JEE Main 2024 — Mathematics Vectors & 3D Geometry

2024mcqhard

Let $\overrightarrow{\mathrm{a}}=2 \hat{i}+5 \hat{j}-\hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}-2 \hat{j}+2 \hat{k}$ and $\overrightarrow{\mathrm{c}}$ be three vectors such that $(\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i})=\vec{a} \times(\vec{c}+\hat{i})$ . If $\vec{a} \cdot \vec{c}=-29$ , then $\vec{c} \cdot(-2 \hat{i}+\hat{j}+\hat{k})$ is equal to:

Official previous-year question

Held on 5 Apr 2024 · Verified 6 Jul 2026.

Options

A
15
B
12
C
10
D
5

Did you get this right?

Solution

Let's assume $\begin{aligned}
\vec{v} & =\vec{a}+\vec{b}+\hat{i} \
& =5 \hat{i}+3 \hat{j}+\hat{k}
\end{aligned} $and$ \overrightarrow{\mathrm{c}}+\hat{\mathrm{i}}=\overrightarrow{\mathrm{p}} $So,$ \begin{aligned}
& \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{p}} \
& \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{0} \
& \overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{a}})=\overrightarrow{0} \
& \Rightarrow \overrightarrow{\mathrm{p}}=\lambda(\overrightarrow{\mathrm{v}}+\overrightarrow{\mathrm{a}}) \
& \overrightarrow{\mathrm{c}}+\mathrm{i}=\lambda(7 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}) \
& \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{a}} \cdot \hat{\mathrm{i}}=\lambda \overline{\mathrm{a}} \cdot(7 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}) \
& -29+2=\lambda(14+40) \
& \lambda=-\frac{1}{2}
\end{aligned}\begin{gathered}\overrightarrow{\mathrm{c}} \cdot(-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\hat{\mathrm{i}} \cdot(-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=\lambda(7 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}) \cdot(-2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \ =-\frac{1}{2}(-14+8)+2=5\end{gathered}$

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