If vec{a} and vec{b} are two unit vectors wit | vedclass

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

Question

(A) Given that $\vec{a}$ and $\vec{b}$ are unit vectors,so $|\vec{a}| = 1$ and $|\vec{b}| = 1$. Given $|\vec{a} - \vec{b}| = 1$,squaring both sides gi

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) Given that $\vec{a}$ and $\vec{b}$ are unit vectors,so $|\vec{a}| = 1$ and $|\vec{b}| = 1$. Given $|\vec{a} - \vec{b}| = 1$,squaring both sides gives $|\vec{a} - \vec{b}|^2 = 1$. $|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} = 1$. $1 + 1 - 2\vec{a} \cdot \vec{b} = 1$,which implies $2\vec{a} \cdot \vec{b} = 1$,so $\vec{a} \cdot \vec{b} = \frac{1}{2}$. Now,$|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = 1 + 1 + 2 	imes \frac{1}{2} = 3$. Thus,$|\vec{a} + \vec{b}| = \sqrt{3}$. Since $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos 	heta = \frac{1}{2}$,we have $\cos 	heta = \frac{1}{2}$. Using the identity $\cos 	heta = 2 \cos^2 \frac{	heta}{2} - 1$,we get $2 \cos^2 \frac{	heta}{2} - 1 = \frac{1}{2}$,so $2 \cos^2 \frac{	heta}{2} = \frac{3}{2}$,which means $\cos^2 \frac{	heta}{2} = \frac{3}{4}$. Therefore,$\cos \frac{	heta}{2} = \frac{\sqrt{3}}{2}$. Finally,$2|\vec{a} + \vec{b}| \cos \frac{	heta}{2} = 2 	imes \sqrt{3} 	imes \frac{\sqrt{3}}{2} = 3$.

If $\vec{a}$ and $\vec{b}$ are two unit vectors with $(\vec{a}, \vec{b}) = \theta$ and $|\vec{a} - \vec{b}| = 1$,then $2|\vec{a} + \vec{b}| \cos \frac{\theta}{2} =$

Similar Questions

In a quadrilateral $ABCD$,if $P$ and $Q$ are the midpoints of $\overline{BC}$ and $\overline{AD}$ respectively,then $\vec{AB} + \vec{DC} = \dots$

If $(a \times b)^{2} + (a \cdot b)^{2} = 144$ and $|a| = 4$,then $|b|$ is equal to

If $\theta$ is the angle between the vectors $4 \hat{i}-\hat{j}+2 \hat{k}$ and $\hat{i}+3 \hat{j}-2 \hat{k}$,then $\sin 2 \theta=$

The value of $\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{k} \times \hat{i}) + \hat{k} \cdot (\hat{i} \times \hat{j})$ is . . . . . . .

In a parallelogram $ABCD$,$|\overline{AB}| = a$,$|\overline{AD}| = b$ and $|\overline{AC}| = c$,then the value of $\overline{DA} \cdot \overline{AB}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module