# Vectors

Intro To Vectors

Tip: This isn't the place to ask a question because the teacher can't reply.

1 of 2 videos by Darshan Senthil

## Key Questions

• To find the direction of a vector, we use Trigonometry and the vector components.

To start, we need the vector components. If you are asked for direction, you have most likely been given the component form of a vector, and not the polar form. Component form looks like this:

$\vec{V} = \left({V}_{y}\right) \hat{i} + \left({V}_{x}\right) \hat{j}$

Where the components are ${V}_{y}$ and ${V}_{x}$, while the directions of the components are $\hat{i}$ and $\hat{j}$.

Now that we have the lengths of the x and y components (also known as the i and j components), we are ready for some Trigonometry.

What we know is that we have the lengths of the x and y components. So what do we need? Well, the x component can be thought of as the adjacent leg of a right triangle, and the y component as the opposite leg. So what we need is a trig function that relates $\theta$ to x and y, which is:

$\tan \theta = \frac{y}{x}$

Thus to find the angle, we need to take the inverse tangent of the ratio of the triangle's height and length.

$\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right)$

Example

$\vec{V} = \left(15.2\right) \hat{i} + \left(11.2\right) \hat{j}$

Find the direction of this vector.

As we can see, we have the length of the y leg and the length of the x leg (15.2 and 11.2 respectively)

So we know that the Tangent function is equal to the ratio of their lengths:

$\tan \theta = \left(\frac{15.2}{11.2}\right)$

Now all we have to do is take the inverse tangent of the ratio to find the direction:

$\theta = {\tan}^{-} 1 \left(\frac{15.2}{11.2}\right) \approx {53.6}^{\circ}$

• Conventionally, vectors are added and subtracted via the addition and subtraction of their corresponding components.

${\vec{V}}_{1} + {\vec{V}}_{2} = \left({V}_{1} y + {V}_{2} y\right) \hat{i} + \left({V}_{1} x + {V}_{2} x\right) \hat{j}$

where ${V}_{1} y$ and ${V}_{2} y$ refer to the vertical (y) components of vector 1 and vector 2 respectively. Accordingly, ${V}_{1} x$ and ${V}_{2} x$ refer to the horizontal (x) components of vector 1 and vector 2 respectively.

$\hat{i}$ simply designates that the component(s) in its parentheses are operating vertically, while $\hat{j}$ designates that the component(s) in its parenthesis are operating horizontally.

Conversly, if you wished to subtract vectors, simply replace every (+) in the previous equation with a (-):

${\vec{V}}_{1} - {\vec{V}}_{2} = \left({V}_{1} y - {V}_{2} y\right) \hat{i} + \left({V}_{1} x - {V}_{2} x\right) \hat{j}$

Note that this implies that the second vector maintains its magnitude, but is reversed in direction. This should make sense, since the minus would simply be distributed through both components of the second vector. Thus it is crucial to keep track of the signs of the vector components when adding or subtracting them.

What is a Vector Component?

Vector components are the horizontal and vertical distance coordinates that a vector possesses. This means that they describe the coordinates of the vector's tip, and subsequently the angle between the vector and the horizontal axis.

Example

The velocity vector of a plane and the velocity vector of the wind is given below:
${\vec{V}}_{p} = \left(4.8\right) \hat{i} + \left(5.6\right) \hat{j}$
${\vec{V}}_{w} = \left(- 1.2\right) \hat{i} + \left(0.5\right) \hat{j}$

What is the final velocity vector of the plane?

To answer this question, we must add the plane's initial vector to that of the wind's. This will give us the plane's final velocity vector. After all, the wind will push the plane, thus making the plane's final velocity vector a combination of the two.

Thus all we must do is add the y components and the x components respectively.

${\vec{V}}_{1} + {\vec{V}}_{2} = \left({V}_{1} y + {V}_{2} y\right) \hat{i} + \left({V}_{1} x + {V}_{2} x\right) \hat{j}$

${\vec{V}}_{p} + {\vec{V}}_{w} = \left({V}_{p} y + {V}_{w} y\right) \hat{i} + \left({V}_{p} x + {V}_{w} x\right) \hat{j}$

${\vec{V}}_{p} + {\vec{V}}_{w} = \left(4.8 + \left(- 1.2\right)\right) \hat{i} + \left(5.6 + 0.5\right) \hat{j} = {\vec{V}}_{f}$

${\vec{V}}_{f} = \left(3.6\right) \hat{i} + \left(6.1\right) \hat{j}$

To convert this to polar form, we simply find the magnitude and direction of the vector.

$M a g n i t u \mathrm{de} = \sqrt{{\left({V}_{y}\right)}^{2} + {\left({V}_{x}\right)}^{2}} = \sqrt{{\left(3.6\right)}^{2} + {\left(6.1\right)}^{2}} = 7.08$

$D i r e c t i o n = T a {n}^{-} 1 \left({V}_{y} / {V}_{x}\right) = T a {n}^{-} 1 \left(\frac{3.6}{6.1}\right) = {30.5}^{\circ}$

${\vec{V}}_{f} = \left(7.08 , {30.5}^{\circ}\right)$

• A vector $\vec{v}$ can be represented as a pointed arrow drawn in space:

The length of the arrow (relative to some kind of reference or scale) represents the relative magnitude of the vector while the arrow head gives us the direction in which the vector operates.

Another way to represent a vector is by giving its components .
Basically, in space, you choose three unit vectors ( of magnitude $1$ and directions the positive ones of the $x , y \mathmr{and} z$ axes) indicated as $\vec{i} , \vec{j} \mathmr{and} \vec{k}$ and then you do a linear combination of these three unit vectors using three components (=numbers such as $5 , 4 \mathmr{and} 6$).

For example, $\vec{u} = 5 \vec{i} + 4 \vec{j} + 6 \vec{k}$

Basically you can think of it as a recipe of a cake where $\vec{i} , \vec{j} \mathmr{and} \vec{k}$ are the ingredients:
To "make" vector $\vec{u}$ you need to mix (=combine linearly):
5 units of ingredient $\vec{i}$;
4 units of ingredient $\vec{j}$;
6 units of ingredient $\vec{k}$.

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