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.

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

$\vec{a} \times \vec{b}$ 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.

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 | = $\sqrt {(b_{1}^2 + b_{2}^2 +b_{3}^2)}$.

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

$\Rightarrow$ 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)

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

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

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 $\frac{\pi}{4}$

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