This is the question. The vectors u and v are such that ‖ u ‖ = 4, ‖ v ‖ = 5 and ( u | v ) = − 12. Compute the length of the vector ( 2 u + 3 v ) × ( 5 v − 4 u )I broke it down into 2 u × ( 5 v − 4 u ) + 3 v ( 5 v − 4 u ) and then further broke it down into ( 2 u × 5 v ) − ( 2 u × 4 u ) + ( 3 v × 5 v ) − ( 3 v × 4 u )But what I don't get is, how do I use ( u | v ) = − 12 here in order to compute the answer? Is it connected to the formula u × v = ‖ v ‖ T ( u ′ )?

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This is the question. The vectors u and v are such that   ‖  u  ‖  =  4,   ‖  v  ‖  =  5 and   (  u      |    v  )  =  −  12. Compute the length of the vector   (  2  u  +  3  v  )  ×  (  5  v  −  4  u  )I broke it down into   2  u  ×  (  5  v  −  4  u  )  +  3  v  (  5  v  −  4  u  ) and then further broke it down into   (  2  u  ×  5  v  )  −  (  2  u  ×  4  u  )  +  (  3  v  ×  5  v  )  −  (  3  v  ×  4  u  )But what I don&#039;t get is, how do I use   (  u      |    v  )  =  −  12 here in order to compute the answer? Is it connected to the formula   u  ×  v  =  ‖  v  ‖  T  (      u    ′    )?

Reinfarktq6 · Accepted Answer

The cross-product of a vector by itself is always 0, and   v  ×  u  =  −  u  ×  v. So,  (  2  u  +  3  v  )  ×  (  5  v  −  4  u  )  =  (  10  +  12  )  (  u  ×  v  )  =  22  u  ×  v  .On the other hand  −  12  =  (  u      |    v  )  =  ‖  u  ‖  ‖  v  ‖  cos  ⁡  θ  =  20  cos  ⁡  θ  ,where   θ is the angle between u and v. So,   cos  ⁡  θ  =  −      3    5  , and therefore   sin  ⁡  θ  =      4    5  . Now, use the fact that   ‖  u  ×  v  ‖  =      |    sin  ⁡  θ      |    ‖  u  ‖  ‖  v  ‖

This is the question. The vectors u and v are such that ∥u∥=4, ∥v∥=5 and (u|v)=−12. Compute the length of the vector (2u+3v) xx (5v−4u)

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