Operations with vectors

Vectors follow the same operational rules as matrices, but they have some special operations.

The magnitude of a vector with $N$ dimensions, is given by:

$\displaystyle \vert\vec{a}\vert = \sqrt{\sum^N_{n=1}\vec{a}_n^2}$

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A vector $\vec{a}$ is a unit vector, if it satisfies $\vert\vec{a}\vert=1$ . The unit vector, or simply put, the direction of a vector, can be found by normalising said vector:

$\begin{indisplay}\begin{split} & \textrm{Let }\vec{b}\textrm{ be a vector with m... ... & \hat{\vec{b}} = \frac{\vec{b}}{\vert\vec{b}\vert} \end{split}\end{indisplay}$

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Note that one may perform the above division, simply by dividing all the elements of vector dividend, by the scalar divisor.

The dot product is effectively a measure of how different two vectors $\vec{a},\vec{b}$ are. (Note that $\vec{a}$ is notation for a vector named $a$ ). It forms a relation between the components of the vector, and the acute angle between them $\theta$ .

$\begin{indisplay}\begin{split} & \textrm{Let }N\textrm{ be the number of dimensi... ...mes\cos(\theta) \\ &= \sum^N_{n=1}\vec{a}_n\vec{b}_n \end{split}\end{indisplay}$

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The cross product, allows you to find the vector which is perpendicular to two vectors. Note that $\vec{n}$ is the unit vector perpendicular to the two vectors $\vec{a}$ and $\vec{b}$ .

$\begin{indisplay}\begin{split} \vec{a}\times\vec{b} & \\ &= \vert\vec{a}\vert\times\vert\vec{b}\vert\times\sin(\theta)\times\vec{n} \end{split}\end{indisplay}$

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You may ask youself, `but Nathan, that's stupid! You're saying that to find the perpendicular vector to two vectors, you need the perpendicular unit vector¡, and you would be correct. The cross product is, for reasons I cannot be asked to understand or explain, only defined to be non-zero in three and seven dimensions. As such, we have the following if both $\vec{a}$ and $\vec{b}$ have three dimensions:

$\displaystyle \vec{a}\times\vec{b} = \left[\begin{matrix} \vec{a}_y\vec{b}_z - ... ...c{a}_x\vec{b}_z \\ \vec{a}_x\vec{b}_y - \vec{a}_y\vec{b}_x \end{matrix}\right]$

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