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If $\bar{a}=\hat{\imath}+5 \hat{k}, \bar{b}=2 \hat{\imath}+3 \hat{k}, \bar{c}=4 \hat{\imath}-\hat{\jmath}+2 \hat{k}$ and $\bar{d}=\hat{\imath}-\hat{\jmath}$,
then $(\bar{c}-\bar{a}) \cdot(\bar{b} \times \bar{d})=$
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then $(\bar{c}-\bar{a}) \cdot(\bar{b} \times \bar{d})=$
Solution:
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Verified Answer
The correct answer is:
$12$
$\bar{b} \times \bar{d}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 3 \\ 1 & -1 & 0\end{array}\right|=\hat{i}(3)-\hat{j}(0-3)+\hat{k}(-2)$
$\bar{b} \times \bar{d}=3 \hat{i}+3 \hat{j}-2 \hat{k}$
and $\bar{c}-\bar{a}=3 \hat{i}-\hat{j}-3 \hat{k}$
then $(\bar{c}-\bar{a}) \cdot(\bar{b} \times \bar{d})=3(3)+(-1)(3)+(-3)(-2)$
$=9-3+6=12$
$\bar{b} \times \bar{d}=3 \hat{i}+3 \hat{j}-2 \hat{k}$
and $\bar{c}-\bar{a}=3 \hat{i}-\hat{j}-3 \hat{k}$
then $(\bar{c}-\bar{a}) \cdot(\bar{b} \times \bar{d})=3(3)+(-1)(3)+(-3)(-2)$
$=9-3+6=12$
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