Let a^,b^ be unit vectors. If c→ be a vector such that the angle between a^ and c→ is (π)/12, and b^=c→+2c→×a^, then 6c→^2 is equal to:

Correct answer: (B) 63+√3. We have, b^=c→+2c→×a^, b^·b^=c→+2c→×a^·c→+2c→×a^ ⇒1=c→·c→+c→·2c→×a^+2c→×a^·c→+4c→×a^·c→×a^ ⇒1=c→^2+0+0+c→×a^^2 ⇒1=c→^2+4c→^2a^^2sin^2(π)/12 ⇒1=c→^2+4c→^2(1-cosπ6)/2 ⇒1=c→^21+21-√32 ⇒1=c→^23-√3 ⇒c→^2=1/(3-√3)=(3+√3)/6 ∴36c→^2=(3+√3)/6×36=63+√3

Q46 / 90 · JEE Main 2022 (24 Jun Shift 1) · Mathematics

Let $\hat{a}, \hat{b}$ be unit vectors. If $\vec{c}$ be a vector such that the angle between $\hat{a}$ and $\vec{c}$ is $\frac{π}{12}$ , and $\hat{b} = \vec{c} + 2 (\vec{c} \times \hat{a})$ , then ${|6 \vec{c}|}^{2}$ is equal to:

A
$6 (3 - \sqrt{3})$
Correct answer
B
$6 (3 + \sqrt{3})$
C
$3 + \sqrt{3}$
D
$6 (\sqrt{3} + 1)$

Solution

We have, $\hat{b} = \vec{c} + 2 (\vec{c} \times \hat{a})$ ,

$\hat{b} \cdot \hat{b} = (\vec{c} + 2 (\vec{c} \times \hat{a})) \cdot (\vec{c} + 2 (\vec{c} \times \hat{a}))$

$\Rightarrow 1 = (\vec{c} \cdot \vec{c}) + \vec{c} \cdot 2 (\vec{c} \times \hat{a}) + 2 (\vec{c} \times \hat{a}) \cdot \vec{c} + 4 (\vec{c} \times \hat{a}) \cdot (\vec{c} \times \hat{a})$

$\Rightarrow 1 = {|\vec{c}|}^{2} + 0 + 0 + {|\vec{c} \times \hat{a}|}^{2}$ $\Rightarrow 1 = {|\vec{c}|}^{2} + 4 {|\vec{c}|}^{2} {|\hat{a}|}^{2} \sin^{2} \frac{π}{12}$

$\Rightarrow 1 = {|\vec{c}|}^{2} + 4 {|\vec{c}|}^{2} (\frac{1 - \cos \frac{π}{6}}{2})$

$\Rightarrow 1 = {|\vec{c}|}^{2} [1 + 2 (1 - \frac{\sqrt{3}}{2})]$ $\Rightarrow 1 = {|\vec{c}|}^{2} [3 - \sqrt{3}]$ $\Rightarrow {|\vec{c}|}^{2} = \frac{1}{3 - \sqrt{3}} = \frac{3 + \sqrt{3}}{6}$

$∴ 36 {|\vec{c}|}^{2} = \frac{3 + \sqrt{3}}{6} \times 36 = 6 (3 + \sqrt{3})$

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

Let a^,b^ be unit vectors. If c→ be a vector such that the angle between a^ and c→ is π12, and b^=c→+2c→×a^, then 6c→2 is equal to:

Solution

Relevant Questions

Let $\hat{a}, \hat{b}$ be unit vectors. If $\vec{c}$ be a vector such that the angle between $\hat{a}$ and $\vec{c}$ is $\frac{π}{12}$ , and $\hat{b} = \vec{c} + 2 (\vec{c} \times \hat{a})$ , then ${|6 \vec{c}|}^{2}$ is equal to: