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

Let $ABC$ be a triangle and $P$ be a point inside $ABC$ such that $\overrightarrow{PA} + 2\overrightarrow{PB} + 3\overrightarrow{PC} = \vec{0}$. The ratio of the area of $\triangle ABC$ to that of $\triangle APC$ is

In a triangle $ABC$ with usual notations,if $|\overline{BC}|=8, |\overline{CA}|=7, |\overline{AB}|=10$,then the projection of $\overline{AB}$ on $\overline{AC}$ is

Find the sine of the angle between the vectors $\vec{a}=3 \hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}=2 \hat{i}-2 \hat{j}+4 \hat{k}$.

The position vectors of the vertices of a triangle $ABC$ are $4i - 2j$,$i + 4j - 3k$,and $-i + 5j + k$ respectively. Then $\angle ABC = $

If $A=(1,-1,2)$, $B=(3,4,-2)$, $C=(0,3,2)$ and $D=(3,5,6)$, then the angle between the lines $\overrightarrow{AB}$ and $\overrightarrow{CD}$ is (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module