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
Show Hint
- \(\frac{2\sqrt{2}}{3}\)
- \(\frac{2}{3}\)
- \(\frac{\sqrt{2}}{3}\)
- \(\frac{1}{3}\)
The Correct Option is A
Solution and Explanation
To solve this problem, we need to understand the vector identity and how it applies to the given equation.
We have the expression \((\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = \frac{1}{3} |\mathbf{b}||\mathbf{c}|\mathbf{a}\).
The vector identity for the vector triple product states:
\((\mathbf{u} \times \mathbf{v}) \times \mathbf{w} = (\mathbf{u} \cdot \mathbf{w})\mathbf{v} - (\mathbf{v} \cdot \mathbf{w})\mathbf{u}\)
Applying this identity, we substitute \(\mathbf{u} = \mathbf{a}\), \(\mathbf{v} = \mathbf{b}\), \(\mathbf{w} = \mathbf{c}\):
\((\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{b} \cdot \mathbf{c})\mathbf{a}\)
The equation becomes:
\((\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{b} \cdot \mathbf{c})\mathbf{a} = \frac{1}{3} |\mathbf{b}||\mathbf{c}|\mathbf{a}\)
Since no two vectors are collinear, this implies the coefficient of \(\mathbf{b}\) should be zero:
\(\mathbf{a} \cdot \mathbf{c} = 0\).
So \(\mathbf{a}\) and \(\mathbf{c}\) are perpendicular.
The equation simplifies to:
\(-(\mathbf{b} \cdot \mathbf{c}) = \frac{1}{3} |\mathbf{b}||\mathbf{c}|\)
or, removing the negative sign:
\(\mathbf{b} \cdot \mathbf{c} = -\frac{1}{3} |\mathbf{b}||\mathbf{c}|\)
We know:
\(\mathbf{b} \cdot \mathbf{c} = |\mathbf{b}||\mathbf{c}|\cos \theta\)
Therefore:
\(|\mathbf{b}||\mathbf{c}|\cos \theta = -\frac{1}{3} |\mathbf{b}||\mathbf{c}|\\)
This implies:
\(\cos \theta = -\frac{1}{3}\)
Since \(\theta\) is acute, \(\cos \theta\) should be positive, our derived cosine being negative tells us about the perpendicularity of components within a rotated system. However, focusing on sine due to acute angle property:
We know the identity:
\(\sin^2 \theta + \cos^2 \theta = 1\)
Substitute \(\cos \theta = -\frac{1}{3}\) directly may mislead due to need of acute properties, observed SIN directly suits:
\(\sin^2 \theta = 1 - \left(-\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9}\)
Thus,
\(\sin \theta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}\)
Hence, the correct option is \(\frac{2\sqrt{2}}{3}\).
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