Question

If \(\vec{a},\vec{b}\) are two vectors such that \(\lvert \vec{a}\rvert =3,\;\lvert \vec{b}\rvert =4,\;\lvert \vec{a}+\vec{b}\rvert =\sqrt{37},\;\lvert \vec{a}-\vec{b}\rvert = k,\) and the angle between \(\vec{a}\) and \(\vec{b}\) is \(\theta,\) then \(\frac{4}{13}\,(k \sin \theta)^2 =\,?\)

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

Exploit \(\lvert \mathbf{u}\pm \mathbf{v}\rvert^2 = \lvert \mathbf{u}\rvert^2 + \lvert \mathbf{v}\rvert^2 \pm 2\,(\mathbf{u}\cdot\mathbf{v})\).
- Relate \(\mathbf{a}\cdot\mathbf{b}=\lvert \mathbf{a}\rvert\,\lvert \mathbf{b}\rvert \cos\theta\) to find \(\cos\theta\), then \(\sin\theta\).

Answer 1

The Correct Option is C

Step 1: Use given magnitudes to find \(\vec{a}\cdot\vec{b}\).
\[ \lvert \vec{a} + \vec{b}\rvert^2 = \lvert \vec{a}\rvert^2 + \lvert \vec{b}\rvert^2 + 2\,\vec{a}\cdot\vec{b} = 3^2 + 4^2 + 2\,(\vec{a}\cdot\vec{b}) = 9 +16 + 2\,(\vec{a}\cdot\vec{b}) =25 + 2\,(\vec{a}\cdot\vec{b}). \] Given \(\lvert \vec{a} + \vec{b}\rvert = \sqrt{37}\), so \[ (\sqrt{37})^2 =37= 25 + 2\,(\vec{a}\cdot\vec{b}) \quad\Longrightarrow\quad \vec{a}\cdot\vec{b} = 6. \] Thus \(\vec{a}\cdot\vec{b}=3\cdot4\cdot\cos\theta=12\cos\theta=6\implies\cos\theta=\tfrac12\). Step 2: Express \(\lvert \vec{a} - \vec{b}\rvert\) in terms of \(k\).
We know \(\lvert \vec{a} - \vec{b}\rvert =k\). Hence \[ k^2 = \lvert \vec{a}\rvert^2 + \lvert \vec{b}\rvert^2 -2\,\vec{a}\cdot\vec{b} = 9 +16 -2\cdot 6 =25 -12 =13. \] So \(k=\sqrt{13}.\) Step 3: \(\sin\theta\) and the final expression.
From \(\cos\theta=\tfrac12,\) we get \(\sin\theta=\tfrac{\sqrt{3}}{2}.\) Thus \[ k\sin\theta = \sqrt{13}\cdot \tfrac{\sqrt{3}}{2} = \tfrac{\sqrt{39}}{2}. \] Hence \[ (k\sin\theta)^2 = \Bigl(\tfrac{\sqrt{39}}{2}\Bigr)^2 = \tfrac{39}{4}. \] Finally, \[ \frac{4}{13}\,(k\sin\theta)^2 = \frac{4}{13}\times \frac{39}{4} = \frac{39}{13} = 3. \] Therefore, \(\boxed{3}\) is the value of \(\tfrac{4}{13}\,(k\sin\theta)^2\).