Question

what is the angle between vector A and vector B if C=B/2​

Answer 1

Answer:

θ=arccos⎛⎝⎜−(|a⃗ |2|b⃗ |2+34)2|a⃗ ||b⃗ |⎞⎠⎟

Explanation:

Let the two vectors be  a⃗   and  b⃗   and the resultant  c⃗ .  

Let the angle between  a⃗   and  b⃗   be  θ.  

The magnitude of the resultant vector is,

|c⃗ |=|a⃗ |2+2|a⃗ ||b⃗ |cosθ+|b⃗ |2−−−−−−−−−−−−−−−−−−−√  

⇒|c⃗ |2=|a⃗ |2+2|a⃗ ||b⃗ |cosθ+|b⃗ |2.  

It is given that  |c⃗ |=12|b⃗ |.  

⇒14|b⃗ |2=|a⃗ |2+2|a⃗ ||b⃗ |cosθ+|b⃗ |2.  

⇒2|a⃗ ||b⃗ |cosθ=−(|a⃗ |2+34|b⃗ |2).  

⇒2|a⃗ ||b⃗ |cosθ=−(|a⃗ |2|b⃗ |2+34).  

⇒cosθ=−(|a⃗ |2|b⃗ |2+34)2|a⃗ ||b⃗ |.  

⇒θ=arccos⎛⎝⎜−(|a⃗ |2|b⃗ |2+34)2|a⃗ ||b⃗ |⎞⎠⎟.  

Thus, if we know the ratio of the magnitudes of the two vectors we can determine the angle between them.

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