# A vector F1 = 500 N due east and another F2 =250 N due north are on a plane. How do you find F2-F1?

Jul 15, 2016

${\vec{F}}_{2} - {\vec{F}}_{1} = \left(500 , 250\right)$ has magnitude $250 \sqrt{5} N$ and it acts at an angle $\theta = \pi - {\cos}^{-} 1 \left(\frac{2}{\sqrt{5}}\right)$ due East.

#### Explanation:

We will denote by ${\vec{F}}_{1}$ the vector F_1.

We are given that, magnitude of ${\vec{F}}_{1} = | | {\vec{F}}_{1} | | = 500 N$, &, its direction is due East, i.e., the $+ v e$ direction of the X-axis. The unit vector along this direction is $\hat{i} = \left(1 , 0\right)$.

Recall that a $\vec{x} \ne \vec{0}$ can completely be described as $\vec{x} = | | \vec{x} | | \hat{x}$, where, $\hat{x}$ is a unit vector in the direction of $\vec{x}$.

Hence, ${\vec{F}}_{1} = | | {\vec{F}}_{1} | | \hat{i} = 500 \left(1 , 0\right) = \left(500 , 0\right)$.

Similarly, ${\vec{F}}_{2} = \left(0 , 250\right)$

Therefore, ${\vec{F}}_{2} - {\vec{F}}_{1} = \vec{F}$, say $= \left(0 , 250\right) - \left(500 , 0\right) = \left(- 500 , 250\right) = 250 \left(- 2 , 1\right)$

$| | \vec{F} | | = 250 \sqrt{{\left(- 2\right)}^{2} + {1}^{2}} = 250 \sqrt{5} N$

To find the direction of $\vec{F}$, suppose that $\vec{F}$ is making an angle $\theta$ due East, i.e., with $\hat{i}$.

Then, $\vec{F} . \hat{i} = | | \vec{F} | | \cdot | | \hat{i} | | \cdot \cos \theta$

$\therefore \left(- 500 , 250\right) . \left(1 , 0\right) = 250 \sqrt{5} \cdot 1 \cdot \cos \theta$

$\therefore - 500 = 250 \sqrt{5} \cdot \cos \theta$

$\therefore \cos \theta = - \frac{500}{250 \sqrt{5}} = - \frac{2}{\sqrt{5}}$

$\therefore \theta = {\cos}^{-} 1 \left(- \frac{2}{\sqrt{5}}\right) = \pi - {\cos}^{-} 1 \left(\frac{2}{\sqrt{5}}\right)$

Thus, ${\vec{F}}_{2} - {\vec{F}}_{1} = \left(- 500 , 250\right)$ has magnitude $250 \sqrt{5} N$ and it acts at an angle $\theta = \pi - {\cos}^{-} 1 \left(\frac{2}{\sqrt{5}}\right)$ due East.

Hope, this will be of Help! Enjoy maths.!