CBSE 2022 · Region 4 · Set 1 · Q9 · 3 marks
If $\displaystyle \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\displaystyle \overrightarrow{\mathrm{c}}$ are mutually perpendicular vectors of equal magnitude, then prove that the vector $\displaystyle (2 \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+2 \overrightarrow{\mathrm{c}})$ is equally inclined to both $\displaystyle \overrightarrow{\mathrm{a}}$ and $\displaystyle \overrightarrow{\mathrm{c}}$. Also, find the angle between $\displaystyle \overrightarrow{\mathrm{a}}$ and $\displaystyle (2 \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+2 \overrightarrow{\mathrm{c}})$.
Marking-scheme solution
Let, \(\displaystyle |\vec{a}|=|\vec{b}|=|\vec{c}|=m\)
\[\begin{aligned}
|2 \vec{a}+\vec{b}+2 \vec{c}|^{2} & =4|\vec{a}|^{2}+|\vec{b}|^{2}+4|\vec{c}|^{2}+4 \vec{a} \cdot \vec{b}+4 \vec{b} \cdot \vec{c}+8 \vec{a} \cdot \vec{c} \\
& =9 m^{2} \\
|2 \vec{a}+\vec{b}+2 \vec{c}| & =3 m
\end{aligned}
\]
Let \(\displaystyle \theta\) be angle between \(\displaystyle 2 \vec{a}+\vec{b}+2 \vec{c}\) and \(\displaystyle \vec{a}\).
\(\displaystyle \cos \theta=\frac{(2 \vec{a}+\vec{b}+2 \vec{c}) \cdot \vec{a}}{|2 \vec{a}+\vec{b}+2 \vec{c}||\vec{a}|}=\frac{2 m^{2}}{3 m \cdot m}=\frac{2}{3}\)
Let, \(\displaystyle \alpha\) be angle between \(\displaystyle 2 \vec{a}+\vec{b}+2 \vec{c}\) and \(\displaystyle \vec{c}\)
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.