CBSE 2022 · Region 5 · Set 3 · Q7 · 3 marks
If $\displaystyle \overrightarrow{\mathrm{a}}$ and $\displaystyle \overrightarrow{\mathrm{b}}$ are two vectors such that $\displaystyle |\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{b}}|$, then prove that $\displaystyle (\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{~b}})$ is perpendicular to $\displaystyle \overrightarrow{\mathrm{a}}$.If $\displaystyle \overrightarrow{\mathrm{a}}$ and $\displaystyle \overrightarrow{\mathrm{b}}$ are unit vectors and $\displaystyle \theta$ is the angle between them, then prove that $\displaystyle \sin \frac{\theta}{2}=\frac{1}{2}|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|$.
If $\displaystyle \overrightarrow{\mathrm{a}}$ and $\displaystyle \overrightarrow{\mathrm{b}}$ are two vectors such that $\displaystyle |\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{b}}|$, then prove that $\displaystyle (\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{~b}})$ is perpendicular to $\displaystyle \overrightarrow{\mathrm{a}}$.
If $\displaystyle \overrightarrow{\mathrm{a}}$ and $\displaystyle \overrightarrow{\mathrm{b}}$ are unit vectors and $\displaystyle \theta$ is the angle between them, then prove that $\displaystyle \sin \frac{\theta}{2}=\frac{1}{2}|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|$.
Marking-scheme solution
\[\begin{aligned}
& |\vec{a}+\vec{b}|=|\vec{b}| \\
& (\vec{a}+\vec{b})^{2}=(\vec{b})^{2} \\
& \vec{a}^{2}+\vec{b}^{2}+2 \vec{a} \cdot \vec{b}=\vec{b}^{2} \\
& \vec{a}^{2}+2 \vec{a} \cdot \vec{b}=0 \\
& (\vec{a}+2 \vec{b}) \cdot \vec{a}=0 \\
& \therefore(\vec{a}+2 \vec{b}) \perp \vec{a}
\end{aligned}
\]
Or
Vector AlgebraProduct of Two VectorsApplyshort_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.