Let hat{a} and hat{b} be unit vectors. If vec | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given that $\hat{a}$ and $\hat{b}$ are unit vectors,so $|\hat{a}| = 1$ and $|\hat{b}| = 1$. Given $\hat{b} = \vec{c} + 2(\vec{c} 	imes \hat{a})$.

Vedclass · Accepted Answer

$6(3+\sqrt{3})$

Vedclass · Answer

(C) Given that $\hat{a}$ and $\hat{b}$ are unit vectors,so $|\hat{a}| = 1$ and $|\hat{b}| = 1$. Given $\hat{b} = \vec{c} + 2(\vec{c} 	imes \hat{a})$. Taking the squared magnitude on both sides: $|\hat{b}|^{2} = |\vec{c} + 2(\vec{c} 	imes \hat{a})|^{2}$ $1 = |\vec{c}|^{2} + 4|\vec{c} 	imes \hat{a}|^{2} + 4\vec{c} \cdot (\vec{c} 	imes \hat{a})$ Since $\vec{c} \cdot (\vec{c} 	imes \hat{a}) = 0$ because the cross product is perpendicular to $\vec{c}$,we have: $1 = |\vec{c}|^{2} + 4|\vec{c}|^{2} |\hat{a}|^{2} \sin^{2}(\frac{\pi}{12})$ $1 = |\vec{c}|^{2} + 4|\vec{c}|^{2} (1) \sin^{2}(\frac{\pi}{12})$ Using $\sin(\frac{\pi}{12}) = \sin(15^{\circ}) = \frac{\sqrt{6}-\sqrt{2}}{4}$,so $\sin^{2}(\frac{\pi}{12}) = \frac{6+2-2\sqrt{12}}{16} = \frac{8-4\sqrt{3}}{16} = \frac{2-\sqrt{3}}{4}$. $1 = |\vec{c}|^{2} [1 + 4(\frac{2-\sqrt{3}}{4})] = |\vec{c}|^{2} (1 + 2 - \sqrt{3}) = |\vec{c}|^{2} (3 - \sqrt{3})$. $|\vec{c}|^{2} = \frac{1}{3-\sqrt{3}} = \frac{3+\sqrt{3}}{9-3} = \frac{3+\sqrt{3}}{6}$. Therefore,$|6\vec{c}|^{2} = 36|\vec{c}|^{2} = 36 	imes \frac{3+\sqrt{3}}{6} = 6(3+\sqrt{3})$.

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

Similar Questions

The vectors $2\,i + 3\,j - 4\,k$ and $a\,i + b\,j + c\,k$ are perpendicular,when

Let $\vec{a}$ and $\vec{b}$ be two unit vectors. If $\vec{c} = \vec{a} + 2\vec{b}$ and $\vec{d} = 5\vec{a} - 4\vec{b}$ are perpendicular to each other,then the angle between $\vec{a}$ and $\vec{b}$ is

If the vectors $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$,$\vec{b} = 2\hat{i} + 4\hat{j} + \hat{k}$,and $\vec{c} = \alpha\hat{i} + \hat{j} + \beta\hat{k}$ are mutually orthogonal,then $(\alpha, \beta) = $

If $\vec{a} = 3\hat{i} - 6\hat{j} - 24\hat{k}$,then which of the following vectors is perpendicular to $\vec{a}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module