Question

The X and Y components of a displacement vector are (15,7)m. Find the magnitude and direction of $ \vec A $ .

Answer 1

Hint : The magnitude of a vector is the length of the vector without consideration of its direction. The vector is given by the line joining the origin of a Cartesian plane to the point given as the components.

Formula used: In this solution we will be using the following formula;
 $\Rightarrow \left| A \right| = \sqrt {{X^2} + {Y^2}} $ where $ \left| A \right| $ is the magnitude of any vector $ A $ , $ X $ is the x component and $ Y $ is the y component of the vector.
 $\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{Y}{X}} \right) $ , where $ \theta $ is the direction in angle.

Complete step by step answer
When the components of a vector are stated as (X,Y), this means that a line joining the point and the origin on a Cartesian plane, in the direction of the point, signifies the vector. The length of the line is the magnitude of the vector while the angle between the vector and the x-axis is considered the direction of the vector.
Hence, to calculate the length of the line, from mathematical principles,
 $\Rightarrow \left| A \right| = \sqrt {{X^2} + {Y^2}} $ . Thus, substituting, the values of X and Y, we have
 $\Rightarrow \left| A \right| = \sqrt {{{15}^2} + {7^2}} $ . Hence, by computation
 $\Rightarrow \left| A \right| = 16.55m $ .
In finding the direction, angle from the x-axis, we have that
 $\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{Y}{X}} \right) $ , hence by substitution again, we have that
 $\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{7}{{15}}} \right) $
 $\Rightarrow \theta = 0.44rad $ which can be converted to degree by multiplying by 180 over pi.
 $\Rightarrow \theta = 0.44 \times \dfrac{{180}}{\pi } = 25^\circ $.

Note
The derivation for the magnitude and direction of a vector can be shown from the diagram below

We see the vector point to (XY) form 0, this is the hypotenuse of the right angled triangle. Hence, the length is given by the Pythagoras theorem, thus
 $\Rightarrow L = \sqrt {{x^2} + {y^2}} $
And the angle is
 $\Rightarrow \tan \theta = \dfrac{y}{x} $
 $\Rightarrow \theta = {\tan ^{ - 1}}\dfrac{y}{x} $ .