Step-by-step 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:
![||v||=√((x_2-x_1)^2+(y_2-y_1)^2)=√((-4-4)^2+(-1-3)^2)\cong8.944.](https://img.qammunity.org/2023/formulas/mathematics/college/9b7c097uzz88wjnfnq8z7yqaa1t5qs9cgf.png)
The angle of the vector is given by:
![\tan\theta^(\prime)=(y_2-y_1)/(x_2-x_1)=(-1-3)/(-4-4)=(-4)/(-8)=(1)/(2)\Rightarrow\theta^(\prime)=\tan^(-1)((1)/(2))\cong26.565.](https://img.qammunity.org/2023/formulas/mathematics/college/xaw36a7jjsbbzlmxv2ffgo2jai0wykmans.png)
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:
![θ\cong26.565\degree+180\degree=206.565\degree.](https://img.qammunity.org/2023/formulas/mathematics/college/1kcgm7aa2bjabkhvcfit7ok5u3wqyojo64.png)
Answer
||v|| = 8.944, θ = 206.565°