Let a→, b→ and c→ be three unit vectors such that a→ ×b→ × c→=(√3)/2b→ + c→. If b→ is not parallel to c→ , then the angle between a→ and b→ is

Correct answer: (B) (5π)/6. Given, a→×(b→×c→ )=√32 (b→+c→ ) ⇒a→·c→b→-a→·b→c→=√32 b→+√32c→ On comparing, we get -a→·b→=√32 ⇒a→·b→=-√32 ⇒a→b→cosθ=-√32 Given a→,b→ are unit vector, a→=b→=1 From equation i cosθ=-√32 ⇒cosθ=cos(5π)/6 ⇒θ=(5π)/6

Q55 / 90 · JEE Main 2016 (03 Apr) · Mathematics

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three unit vectors such that $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\sqrt{3}}{2} (\vec{b} + \vec{c}) .$ If $\vec{b}$ is not parallel to $\vec{c}$ , then the angle between $\vec{a}$ and $\vec{b}$ is

A
$\frac{2 π}{3}$
Correct answer
B
$\frac{5 π}{6}$
C
$\frac{3 π}{4}$
D
$\frac{π}{2}$

Solution

Given,
$\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\sqrt{3}}{2} (\vec{b} + \vec{c})$
$\Rightarrow (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \frac{\sqrt{3}}{2} \vec{b} + \frac{\sqrt{3}}{2} \vec{c}$
On comparing, we get
$- (\vec{a} \cdot \vec{b}) = \frac{\sqrt{3}}{2}$
$\Rightarrow (\vec{a} \cdot \vec{b}) = - \frac{\sqrt{3}}{2}$
$\Rightarrow |\vec{a}| |\vec{b}| \cos θ = - \frac{\sqrt{3}}{2}$
Given $\vec{a}, \vec{b}$ are unit vector,
$|\vec{a}| = |\vec{b}| = 1$
From equation $(i)$
$\cos θ = - \frac{\sqrt{3}}{2}$
$\Rightarrow \cos θ = \cos \frac{5 π}{6}$
$\Rightarrow θ = \frac{5 π}{6}$

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

Let a→, b→ and c→ be three unit vectors such that a→ ×b→ × c→=32b→ + c→. If b→ is not parallel to c→ , then the angle between a→ and b→ is

Solution

Relevant Questions

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three unit vectors such that $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\sqrt{3}}{2} (\vec{b} + \vec{c}) .$ If $\vec{b}$ is not parallel to $\vec{c}$ , then the angle between $\vec{a}$ and $\vec{b}$ is