# Given Vector A: 15cm at zero degrees and Vector B: 30cm at 60 degrees, How do you add and subtract these two vectors?

Jan 16, 2017

$\vec{A} + \vec{B}$ is $15 \sqrt{7}$ cm long, at ${40.89}^{o}$ and
$\vec{B} - \vec{A}$ is $15 \sqrt{3}$ cm long, at ${90}^{o}$.

#### Explanation:

The vector with modulus ( length ) r, and in the direction inclined at

${\theta}^{o}$ to x-axis in the anticlockwise ( + ) sense, has

(r, theta)

$a s t h e p o l a r$(|..|, angle) form,

but component-wise, it is

$r < \cos \theta , \sin \theta > = < x , y >$.

Here,

$\vec{A} = \left(15 , {0}^{o}\right) = 15 < \cos {0}^{o} , 5 \sin {0}^{0} \ge 15 < 1 , 0 >$ and

$\vec{B} = \left(30 , {60}^{o}\right) = < 30 \cos {60}^{o} , 30 \sin {60}^{o} >$

$= < 15 , 15 \sqrt{3} \ge 15 < 1 , \sqrt{3} >$.

Now, with component-wise addition and subtraction,

$\vec{A} + \vec{B} = 15 < 1 , 0 > + 15 < 1 , \sqrt{3} \ge 15 < 2 , \sqrt{3} >$ and

$\vec{B} - \vec{A} = 15 < 1 , \sqrt{3} > - 15 < 1 , 0 > = 15 < 0 , \sqrt{3} >$

The polar $\left(| . . | , \angle\right)$ forms are

$\vec{A} + \vec{B} = \left(15 \sqrt{7} , {40.89}^{o}\right)$, nearly, and

$\vec{B} - \vec{A} = \left(15 \sqrt{3} , {90}^{o}\right)$.