JEE MAIN - Mathematics (2024 - 31st January Evening Shift - No. 21)
Let $$\vec{a}=3 \hat{i}+2 \hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$$ and $$\vec{c}$$ 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 ________.
Answer
38
Explanation
$$\begin{aligned}
& (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=2(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+24 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} \\
& (5 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times \overrightarrow{\mathrm{c}}=2(7 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})+24 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} \\
& \left|\begin{array}{lrr}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
5 & 1 & 4 \\
x & y & \mathrm{z}
\end{array}\right|=14 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-20 \hat{\mathrm{k}} \\
& \Rightarrow \hat{\mathrm{i}}(\mathrm{z}-4 \mathrm{y})-\hat{\mathrm{j}}(5 \mathrm{z}-4 \mathrm{x})+\hat{\mathrm{k}}(5 \mathrm{y}-\mathrm{x})=14 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-20 \hat{\mathrm{k}} \\
& \mathrm{z}-4 \mathrm{y}=14,4 \mathrm{x}-5 \mathrm{z}=10,5 \mathrm{y}-\mathrm{x}=-20 \\
& (\mathrm{a}-\mathrm{b}+\mathrm{i}) \cdot \overrightarrow{\mathrm{c}}=-3 \\
& (2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \cdot \overrightarrow{\mathrm{c}}=-3 \\
& 2 x+3 y-2 z=-3 \\
& \therefore x=5, y=-3, z=2 \\
& |\overrightarrow{\mathrm{c}}|^2=25+9+4=38
\end{aligned}$$
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