Given:
Two vectors u = 4i – 3j and v = 8i + 2j.
Required:
To find the angle between the given vectors.
Explanation:
The dot product between two vectors is given by the formula:
[tex]a.b=\lvert{a}\rvert\lvert{b}\rvert cos\theta[/tex]Now first we will find the dot product of the given vectors.
[tex]\begin{gathered} u.v=(4i-3j).(8i+2j) \\ u.v=(4\times8)-(3\times2) \\ u.v=32-6 \\ u.v=26 \end{gathered}[/tex][tex]\begin{gathered} \lvert{u}\rvert=\sqrt{(4)^2+(-3)^2} \\ \lvert{u}\rvert=\sqrt{16+9} \\ \lvert{u}\rvert=\sqrt{25} \\ \lvert{u}\rvert=5 \end{gathered}[/tex][tex]\begin{gathered} \lvert{v}\rvert{=\sqrt{(8)^2+(2)^2}}\rvert \\ \lvert{v}\rvert=\sqrt{64+4} \\ \lvert{v}\rvert=\sqrt{68} \end{gathered}[/tex]Now put these values in the dot product formula:
[tex]\begin{gathered} 26=5\times\sqrt{65}\times cos\theta \\ cos\theta=\frac{26}{5\sqrt{68}} \\ \theta=cos^{-1}(\frac{26}{5\sqrt{68}}) \end{gathered}[/tex]Final Answer:
Thus the angle between the given vectors is:
[tex]\theta=cos^{-1}(\frac{26}{5\sqrt{68}})[/tex]