JEE MAIN - Mathematics (2024 - 1st February Morning Shift - No. 8)
Let $\overrightarrow{\mathrm{a}}=-5 \hat{i}+\hat{j}-3 \hat{k}, \overrightarrow{\mathrm{b}}=\hat{i}+2 \hat{j}-4 \hat{k}$ and
$\overrightarrow{\mathrm{c}}=(((\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \hat{i}) \times \hat{i}) \times \hat{i}$. Then $\vec{c} \cdot(-\hat{i}+\hat{j}+\hat{k})$ is equal to :
$\overrightarrow{\mathrm{c}}=(((\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \hat{i}) \times \hat{i}) \times \hat{i}$. Then $\vec{c} \cdot(-\hat{i}+\hat{j}+\hat{k})$ is equal to :
-12
-10
-13
-15
Explanation
$\begin{aligned} & \vec{a}=-5 \cdot \hat{i}+\hat{j}-3 \hat{k}, \vec{b}=\hat{i}+2 \hat{j}-4 \hat{k} \\\\ & \vec{c}=(((\vec{a} \times \vec{b}) \times \hat{i}) \times \hat{i}) \times \hat{i} \\\\ & =(((\vec{a} \cdot \hat{i}) \vec{b}-(\vec{b} \cdot \hat{i}) \vec{a}) \times \hat{i}) \times \hat{i} \\\\ & =((-5 \vec{b}-\vec{a}) \times \hat{i}) \times \hat{i}\end{aligned}$
$\begin{aligned} & =((-11 \hat{j}+23 \hat{k}) \times \hat{i}) \times \hat{i} \\\\ & = (11 \hat{k}+23 \hat{j}) \times \hat{i} \\\\ & = (11 \hat{j}-23 \hat{k})\end{aligned}$
$\begin{aligned} & \vec{c} \cdot(-\hat{i}+\hat{j}+\hat{k})=0+11-23 \\\\ & =-12\end{aligned}$
$\begin{aligned} & =((-11 \hat{j}+23 \hat{k}) \times \hat{i}) \times \hat{i} \\\\ & = (11 \hat{k}+23 \hat{j}) \times \hat{i} \\\\ & = (11 \hat{j}-23 \hat{k})\end{aligned}$
$\begin{aligned} & \vec{c} \cdot(-\hat{i}+\hat{j}+\hat{k})=0+11-23 \\\\ & =-12\end{aligned}$
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