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If $\vec{a}=\hat{i}-2 \hat{j}+5 \hat{k}, \quad \vec{b}=2 \hat{i}+\hat{j}-3 \hat{k}$, then what is
$(\vec{b}-\vec{a}) \cdot(3 \vec{a}+\vec{b})$ equal to?
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$(\vec{b}-\vec{a}) \cdot(3 \vec{a}+\vec{b})$ equal to?
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Verified Answer
The correct answer is:
$-106$
Given, $\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$
and $\overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}}$
$\therefore \quad \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}}-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-8 \hat{\mathbf{k}}$
and $(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})=(3 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+15 \hat{\mathbf{k}})+(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}})$
$\begin{aligned}=5 \hat{i}-5 \hat{j}+12 \hat{k} & \\ \text { Hence, }(\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}}) \cdot(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) &=(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-8 \hat{\mathbf{k}}) \cdot(5 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}) \\ &=5-15-96 \\ &=-106 \end{aligned}$
and $\overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}}$
$\therefore \quad \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}}-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-8 \hat{\mathbf{k}}$
and $(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})=(3 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+15 \hat{\mathbf{k}})+(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}})$
$\begin{aligned}=5 \hat{i}-5 \hat{j}+12 \hat{k} & \\ \text { Hence, }(\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}}) \cdot(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) &=(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-8 \hat{\mathbf{k}}) \cdot(5 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}) \\ &=5-15-96 \\ &=-106 \end{aligned}$
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