Prove the following

Question:

Let $\vec{c}$ be a vector perpendicular to the vectors

$\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$

If $\overrightarrow{\text { c. }}(\hat{i}+\hat{j}+3 \hat{k})=8$ then the value of $\vec{c} \cdot(\vec{a} \times \vec{b})$

is equal to_________.

Solution:

$\vec{c}=\lambda(\vec{a} \times \vec{b})$

$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 1 & -1 \\ 1 & 2 & 1\end{array}\right|$

$(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$

$\vec{c} \cdot(\hat{i}+\hat{j}+3 \hat{k})=\lambda(3 \hat{i}-2 \hat{j}+\hat{k}) \cdot(\hat{i}+\hat{j}+3 \hat{k})$

$\Rightarrow \lambda(4)=8 \Rightarrow \lambda=2$

$\overrightarrow{\mathrm{c}}=2(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})$

$\vec{c} \cdot(\vec{a} \times \vec{b})=2|\vec{a} \times \vec{b}|^{2}=28$

 

 

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