Let vec a , vec b and vec c be three non-zero vectors such that no two of them are collinear and (vec a times vec b ) times vec c = frac 1 3 |vec b ||vec c |vec a . If θ is the angle between vectors vec b and vec c , then a value of sin θ is:

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EEJ MAIN Mathematics QUESTION #7436

Question 1

Let \(\vec{a}, \vec{b}\) and \(\vec{c}\) be three non-zero vectors such that no two of them are collinear and \((\vec{a} \times \vec{b}) \times \vec{c} = \frac{1}{3}|\vec{b}||\vec{c}|\vec{a}\). If θ is the angle between vectors \(\vec{b}\) and \(\vec{c}\), then a value of sin θ is:

\(\frac{2\sqrt{2}}{3}\)✔️
\(\frac{-\sqrt{2}}{3}\)
\(\frac{2}{3}\)
\(\frac{-2\sqrt{3}}{3}\)

Correct Answer Explanation

Using the vector triple product formula: \((\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}\). Given this equals \(\frac{1}{3}|\vec{b}||\vec{c}|\vec{a}\), we have \((\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a} = \frac{1}{3}|\vec{b}||\vec{c}|\vec{a}\). Since \(\vec{a}\) and \(\vec{b}\) are non-collinear, comparing coefficients: \(\vec{b} \cdot \vec{c} = -\frac{1}{3}|\vec{b}||\vec{c}|\), which gives \(\cos\theta = -\frac{1}{3}\). Therefore, \(\sin\theta = \sqrt{1-\cos^2\theta} = \sqrt{1-\frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}\).

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