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Let u, v, and w be vectors satisfying \(\mathbf{u}\bullet \mathbf{v} = 3, \mathbf{u} \bullet \mathbf{w} = 4, \mathbf{v} \bullet \mathbf{w} = 5.\)Then what are \((\mathbf{u} + 2 \mathbf{v})\bullet \mathbf{w}, (\mathbf{w} - \mathbf{u})\bullet \mathbf{v}, (3\mathbf{v} - 2 \mathbf{w})\bullet \mathbf{u}\) equal to?

 Nov 3, 2019
 #1
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(u + 2v)*w = 14, (w - u)*v = 7, (3v - 2w)*u = -5

 Nov 3, 2019
 #2
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it's dot product not normal product

Guest Nov 3, 2019
 #3
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Now I'm getting (u + 2v)*w = 10, (w - u)*v = 7, (3v - 2w)*u = -3.

Guest Nov 3, 2019
 #4
avatar+23516 
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Let u, v, and w be vectors satisfying \(\mathbf{u}\bullet \mathbf{v} = 3,\ \mathbf{u} \bullet \mathbf{w} = 4,\ \mathbf{v} \bullet \mathbf{w} = 5 \).
Then what are \((\mathbf{u} + 2 \mathbf{v})\bullet \mathbf{w},\ (\mathbf{w} - \mathbf{u})\bullet \mathbf{v},\ (3\mathbf{v} - 2 \mathbf{w})\bullet \mathbf{u}\) equal to?

 

\(\begin{array}{|rcll|} \hline \mathbf{u}\bullet \mathbf{v} = \mathbf{v}\bullet \mathbf{u} &=& 3 \\ \mathbf{u} \bullet \mathbf{w}=\mathbf{w} \bullet \mathbf{u} &=& 4 \\ \mathbf{v} \bullet \mathbf{w}=\mathbf{w} \bullet \mathbf{v} &=& 5 \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline && (\mathbf{u} + 2 \mathbf{v})\bullet \mathbf{w} \\ &=& \mathbf{u} \bullet \mathbf{w}+2\mathbf{v} \bullet \mathbf{w} \\ &=& 4+2\cdot 5 \\ &=&\mathbf{ 14 } \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline && (\mathbf{w} - \mathbf{u})\bullet \mathbf{v} \\ &=& \mathbf{w} \bullet \mathbf{v} - \mathbf{u}\bullet \mathbf{v} \\ &=& 5-3 \\ &=&\mathbf{ 2 } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline && (3\mathbf{v} - 2 \mathbf{w})\bullet \mathbf{u} \\ &=& 3\mathbf{v}\bullet \mathbf{u} - 2\mathbf{w} \bullet \mathbf{u} \\ &=& 3 \cdot 3 -2\cdot 4 \\ &=& 9-8 \\ &=&\mathbf{ 1 } \\ \hline \end{array}\)

 

laugh

 Nov 4, 2019

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