# Vectors in Space

## Key Questions

• There is a simple way to do it, which derives from the two dimensional $\cos \left(\theta\right) = \frac{v \cdot u}{| v | | u |} \therefore \theta = \arccos \left(\frac{v \cdot u}{| v | | u |}\right)$.

• Two vectors are orthogonal if they are perpendicular, i.e., they form a right angle.

This relationship can be verified mathematically if the inner product (In Euclidean spaces this is the dot product) of the vectors is zero.

• 3-D vector means it encompasses all the three co-ordinate axes, i.e. , the x , y and z axes.

We represent the unit vectors along these three axes by $\hat{i}$ , $\hat{j}$ and $\hat{k}$ respectively.

Unit vectors are vectors that have a direction and their magnitude is 1.

Now, we know that in order to find the dot product of two vectors, we multiply their magnitude by the cosine of the angle included between the vectors.

Now consider this:

$\hat{i}$.$\hat{i}$ = |$\hat{i}$| |$\hat{i}$| cos 0
=>$\hat{i}$ . $\hat{i}$= 1 1 1 = 1

Again:

$\hat{i}$.$\hat{j}$= |$\hat{i}$| |$\hat{j}$| cos 90
=> $\hat{i}$.$\hat{j}$= 1 1 0 = 0

Similarly, when the third co-ordinate axis is considered, we find the scalar product using the same rules that are used for finding out the scalar product of 2-D vectors.

Therefore:

$\hat{i}$.$\hat{k}$ = |$\hat{i}$| * |$\hat{k}$| * cos 90
=> $\hat{i}$. $\hat{k}$= 1 * 1* 0 = 0

Thus, for two vectors

$\vec{A}$ = a1 $\hat{i}$ + a2 $\hat{j}$ + a3 $\hat{k}$ and
$\vec{B}$ = b1 $\hat{i}$ + b2 $\hat{j}$ + b3 $\hat{k}$

we have:

$\vec{A}$. $\vec{B}$= (a1 $\hat{i}$ + a2 $\hat{j}$ + a3 $\hat{k}$) * ( b1 $\hat{i}$ + b2 $\hat{j}$ + b3 $\hat{k}$)

=> $\vec{A}$.$\vec{B}$= a1b1 + a2b2 + a3*b3

• Vectors don't normally represent a position, rather they represent a direction and magnitude. You can think of it as an offset or change. So if you subtract a point from another point, you get a vector. If you add a point and a vector, you get a point. You can add 2 vectors and get a vector. Adding points doesn't give you anything. Vectors are normally represented with <> and points are represented by ().

A position vector can represent a point in space. Suppose we have a vector $\vec{v} = \left(a , b , c\right)$ (sorry, can't seem to get <> working). You simply add the vector to the origin, which is a point. Since the origin is $\left(0 , 0 , 0\right)$, the position vector is $\left(a + 0 , b + 0 , c + 0\right)$ or simply $\left(a , b , c\right)$.