Question

How do you find the component form and magnitude of the vector v given initial point $\left( { - 1,5} \right)$ and the terminal point $\left( {15,12} \right)$ ?

Answer 1

In this question we will use the concept of the vector that is the component form of the vector is the ordered pair that shows change in the value of the x and the y, this is obtained by subtracting the coordinates of x and y respectively. The length of the vector is defined as the magnitude of the vector or the magnitude of the vector is the distance of the between the initial point and the terminal point of the vector.

Complete Step by Step solution:
In this question we have given the initial point is $\left( { - 1,5} \right)$ and the final point is$\left( {15,12} \right)$.
Consider the initial vector is of the form $\left( {{x_1},{y_1}} \right)$ and the final point is of the form$\left( {{x_2},{y_2}} \right)$.
The component form of the vector is the difference of $\left( {{x_2},{y_2}} \right)$ and $\left( {{x_1},{y_1}} \right)$. This is done as,
$\left( {{x_2} - {x_1},{y_2} - {y_1}} \right)$
That is given by,
$\left( {15 - \left( { - 1} \right),12 - 5} \right) = \left( {16,7} \right)$
Thus, the component form is $\left( {16,7} \right)$.
The magnitude of the vector is denoted as $\left| {\left| v \right|} \right|$ and defines the distance between initial point the terminal point, the formula for it is given by,
$\Rightarrow \left| {\left| v \right|} \right| = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$
The above formula is also called the distance formula.
Substitute the values in the above equation as,
$\Rightarrow \left| {\left| v \right|} \right| = \sqrt {{{\left( {16} \right)}^2} + {{\left( 7 \right)}^2}}$
Now we will calculate the squares of $16$ and $7$ as,
$\Rightarrow \left| {\left| v \right|} \right| = \sqrt {256 + 49}$
Now we will simplify the above expression as,
$\Rightarrow \left| {\left| v \right|} \right| = \sqrt {305}$
After calculation we will obtain,
$\therefore \left| {\left| v \right|} \right| = 17.46$

Thus, the magnitude of the vector or the distance between the initial and the coterminal point is $17.46$.

Note:
The vector is the quantity that has magnitude and direction both. The vector is displayed by the arrow. The vector is represented by the arrow that has direction and has length that is proportional to the magnitude. The example for the vector quantity is displacement, velocity, position, force and the torque.