Question

Find angle between the two vectors, A=5i^+j^
B=2i^+4j^

Answer 1

To find :

• Angle between vectors $\sf{\vec{A}}$ = 5 $\sf{\hat{i}}$ + $\sf{\hat{j}}$  and $\sf{\vec{B}}$ = 2 $\sf{\hat{i}}$ + 4 $\sf{\hat{j}}$

Answer :

▶ Dot product of $\sf{\vec{A}}$ and $\sf{\vec{B}}$

$:\implies\sf{\vec{A}\;.\;\vec{B}= (5\;\hat{i} + \hat{j} ) \;.\; ( 2\;\hat{i} + 4\;\hat{j})}$

$:\implies\sf{\vec{A}\;.\;\vec{B}= (5) (2) + (1) (4) }$

$:\implies\sf{\vec{A}\;.\;\vec{B}= 14}$

▶ Magnitude of vectors

$:\implies\sf{| \vec{A} | = \sqrt{ (5)^2 + (1)^2 } }$

$:\implies\sf{| \vec{A} | = \sqrt{26} }$

and

$:\implies\sf{|B|=\sqrt{(2)^2+ (4)^2}}$

$:\implies\sf{|B|=\sqrt{20}}$

▶ Let, Angle between vectors $\sf{\vec{A}}$ and $\sf{\vec{B}}$ be $\theta$

then,

$:\implies\sf{cos\;\theta=\dfrac{\vec{A} \;.\;\vec{B}}{|\vec{A}|\;.\;|\vec{B}|}}$

$:\implies\sf{cos\;\theta=\dfrac{14}{\sqrt{26}\;.\;\sqrt{20}}}$

$:\implies\sf{cos\;\theta=\dfrac{14}{\sqrt{520}}}$

$:\implies\sf{cos\;\theta=\dfrac{14}{2\sqrt{130}}}$

$:\implies\sf{cos\;\theta=\dfrac{7}{\sqrt{130}}}$

$:\implies\sf{cos\;\theta=\dfrac{7\sqrt{130}}{130}}$

$:\implies\sf{\theta=cos^{-1}\bigg(\dfrac{7\sqrt{130}}{130}\bigg)}$

$:\implies\boxed{\boxed{\large{\sf{\;\;\theta=52.125\;^{\circ}\;}}}}$

Therefore,

• Angle between vectors $\sf{\vec{A}}$ and $\sf{\vec{B}}$ will be 52.125°. (Approx.)