Question

A vector starts at point ( 8, 7) and ends at point (5, 4) what is the magnitude of the vector, answer to two decimal places.

Answer 1

The points are given below,

$\begin{gathered} A\rightarrow(x_1,y_1)\rightarrow(8,7) \\ B\rightarrow(x_2,y_2)\rightarrow(5,4) \end{gathered}$

The formula for the magnitude of the vector AB is,

$|\vec{AB}|=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2_{}}$

Solving for the magnitude of the vector AB by substituting the values of x₁ = 8, x₂= 5, y₁ = 7, y₂ = 4.

$|\vec{AB}|=\sqrt[]{(5-8)^2+(4-7)^2}$

Therefore,

$\begin{gathered} |\vec{AB}|=\sqrt[]{(-3)^2+(-3)^2_{}^{}} \\ |\vec{AB}|=\sqrt[]{9+9} \end{gathered}$$\begin{gathered} |\vec{AB}|=\sqrt[]{18}=4.242640687\approx4.24(nearest\text{ 2decimal places)} \\ |\vec{AB}|=4.24 \end{gathered}$

Hence, the magnitude of the vector AB is 4.24.