Question

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Answer 1

Given :

$\bullet\;\sf{\vec{F}=3\hat{i}+2\hat{j}}$

$\bullet\;\sf{\vec{s}=2\hat{i}+3\hat{j}}$

To find :

$\bullet\;\sf{angle\:between\:\vec{F}\:and\:\vec{s}\:,\:\theta=?}$

Knowledge required :

• magnitude of a vector with orthogonal notations :

$\sf{Let\:\vec{A}=a\hat{i}+b\hat{j}\:\:then,}$

$\red{\bigstar}\;\boxed{\sf{ \:\mid\vec{ A}\mid=\sqrt{a^2+b^2}}}$

• Dot product of two vectors with orthogonal notations :

$\sf{Let\,\:\vec{A}=a\hat{i}+b\hat{j}\:and\:\vec{B}=c\hat{i}+d\hat{j}\:then,}$

$\red{\bigstar}\;\boxed{\sf{\vec{A}\;.\;\vec{B}=ac\;+\;bd}}$

• Dot product of any two vectors :

$\sf{Let\:\vec{A}\:and\:\vec{B}\:are\:two\:given\:vectors\:and\:\theta\:is\:angle\:b/w\:them\:then,}$

$\red{\bigstar}\;\boxed{\sf{\vec{A}\;.\;\vec{B}=| \vec{A} | .|\vec{B}| \;cos\;\theta}}$

Solution :

Calculating magnitude of vector F

$\implies\sf{\mid \vec{F}\mid=\sqrt{(3)^2+(2)^2}}$

$\implies\sf{\mid \vec{F}\mid=\sqrt{9+4}}$

$\underline{\underline{\red{\sf{\implies\mid \vec{F} \mid = \sqrt{13}}}}}$

Calculating magnitude of vector s

$\implies\sf{\mid \vec{s}\mid=\sqrt{(2)^2+(3)^2}}$

$\implies\sf{\mid \vec{s}\mid=\sqrt{4+9}}$

$\underline{\underline{\red{\sf{\implies\mid \vec{s} \mid = \sqrt{13}}}}}$

Calculating Dot product of vector F and vector s

$\implies\sf{\vec{F}\;.\;\vec{s}=(3)(2)+(2)(3)}$

$\implies\sf{\vec{F}\;.\;\vec{s}=(6)+(6)}$

$\underline{\underline{\red{\sf{\implies \vec{F}\;.\;\vec{s} = 12}}}}$

Calculating the angle between vector F and vector s (θ)

$\implies\sf{\vec{F}\;.\;\vec{s}=|\vec{F}|.|\vec{s}|\;cos\theta}$

putting values

$\implies\sf{12=\sqrt{13}\;.\;\sqrt{13}\;\;cos\;\theta}$

$\implies\sf{12=13\;\;cos\;\theta}$

$\implies\sf{cos\;\theta=\dfrac{12}{13}}$

$\implies\underline{\underline{\boxed {\red{\sf{ \theta=cos^{-1}\left(\dfrac{12}{13}\right)}}}}}$

Therefore,

Angle between vector F and vector s is

$\red{\sf{\theta=cos^{-1}\left(\dfrac{12}{13}\right)}}$