Question

If \( \vec{a} \) and \( \vec{b} \) are two vectors such that \( |\vec{a}| = \sqrt{14} \), \( |\vec{b}| = \sqrt{14} \), and \( \vec{a} \cdot \vec{b} = -7 \), then \[ \frac{|\vec{a} \times \vec{b}|}{|\vec{a} \cdot \vec{b}|} \] is:

• A. \( \frac{7}{\sqrt{3}} \)
• B. \( \sqrt{3} \)
• C. \( \frac{49}{\sqrt{3}} \)
• D. \( \frac{\sqrt{3}}{7} \)

Hint:

For vector problems, use trigonometric identities and vector product formulas to relate the given quantities and solve for unknowns.

Answer 1

The Correct Option is B

We are given: \[ |\vec{a}| = \sqrt{14}, \quad |\vec{b}| = \sqrt{14}, \quad \vec{a} \cdot \vec{b} = -7 \] We need to find the value of: \[ \frac{|\vec{a} \times \vec{b}|}{|\vec{a} \cdot \vec{b}|} \] Step 1: Use the formula for the magnitude of the cross product: \[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \] where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \). Step 2: Use the formula for the dot product: \[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \] Substitute the given values: \[ -7 = \sqrt{14} \times \sqrt{14} \cos \theta = 14 \cos \theta \] So: \[ \cos \theta = -\frac{1}{2} \] Step 3: Using \( \sin^2 \theta + \cos^2 \theta = 1 \), we find: \[ \sin^2 \theta = 1 - \left( -\frac{1}{2} \right)^2 = \frac{3}{4} \] Thus: \[ \sin \theta = \frac{\sqrt{3}}{2} \] Step 4: Now substitute \( \sin \theta \) into the formula for the cross product: \[ |\vec{a} \times \vec{b}| = \sqrt{14} \times \sqrt{14} \times \frac{\sqrt{3}}{2} = 14 \times \frac{\sqrt{3}}{2} = 7\sqrt{3} \] Step 5: Finally, calculate the ratio: \[ \frac{|\vec{a} \times \vec{b}|}{|\vec{a} \cdot \vec{b}|} = \frac{7\sqrt{3}}{7} = \sqrt{3} \] % Final Answer \[ \boxed{\sqrt{3}} \]

Answer 2

Given:
- \( |\vec{a}| = \sqrt{14} \)
- \( |\vec{b}| = \sqrt{14} \)
- \( \vec{a} \cdot \vec{b} = -7 \)

Step 1: Use identity for cross product magnitude
The magnitude of the cross product is given by:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta \]
Also, the dot product is:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta \]
So:
\[ \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{-7}{\sqrt{14} \cdot \sqrt{14}} = \frac{-7}{14} = -\frac{1}{2} \]

Step 2: Find sinθ using identity
We use the identity:
\[ \sin^2\theta = 1 - \cos^2\theta = 1 - \left( -\frac{1}{2} \right)^2 = 1 - \frac{1}{4} = \frac{3}{4} \Rightarrow \sin\theta = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

Step 3: Compute the value of the expression
\[ \frac{|\vec{a} \times \vec{b}|}{|\vec{a} \cdot \vec{b}|} = \frac{|\vec{a}||\vec{b}| \sin\theta}{|\vec{a} \cdot \vec{b}|} = \frac{\sqrt{14} \cdot \sqrt{14} \cdot \frac{\sqrt{3}}{2}}{| -7 |} = \frac{14 \cdot \frac{\sqrt{3}}{2}}{7} = \frac{7 \sqrt{3}}{7} = \sqrt{3} \]

Final Answer:
\[ \boxed{\sqrt{3}} \]