CBSE 2024 · Region 5 · Set 2 · Q30 · 3 marks
Given $\displaystyle \vec{a}=2 \hat{i}-\hat{j}+\hat{k}, \vec{b}=3 \hat{i}-\hat{k}$ and $\displaystyle \vec{\mathrm{c}}=2 \hat{i}+\hat{j}-2 \hat{k}$. Find a vector $\displaystyle \vec{\mathrm{d}}$ which is perpendicular to both $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$ and $\displaystyle \vec{\mathrm{c}} \cdot \vec{\mathrm{d}}=3$.
Marking-scheme solution
Since $\displaystyle \overrightarrow{\mathbf{d}} \perp \overrightarrow{\mathbf{a}}$ and $\displaystyle \overrightarrow{\mathbf{d}} \perp \overrightarrow{\mathbf{b}} \Rightarrow \overrightarrow{\mathbf{d}}=\lambda(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})$\overrightarrow{\mathbf{d}}=\lambda(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})$\displaystyle \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}}=3 \Rightarrow 2 \lambda+5 \lambda-6 \lambda=3 \Rightarrow \lambda=3$
$\displaystyle \Rightarrow \overrightarrow{\mathbf{d}}=3 \hat{\mathbf{i}}+\mathbf{1 5} \hat{\mathbf{j}}+\mathbf{9} \hat{\mathbf{k}}$
Vector AlgebraProduct of Two VectorsApplyshort_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.