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