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If $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{b}=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$, then find $(a+3 b) \cdot(2 a-b) .$
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Verified Answer
The correct answer is:
$-15$
Here, $\quad \mathbf{a}+3 \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}+3(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})$
$=10 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
and $2 \mathbf{a}-\mathbf{b}=2(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})-(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})=-\hat{\mathbf{i}}+5 \hat{\mathbf{k}}$
$(\mathbf{a}+3 \mathbf{b}) \cdot(2 \mathbf{a}-\mathbf{b})=(10 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \cdot(-\hat{\mathbf{i}}+5 \hat{\mathbf{k}})$
$=10 \times(-1)+7 \times 0+-1 \times 5=-15$
$=10 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
and $2 \mathbf{a}-\mathbf{b}=2(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})-(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})=-\hat{\mathbf{i}}+5 \hat{\mathbf{k}}$
$(\mathbf{a}+3 \mathbf{b}) \cdot(2 \mathbf{a}-\mathbf{b})=(10 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \cdot(-\hat{\mathbf{i}}+5 \hat{\mathbf{k}})$
$=10 \times(-1)+7 \times 0+-1 \times 5=-15$
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