Let a→, b→ and c→ be three non - zero vectors such that no two of them are collinear and a→×b→×c→=1/3b→c→a→. If θ is the angle between vectors b→ and c→, then a value of sinθ is

Correct answer: (B) (2 √2)/3. Given a→ × b→ × c→= 1/3 b→ c→a→ a→⋅c→ b→- b→⋅c→ a→=1/3 b→c→ a→ ⇒ a→⋅c→=0 and b→⋅c→= (-1)/3 b→c→ b→c→cosθ= (-1)/3 b→c→ cosθ= (-1)/3 sinθ= √(1-19)= √89 = (2√2)/3

Q60 / 90 · JEE Main 2015 (04 Apr) · Mathematics

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

A
$\frac{- 2 \sqrt{3}}{3}$
Correct answer
B
$\frac{2 \sqrt{2}}{3}$
C
$\frac{- \sqrt{2}}{3}$
D
$\frac{2}{3}$

Solution

Given

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

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

\Rightarrow \vec{a} \cdot \vec{c} = 0 and \vec{b} \cdot \vec{c} = \frac{- 1}{3} |\vec{b}| |\vec{c}|

|\vec{b}| |\vec{c}| \cos θ = \frac{- 1}{3} |\vec{b}| |\vec{c}|

\cos θ = \frac{- 1}{3}

\sin θ = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2 \sqrt{2}}{3}

Hence, the correct answer is option: 'B'.

Let a→, b→ and c→ be three non - zero vectors such that no two of them are collinear and a→×b→×c→=13b→c→a→. If θ is the angle between vectors b→ and c→, then a value of sinθ is

Solution

Relevant Questions

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