CBSE 2022 · Region 2 · Set 1 · Q3 · 2 marks
$\displaystyle \overrightarrow{\mathrm{a}}$ and $\displaystyle \overrightarrow{\mathrm{b}}$ are two unit vectors such that $\displaystyle |2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{~b}}|=|3 \overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{~b}}|$. Find the angle between $\displaystyle \overrightarrow{\mathrm{a}}$ and $\displaystyle \overrightarrow{\mathrm{b}}$.
Marking-scheme solution
\[\begin{aligned}
& |2 \vec{a}+3 \vec{b}|=|3 \vec{a}-2 \vec{b}| \\
& \Rightarrow|2 \vec{a}+3 \vec{b}|^{2}=|3 \vec{a}-2 \vec{b}|^{2} \\
& \Rightarrow 4|\vec{a}|^{2}+12 \vec{a} \cdot \vec{b}+9|\vec{b}|^{2}=9|\vec{a}|^{2}-12 \vec{a} \cdot \vec{b}+4|\vec{b}|^{2}
\end{aligned}
\]
As \(\displaystyle |\vec{a}|=|\vec{b}|=1\)
\(\displaystyle \therefore 24 \vec{a} \cdot \vec{b}=5|\vec{a}|^{2}-5|\vec{b}|^{2}=0 \Longrightarrow \vec{a} \cdot \vec{b}=0\)
So, \(\displaystyle \vec{a} \perp \vec{b}\) or Angle between them is \(\displaystyle \frac{\pi}{2}\)
Vector AlgebraProduct of Two VectorsApplyvery_short_answermedium
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