Question

If \( \vec{a} \) and \( \vec{b} \) are two vectors such that \( |\vec{a}|=|\vec{b}|=\sqrt{6} \) and \( \vec{a} \cdot \vec{b} = -1 \), then \( |\vec{a} \times \vec{b}| \sin(\vec{a}, \vec{b}) = \)

• A. \( (|\vec{a}|^2-1)(|\vec{b}|^2+1) \)
• B. \( \frac{1}{6} \)
• C. \( (|\vec{a}|^2-1)(1+\frac{1}{|\vec{b}|^2}) \)
• D. \( \frac{\sqrt{35}}{6} \)

Hint:

Remember the fundamental relationships between dot product, cross product, and the angle \( \theta \) between vectors: \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \) and \( |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta \). These allow you to switch between geometric and algebraic representations.

Answer 1

The Correct Option is C

Let \( \theta \) be the angle between vectors \( \vec{a} \) and \( \vec{b} \).
The expression we need to evaluate is \( |\vec{a} \times \vec{b}| \sin(\theta) \).
By definition, the magnitude of the cross product is \( |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin(\theta) \).
Substituting this into the expression, we get:
\( (|\vec{a}||\vec{b}|\sin(\theta)) \sin(\theta) = |\vec{a}||\vec{b}|\sin^2(\theta) \).
We can find \( \cos(\theta) \) from the dot product formula: \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos(\theta) \).
Given \( \vec{a} \cdot \vec{b} = -1 \) and \( |\vec{a}|=|\vec{b}|=\sqrt{6} \).
\( -1 = (\sqrt{6})(\sqrt{6})\cos(\theta) \implies -1 = 6\cos(\theta) \implies \cos(\theta) = -\frac{1}{6} \).
Using the Pythagorean identity, \( \sin^2(\theta) = 1 - \cos^2(\theta) \).
\( \sin^2(\theta) = 1 - (-\frac{1}{6})^2 = 1 - \frac{1}{36} = \frac{35}{36} \).
Now, substitute the values back into our target expression:
\( |\vec{a}||\vec{b}|\sin^2(\theta) = (\sqrt{6})(\sqrt{6})(\frac{35}{36}) = 6 \cdot \frac{35}{36} = \frac{35}{6} \).
Now we must check which option gives the same numerical value. Let's evaluate option (C).
\( (|\vec{a}|^2-1)(1+\frac{1}{|\vec{b}|^2}) \).
Substitute \( |\vec{a}|^2 = (\sqrt{6})^2 = 6 \) and \( |\vec{b}|^2 = (\sqrt{6})^2 = 6 \).
\( (6-1)(1+\frac{1}{6}) = (5)(\frac{7}{6}) = \frac{35}{6} \).
Since the numerical result matches, option (C) is the correct answer.