JEE MAIN - Mathematics (2025 - 24th January Evening Shift - No. 13)
Let $\overrightarrow{\mathrm{a}}=3 \hat{i}-\hat{j}+2 \hat{k}, \overrightarrow{\mathrm{~b}}=\overrightarrow{\mathrm{a}} \times(\hat{i}-2 \hat{k})$ and $\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}} \times \hat{k}$. Then the projection of $\overrightarrow{\mathrm{c}}-2 \hat{j}$ on $\vec{a}$ is :
$2 \sqrt{7}$
$3 \sqrt{7}$
$\sqrt{14}$
$2 \sqrt{14}$
Explanation
$$\begin{aligned}
&\begin{aligned}
& \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{a}} \times(\hat{\mathrm{i}}-3 \hat{\mathrm{k}}) \\
& =\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
3 & -1 & 2 \\
1 & 0 & -2
\end{array}\right|=2 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\
& \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}} \times \hat{\mathrm{k}}=8 \hat{\mathrm{i}}-2 \hat{\mathrm{j}} \\
& \overrightarrow{\mathrm{c}}-2 \hat{\mathrm{j}}=8 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}
\end{aligned}\\
&\text { Projection of }(\hat{i}-2 \hat{j}) \text { on } \vec{a}\\
&\begin{aligned}
& (\overrightarrow{\mathrm{c}}-2 \hat{\mathrm{j}}) \cdot \hat{\mathrm{a}}=\frac{\langle 8,-4,0\rangle \cdot\langle 3,-1,2\rangle}{\sqrt{14}} \\
& =\frac{28}{\sqrt{14}}=2 \sqrt{14}
\end{aligned}
\end{aligned}$$
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