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.)


Discussion

No Comment Found

Related InterviewSolutions