Explanation
We consider a vector with:
• initial point (x₁, y₁) = (4, 3),
,
• final point (x₂, y₂) = (-4, -1).
The magnitude of the vector is given by:
[tex]||v||=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(-4-4)^2+(-1-3)^2}\cong8.944.[/tex]
The angle of the vector is given by:
[tex]\tan\theta^{\prime}=\frac{y_2-y_1}{x_2-x_1}=\frac{-1-3}{-4-4}=\frac{-4}{-8}=\frac{1}{2}\Rightarrow\theta^{\prime}=\tan^{-1}(\frac{1}{2})\cong26.565.[/tex]
We have obtained a positive value of the angle θ'. But we see that our vector points in the negative direction. To take into account this, we must sum 180° to this result:
[tex]θ\cong26.565\degree+180\degree=206.565\degree.[/tex]Answer
||v|| = 8.944, θ = 206.565°
