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

1.	Find angle between the two vectors, A=5i^+j^B=2i^+4j^
Answer» 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.)

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