Question

Let \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) be non-zero vectors such that no two are collinear and \((\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = \frac{1}{3} |\mathbf{b}||\mathbf{c}|\mathbf{a}\). If \(\theta\) is the acute angle between the vectors \(\mathbf{b}\) and \(\mathbf{c}\), then \(\sin \theta\) equals

• A. \(\frac{2\sqrt{2}}{3}\)
• B. \(\frac{2}{3}\)
• C. \(\frac{\sqrt{2}}{3}\)
• D. \(\frac{1}{3}\)

Hint:

Vector triple product identity: \((\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{b} \cdot \mathbf{c})\mathbf{a}\).

Answer 1

The Correct Option is A

To solve the problem, we need to figure out the value of \(\sin \theta\), where \(\theta\) is the acute angle between vectors \(\mathbf{b}\) and \(\mathbf{c}\). We have the equation:

\((\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = \frac{1}{3} |\mathbf{b}||\mathbf{c}|\mathbf{a}\)

We can use the vector triple product identity, which states:

\((\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{b} \cdot \mathbf{c})\mathbf{a}\)

Equate this with the given:

\((\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{b} \cdot \mathbf{c})\mathbf{a} = \frac{1}{3} |\mathbf{b}||\mathbf{c}|\mathbf{a}\)

As no two vectors are collinear, the vector \(\mathbf{a}\) on both sides suggests:

\((\mathbf{b} \cdot \mathbf{c}) = \frac{1}{3} |\mathbf{b}||\mathbf{c}|\)

This implies:

\(\cos \theta = \frac{1}{3}\)

We use the identity \(\sin^2 \theta + \cos^2 \theta = 1\) to find \(\sin \theta\):

\(\sin^2 \theta = 1 - \cos^2 \theta\) \(\sin^2 \theta = 1 - \left(\frac{1}{3}\right)^2\) \(\sin^2 \theta = 1 - \frac{1}{9}\) \(\sin^2 \theta = \frac{8}{9}\)

Therefore, \(\sin \theta\) is:

\(\sin \theta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}\)

The solution is the correct answer:

The value of \(\sin \theta\) is \(\frac{2\sqrt{2}}{3}\).