CBSE 2026 · Region 5 · Set 1 · Q24 · 2 marks
Three honey bees were found flying along the vectors \[\vec{\mathrm{a}}=2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}, \vec{\mathrm{b}}=4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}} \text { and } \vec{\mathrm{c}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{k}} \text { respectively. } \] Find the value of $\displaystyle \lambda$ such that the path for $\displaystyle \overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}$ is perpendicular to $\displaystyle \overrightarrow{\mathrm{c}}$.If A, B and C be three non-collinear points such that $\displaystyle \overrightarrow{\mathrm{AB}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\displaystyle \overrightarrow{\mathrm{AC}}=2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}$, then find the area of $\displaystyle \triangle \mathrm{ABC}$.
Three honey bees were found flying along the vectors \[\vec{\mathrm{a}}=2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}, \vec{\mathrm{b}}=4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}} \text { and } \vec{\mathrm{c}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{k}} \text { respectively. } \] Find the value of $\displaystyle \lambda$ such that the path for $\displaystyle \overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}$ is perpendicular to $\displaystyle \overrightarrow{\mathrm{c}}$.
If A, B and C be three non-collinear points such that $\displaystyle \overrightarrow{\mathrm{AB}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\displaystyle \overrightarrow{\mathrm{AC}}=2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}$, then find the area of $\displaystyle \triangle \mathrm{ABC}$.
Marking-scheme solution
As $\displaystyle \vec{a}+\lambda \vec{b}$ is perpendicular to $\displaystyle \vec{c}$
$\displaystyle \therefore(\vec{a}+\lambda \vec{b}) \cdot \vec{c}=0$
i.e. $\displaystyle \vec{a} \cdot \vec{c}+\lambda \vec{b} \cdot \vec{c}=0$
which gives $\displaystyle 8+\lambda(-4)=0$
$\displaystyle \lambda=2$
Area of $\displaystyle \triangle ABC=\dfrac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|$
$\displaystyle =\dfrac{1}{2}|-3 \hat{i}-2 \hat{j}-7 \hat{k}|$
$\displaystyle =\dfrac{\sqrt{62}}{2}$
Vector AlgebraProduct of Two VectorsApplyvery_short_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.