2 = v 3 = V
This implies that 2 V 2 c o s ( V 1 , V 2 ) = V 2
Landon Wolf

Landon Wolf

Beginner2022-08-20Added 4 answers

You cannot "square" a vector: the statement " ( V 1 + V 2 ) 2 = V 3 2 " makes no sense.
I sense that you are trying to use the dot product. In that case, you will instead obtain:
| V 3 | 2 = V 3 V 3 = ( V 1 + V 2 ) ( V 1 + V 2 ) = | V 1 | 2 + 2 V 1 V 2 + | V 2 | 2
Now we continue with your method. We have from
| V 1 | 2 = 2 V 1 V 2
cos ( a ) = V 1 V 2 | V 1 | | V 2 | = V 1 V 2 | V 1 | 2 = 1 2

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