Q71 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics
Let $|\vec{a}|=2,|\vec{b}|=3$ and the angle between $\vec{a}$ and $\vec{b}$ be $\frac{\pi}{3}$. If a parallelogram is constructed with adjacent sides $2 \vec{a}+3 \vec{b}$ and $\vec{a}-\vec{b}$, then its shorter diagonal is of length
- A$108$
- B$172$
- Correct answerC$6 \sqrt{3}$
- D$2 \sqrt{43}$
Solution
$\begin{gathered}
|\vec{a}|=2,|\vec{b}|=3, \theta=\frac{\pi}{3} \\
\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta=6 \times \frac{1}{2}=3
\end{gathered}$
Diagonals are $\vec{p}+\vec{q}$ and $\vec{p}-\vec{q}$ where $\begin{aligned} & \vec{p}=2 \vec{a}+3 \vec{b} \text { and } \vec{q}=\vec{a}-\vec{b} \\ & |\vec{p}+\vec{q}|^2=|3 \vec{a}+2 \vec{b}|^2=9|\vec{a}|^2+4|\vec{b}|^2+12 a . b \\ & =9 \times 4+4 \times 9+12 \times 3=108 \\ & |\vec{p}+\vec{q}|=6 \sqrt{3} \end{aligned}$
Diagonals are $\vec{p}+\vec{q}$ and $\vec{p}-\vec{q}$ where $\begin{aligned} & \vec{p}=2 \vec{a}+3 \vec{b} \text { and } \vec{q}=\vec{a}-\vec{b} \\ & |\vec{p}+\vec{q}|^2=|3 \vec{a}+2 \vec{b}|^2=9|\vec{a}|^2+4|\vec{b}|^2+12 a . b \\ & =9 \times 4+4 \times 9+12 \times 3=108 \\ & |\vec{p}+\vec{q}|=6 \sqrt{3} \end{aligned}$
Hence, the correct answer is option: 'C'.