Answer:
Angle between u an v i.e θ is 6°.
Step-by-step explanation:
Given:
[tex]u=2\vec {i}-4\vec{j}\\\\v=3\vec {i}-8\vec{j}[/tex]
To Find:
Angle between u an v i.e θ = ?
Solution:
If u bar and be v bar are two vectors inclined at an angle θ then their scalar product is defined by,
[tex]\vec{u}.\vec{v}=|\vec{u}||\vec{v}|\cos \theta[/tex]
Substituting the given values we get
[tex](2\vec {i}-4\vec{j}).(3\vec {i}-8\vec{j})=\sqrt{2^{2}+(-4)^{2}}\times \sqrt{3^{2}+(-8)^{2}}\cos \theta[/tex]
[tex]2\times 3 + -4\times -8 =\sqrt{20}\times \sqrt{73}\cos \theta \\\\\cos \theta=\dfrac{\sqrt{1460}}{38} =\frac{38}{38.209}=0.9945\\ \\\therefore \theta=\cos^{-1}(0.9945)=6.01\°\\\\\therefore \theta=6\°[/tex]
Angle between u an v i.e θ is 6°.
