Question

Question #bb479

Answer 1

Since `(8vec{u}-vec{v}) _|_ (4vec{u}+3vec{v})`,

`(8vec{u}-vec{v})cdot(4vec{u}+3vec{v})=0`

by multiplying out,

`=> 32|vec{u}|^2+20vec{u}cdot vec{v}-3|vec{v}|^2=0`

by solving for `20vec{u}cdot vec{v}`,

`=> 20vec{u}cdot vec{v}=3|vec{v}|^2-32|vec{u}|^2`

by `vec{u}cdot vec{v}=|vec{u}||vec{v}|cos theta=2|vec{u}|^2cos theta` and `|vec{v}|=2|vec{u}|`,

`=> 40|vec{u}|^2cos theta=-20|vec{u}|^2`

by dividing by `40|vec{u}|`,

`=>cos theta=-1/2 => theta={2pi}/3=120^circ`

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I hope that this was helpful.