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Let \(\mathbf{a}\) be a vector of length 2, and let \(\mathbf{b}\) and \(\mathbf{c}\) be vectors such that \(\mathbf{a} \bullet \mathbf{b} = 3, \mathbf{a} \bullet \mathbf{c}  = 4, \mathbf{b} \bullet \mathbf{c} = 5. \) Find\(\mathbf{a}\bullet( \mathbf{b} + \mathbf{c}), \mathbf{b}\bullet(\mathbf{a} +\mathbf{b} + \mathbf{c}), (\mathbf{a} + 2\mathbf{b})\bullet(-3\mathbf{a} + 3\mathbf{c})\). If any of these cannot be uniquely determined from the information given, enter a question mark.

 

Could someone help or point me in the right direction? Thank you!

 Feb 24, 2020
 #1
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Let \(\mathbf{a}\) be a vector of length 2, and let \(\mathbf{b}\) and \(\mathbf{c} \) be vectors such that
\(\mathbf{a} \bullet \mathbf{b} = 3,\ \mathbf{a} \bullet \mathbf{c} = 4,\ \mathbf{b} \bullet \mathbf{c} = 5\).
Find \(\mathbf{a}\bullet( \mathbf{b} + \mathbf{c}),\ \mathbf{b}\bullet(\mathbf{a} +\mathbf{b} + \mathbf{c}),\ (\mathbf{a} + 2\mathbf{b})\bullet(-3\mathbf{a} + 3\mathbf{c})\).
If any of these cannot be uniquely determined from the information given, enter a question mark.

 

\(\mathbf{a}\bullet( \mathbf{b} + \mathbf{c})\\ \begin{array}{|rcll|} \hline && \mathbf{a}\bullet( \mathbf{b} + \mathbf{c}) \\ &=& \mathbf{a}\bullet\mathbf{b}+\mathbf{a}\bullet \mathbf{c} \quad | \quad \mathbf{a} \bullet \mathbf{b} = 3,\ \mathbf{a} \bullet \mathbf{c} = 4 \\ &=& 3+4 \\ &=& \mathbf{7} \\ \hline \end{array} \)

 

\(\mathbf{b}\bullet(\mathbf{a} +\mathbf{b} + \mathbf{c}) \\ \begin{array}{|rcll|} \hline && \mathbf{b}\bullet(\mathbf{a} +\mathbf{b} + \mathbf{c}) \\ &=& \mathbf{b}\bullet \mathbf{a} + \mathbf{b}\bullet\mathbf{b} + \mathbf{b}\bullet\mathbf{c} \\ &=& \mathbf{a}\bullet \mathbf{b} + \mathbf{b}\bullet\mathbf{b} + \mathbf{b}\bullet\mathbf{c} \\ &=& 3 + \mathbf{b}\bullet\mathbf{b} + 5 \\ &=& 8 + \mathbf{b}\bullet\mathbf{b} \\ &=& ? \\ \hline \end{array} \)

 

\((\mathbf{a} + 2\mathbf{b})\bullet(-3\mathbf{a} + 3\mathbf{c}) \\ \begin{array}{|rcll|} \hline && (\mathbf{a} + 2\mathbf{b})\bullet(-3\mathbf{a} + 3\mathbf{c}) \\ &=& (-3)\mathbf{a}\bullet\mathbf{a}+3\mathbf{a}\bullet\mathbf{c}-6\mathbf{b}\bullet\mathbf{a}+6\mathbf{b}\bullet\mathbf{c} \\ &=& (-3)\mathbf{a}\bullet\mathbf{a}+3\mathbf{a}\bullet\mathbf{c}-6\mathbf{a}\bullet\mathbf{b}+6\mathbf{b}\bullet\mathbf{c} \quad | \quad \mathbf{a}\bullet\mathbf{a} = 2^2 \\ &=& (-3)*2^2+3*4-6*3+6*5 \\ &=& -12+12-18+30 \\ &=& -18+30 \\ &=& \mathbf{12} \\ \hline \end{array} \)

 

laugh

 Feb 25, 2020
 #2
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Thank you so much for your answer, it really helped me!! Really appreciate it :)

Guest Mar 1, 2020

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