Question

If \(\bar{a}, \bar{b}, \bar{c}\) are three vectors such that \(|\overrightarrow{\mathrm{a}}|=\sqrt{31}, 4|\overrightarrow{\mathrm{~b}}|=|\overrightarrow{\mathrm{c}}|=2\) and \(2(\overline{\mathrm{a}} \times \overline{\mathrm{b}})=3(\overline{\mathrm{c}} \times \overline{\mathrm{a}})\) and if the angle between \(\overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\) is \(\frac{2 \pi}{3}\) then \(\left|\frac{\bar{a} \times \bar{c}}{\bar{a} \cdot \bar{b}}\right|^2=\)

• A. 1
• B. 2
• C. 3
• D. 11

Hint:

Rewrite the cross product relation to show that a is parallel to 2b + 3c, then compute both products.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
\(2(\bar{a}\times\bar{b}) = 3(\bar{c}\times\bar{a}) = -3(\bar{a}\times\bar{c})\), so \(\bar{a}\times(2\bar{b} + 3\bar{c}) = 0\). Hence \(\bar{a}\) is parallel to \(2\bar{b} + 3\bar{c}\).

Step 2: Magnitudes
\(|\bar{b}| = \dfrac{1}{2}\), \(|\bar{c}| = 2\), \(\bar{b}\cdot\bar{c} = \dfrac{1}{2}\cdot 2\cos 120^{\circ} = -\dfrac{1}{2}\).
\(|2\bar{b} + 3\bar{c}|^2 = 1 + 36 - 6 = 31 = |\bar{a}|^2\), so \(\bar{a} = \pm(2\bar{b} + 3\bar{c})\).

Step 3: Dot product
\(\bar{a}\cdot\bar{b} = \pm(2|\bar{b}|^2 + 3\bar{b}\cdot\bar{c}) = \pm\left(\dfrac{1}{2} - \dfrac{3}{2}\right) = \mp 1\), so \(|\bar{a}\cdot\bar{b}| = 1\).

Step 4: Cross product
\(\bar{a}\times\bar{c} = \pm 2(\bar{b}\times\bar{c})\), and \(|\bar{b}\times\bar{c}| = \dfrac{1}{2}\cdot 2\cdot\dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}}{2}\).
So \(|\bar{a}\times\bar{c}| = \sqrt{3}\), and the required value is \(\dfrac{3}{1} = 3\).

Final Answer:
The value is 3. \[ \boxed{\text{(C) }3} \]