Q72 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics
If $\vec{a}, \vec{b}$ are two vectors such that $|\vec{a}|=3,|\vec{b}|=4,|\vec{a}+\vec{b}|=\sqrt{37},|\vec{a}-\vec{b}|=k$ and $(\vec{a}, \vec{b})=\theta$, then $\frac{4}{13}(k \sin \theta)^2=$
- A1
- B2
- Correct answerC3
- D4
Solution
$\because|\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b}$
$\begin{aligned}
& \therefore 37=9+16+2 \vec{a} \cdot \vec{b} \\
& \Rightarrow 12=2 \vec{a} \cdot \vec{b} \Rightarrow \vec{a} \cdot \vec{b}=6 \\
& \Rightarrow|\vec{a}||\vec{b}| \cos \theta=6 \\
& \Rightarrow 3 \cdot 4 \cos \theta=6 \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=60^{\circ} \\
& \therefore|\vec{a}-\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2-2 \vec{a} \cdot \vec{b} \\
& \Rightarrow k^2=9+16-12=13
\end{aligned}$
Now, $\frac{4}{13}(k \sin \theta)^2=\frac{4}{13} \cdot k^2 \sin ^2 \theta$ $=\frac{4}{13} \cdot 13 \sin ^2 60^{\circ}=4 \cdot \frac{3}{4}=3 .$
Now, $\frac{4}{13}(k \sin \theta)^2=\frac{4}{13} \cdot k^2 \sin ^2 \theta$ $=\frac{4}{13} \cdot 13 \sin ^2 60^{\circ}=4 \cdot \frac{3}{4}=3 .$
Hence, the correct answer is option: 'C'.