let v be a vector with initial point (-5, -11) and terminal point (6, 2). what is |v|?\n√24\n√290\n24\n290

Answer:

B. $\sqrt{290}$

Explanation:

Step1: Find vector components

If the initial point is $(x_1,y_1)=(-5,-11)$ and the terminal point is $(x_2,y_2)=(6,2)$, then the components of the vector $\vec{v}$ are $v_x=x_2 - x_1$ and $v_y=y_2 - y_1$. $v_x=6-(-5)=11$ $v_y=2 - (-11)=13$

Step2: Calculate the magnitude

The magnitude of a 2 - dimensional vector $\vec{v}=\langle v_x,v_y\rangle$ is given by $|\vec{v}|=\sqrt{v_x^{2}+v_y^{2}}$. $|\vec{v}|=\sqrt{11^{2}+13^{2}}=\sqrt{121 + 169}=\sqrt{290}$