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JEE Main 24-Jan-2025 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 24-Jan-2025 Paper - Shift 24 Marks
MCQ
Let $\vec{a}=3 \hat{i}-\hat{j}+2 \hat{k}, \vec{b}=\vec{a} \times(\hat{i}-2 \hat{k})$ and $\vec{c}=\vec{b} \times \hat{k}$.
Then the projection of $\overrightarrow{\mathbf{c}}-2 \hat{\mathrm{j}}$ on $\overrightarrow{\mathrm{a}}$ is:
  • A
    $3 \sqrt{7}$
  • B
    $\sqrt{14}$
  • ✓
    $2 \sqrt{14}$
  • D
    $2 \sqrt{7}$
✓

Answer

Correct option: C.
$2 \sqrt{14}$
(C)
Sol. $\dot{b}=\vec{a} \times(\hat{i}-3 \hat{k})$
$=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 2 \\ 1 & 0 & -2\end{array}\right|=2 \hat{i}+8 \hat{j}+\hat{k}$
$\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}} \times \hat{\mathrm{k}}=8 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}$
$\overrightarrow{\mathrm{c}}-2 \hat{\mathrm{j}}=8 \hat{\mathbf{i}}-4 \hat{\mathrm{j}}$
Projection of $(\hat{\mathrm{i}}-2 \hat{\mathrm{j}})$ on $\overrightarrow{\mathrm{a}}$
$(\overrightarrow{\mathrm{c}}-2 \hat{\mathrm{j}}) \cdot \hat{\mathrm{a}}=\frac{\langle 8,-4,0\rangle \cdot\langle 3,-1,2\rangle}{\sqrt{14}}$
$=\frac{28}{\sqrt{14}}=2 \sqrt{14}$

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