Let a =3 i +2 j + k , b =2 i - j +3 k and c be a vector such that… - Vidyadip

Question

Let $\overrightarrow{\mathrm{a}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}}$ be a vector such that $(\vec{a}+\vec{b}) \times \vec{c}=2(\vec{a} \times \vec{b})+24 \hat{j}-6 \hat{k}$ and $(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}+\hat{\mathrm{i}}) \cdot \overrightarrow{\mathrm{c}}=-3$. Then $|\overrightarrow{\mathrm{c}}|^2$ is equal to_______.

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Answer

b
$\begin{aligned} & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=2(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+24 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} \\ & (5 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times \overrightarrow{\mathrm{c}}=2(7 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})+24 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}\end{aligned}$

$\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 5 & 1 & 4 \\ x & y & z\end{array}\right|=14 \hat{i}+10 \hat{j}-20 \hat{k}$

$\begin{aligned} & \Rightarrow \hat{i}(z-4 y)-\hat{j}(5 z-4 x)+\hat{k}(5 y-x)=14 \hat{i}+10 \hat{j}-20 \hat{k} \\ & z-4 y=14,4 x-5 z=10,5 y-x=-20 \\ & (a-b+i) \cdot \vec{c}=-3 \\ & (2 \hat{i}+3 \hat{j}-2 \hat{k}) \cdot \vec{c}=-3 \\ & 2 x+3 y-2 z=-3 \\ & \therefore x=5, y=-3, z=2 \\ & |\vec{c}|^2=25+9+4=38\end{aligned}$

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