Vector Algebra Question 15
Question 15 - 2024 (31 Jan Shift 2)
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 $(\vec{a}-\vec{b}+\hat{i}) \cdot \vec{c}=-3$. Then $|\vec{c}|^{2}$ is equal to
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Answer (38)
Solution
$(\vec{a}+\vec{b}) \times \vec{c}=2(\vec{a} \times \vec{b})+24 \hat{j}-6 \hat{k}$
$(5 \hat{i}+\hat{j}+4 \hat{k}) \times \vec{c}=2(7 \hat{i}-7 \hat{j}-7 \hat{k})+24 \hat{j}-6 \hat{k}$
$\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ 5 & 1 & 4 \ x & y & z\end{array}\right|=14 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-20 \hat{\mathrm{k}}$
$\Rightarrow \hat{\mathrm{i}}(\mathrm{z}-4 \mathrm{y})-\hat{\mathrm{j}}(5 \mathrm{z}-4 \mathrm{x})+\hat{\mathrm{k}}(5 \mathrm{y}-\mathrm{x})=14 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-20 \hat{\mathrm{k}}$
$z-4 y=14,4 \mathrm{x}-5 \mathrm{z}=10,5 y-x=-20$
$(\mathrm{a}-\mathrm{b}+\mathrm{i}) \cdot \overrightarrow{\mathrm{c}}=-3$
$(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \cdot \overrightarrow{\mathrm{c}}=-3$
$2 x+3 y-2 z=-3$
$\therefore x=5, y=-3, z=2$
$|\overrightarrow{\mathrm{c}}|^{2}=25+9+4=38$
