CBSE 2022 · Region 1 · Set 3 · Q6 · 2 marks
If $\displaystyle \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\displaystyle \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ are two vectors for which the vector $\displaystyle (\vec{\mathrm{a}}+\vec{\mathrm{b}})$ is perpendicular to the vector $\displaystyle (\vec{\mathrm{a}}-\vec{\mathrm{b}})$, then find all the possible values of y.
Marking-scheme solution
\[\begin{aligned}
& (\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=0 \\
& \quad \Rightarrow|\vec{a}|^{2}-|\vec{b}|^{2}=0 \\
& \quad \Rightarrow y^{2}+5-14=0 \\
& \quad \Rightarrow y=+3 \quad \text { or }-3
\end{aligned}
\]
Vector AlgebraProduct of Two VectorsApplyvery_short_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.