A quantity that has some magnitude and a direction associated with it, is termed as a vector. It is represented by an arrow over the variable such as $\vec{P}$ or by placing a arrow over the endpoints of a vector $\vec{AB}$.

**Geometrically, a vector is denoted by an arrow whose length represents its magnitude, while the head of the arrow shows the direction of the vector as below:**In a two dimensional plane, we simply represent a vector as (x, y) or ($x_{1}, y_{1}$) and for three dimensional plane, we use either ($x, y, z$) or at times ($x_{1}, x_{2}, x_{3}$). For a multi-dimensional plane we represent any vector as ($x_{1}, x_{2}, ......, x_{n}$). Though, we focus more on two and three dimensional vectors in general.

Even 0 can be a vector and being a vector, it cannot be neglected as there is some direction associated to it which is also a useful piece of information no matter that the magnitude of the vector is zero.

## Formula

**We can perform various operations on given vectors. Let us take $\vec{a}$ = $a_1 i + a_2 j + a_3 k$ and $\vec{b}$ = $b_1 i + b_2 j + b_3 k$**

two vectors:

**1)**

**Scalar multiplication :**If ‘n’ is a scalar quantity, then n . b = n b which is also a vector quantity. This means n . b =n b

**2)**

**Vector addition and subtraction:**If a and b are two vectors, then $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ both result in a vector again.

Simply we can add or subtract the respective components as well to perform the given operation.

**3)**

**Dot product:**$\vec{a} . \vec{b} = |a |b| cos \theta $ where, |a|, |b| are the magnitudes of vectors ‘$\vec{a}’$ and ‘$\vec{b}$’ respectively, while $\theta$ is the angle between the given two vectors.Dot product always outputs a scalar number.

**4) Cross product:**

$\vec{a} \times \vec{b}$ Cross product always outputs a vector quantity.In three dimensions, we can find $\vec{a} \times \vec{b}$ by

$\begin{bmatrix}i & j & k\\ a_{1}& a_{2} & a_{3}\\ b_{1}& b_{2} & b_{3} \end{bmatrix}$

= $(a_{2} b_{3} - a_{3} b_{2}) i - (a_{1} b_{3} - a_{3} b_{1})j + (a_{1}b_{2} - a_{2}b_{1})k$

Similarly we can solve for three-dimensional vectors.

$\begin{bmatrix}i & j & k\\ a_{1}& a_{2} & a_{3}\\ b_{1}& b_{2} & b_{3} \end{bmatrix}$

= $(a_{2} b_{3} - a_{3} b_{2}) i - (a_{1} b_{3} - a_{3} b_{1})j + (a_{1}b_{2} - a_{2}b_{1})k$

Similarly we can solve for three-dimensional vectors.

## Steps

Every method above has its steps associated specially the dot product and cross product operation as the rest of them are simple and straight forward to be apply to. To obtain correct answer in case of dot and cross product one should be following proper steps.

For solving the cross product knowledge of solving a determinant is a must because the determinant is used to solve cross product if the vectors are given in component form.

In general, to find the magnitude of a vector we simply find the square root of the sum of squares of all components likewise, if b = <$b_{1}, b_{2}, b_{3}$> then magnitude of b represented by | b | = $\sqrt {(b_{1}^2 + b_{2}^2 +b_{3}^2)}$.

## Examples

**Example 1:**Find the cross product of a = <2, 3, -4> and b =<1, 0, 4> and thus the magnitude of the resultant vector.

**Solution:**Let c = a x b

$\begin{bmatrix} i & j & k\\ 2& 3 & -4\\ 1& 0 & 4 \end{bmatrix}$

$\Rightarrow$ c = (12 - 0) i - (8 + 4) j + (0 - 3) k

$\Rightarrow$ c = 12 i - 12 j - 3 k

The magnitude of vector c = | c|

| c | = $\sqrt {(12^{2} + (-12)^{2} + (-3)^{2})}$

= $\sqrt {(144 + 144 + 9)}$

= $\sqrt {297}$

= 17.23 (approx)

**Example 2:**Calculate the angle between two vectors given by $\vec{P}$ = 2i + 2j and $\vec{Q}$ = 3j.

**Solution:**By the formula of dot product, we can find a relation for calculating angle between two vectors.

$\vec{P} . \vec{Q} = |P|\ |Q|\ cos \theta$

$cos \theta$ = $\frac{\vec{P} . \vec{Q}}{|P|\ |Q|}$

Here, $\vec{P} . \vec{Q}$ = (2i + 2j). (3j) = (2i + 2j). (0i + 3j) = 0 + 6 = 6

|P| = $\sqrt {(2^{2} + 2^{2})}$

= $\sqrt {(4 + 4)}$

= $\sqrt {8}$ = $2\sqrt {2}$

|Q| = $\sqrt {3^{2}}$

= $\sqrt {9}$ = 3

So, $cos \theta$ = $\frac{\vec{P} . \vec{Q}}{|P|\ |Q|}$

$cos \theta$ = $\frac{6}{2\sqrt {2} \times 3}$

$cos \theta$ = $\frac{1}{\sqrt {2}}$

$cos \theta$ = cos $\frac{\pi}{4}$

$\theta$ = $\frac{\pi}{4}$