Question

How do you find the component form and the magnitude of the vector v given initial point (1, 3) and terminal point (- 8, - 9)?

Answer 1

Hint: Now we want to find the component form and vector form of the given points. We know that the component form is nothing but $\left( {{x}_{2}}-{{x}_{1}},{{y}_{2}}-{{y}_{1}} \right)$ and the magnitude is given by $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ where $\left( {{x}_{1}},{{y}_{1}} \right)$ is initial point and $\left( {{x}_{2}},{{y}_{2}} \right)$ is the terminal point.

Complete step-by-step solution:
Now first let us understand the concept of vectors and scalars.
Scalars are quantities which have just magnitude. For example weight, distance all these quantities are scalar quantities.
Vectors are the quantities which have magnitude as well as direction. For example velocity is a vector quantity since the direction is also defined.
Now to write the vector in the component we must know the initial and the terminal point.
If $\left( {{x}_{1}},{{y}_{1}} \right)$ is initial point and $\left( {{x}_{2}},{{y}_{2}} \right)$ is the terminal point then the component form of the vector is given by $\left( {{x}_{2}}-{{x}_{1}},{{y}_{2}}-{{y}_{1}} \right)$ and the magnitude is given by $\sqrt{{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}}$.
Now consider the given initial point and terminal point.
Hence we have the component form of the vector is $\left( -8-1,-9-3 \right)=\left( -9,-12 \right)$
And the magnitude of the vector is given by $\sqrt{{{\left( -9 \right)}^{2}}+{{\left( -12 \right)}^{2}}}=\sqrt{81+144}=\sqrt{225}=15$ .
Hence the component form of the vector is (- 9, -12) and the magnitude is 15.

Note: Note that the magnitude is nothing but the distance of the point from the origin and hence it is easily obtained by using the Pythagoras theorem. Also note that since magnitude is distance and distance is always positive we ignore the negative solution while taking square root.