Before we discuss about orthogonal vectors we need to know about what exactly orthogonal mean. Orthogonality is derived from the Greek word orthogonias (orthos means straight and gonia referred as "angle" ) which gives a meaning of "straight angle".

It is used in many fields such as Mathematics, physics, chemistry, etc. It has a wide application in the area of applied Mathematics. Particularly linear algebra, Euclidean geometry and spherical trigonometry. Orthogonality  which takes the meaning of 'Perpendicular'. When two vectors that perpendicular to each other are called "orthogonal vectors".

## Definition

A vector is a quantity that has both magnitude and direction. We can represent a vector as a directed line segment, whose length is the magnitude of the vector and with an arrow indicating the direction.

We denote the magnitude of the vector a by $\left \| u \right \|$.

Two vectors are said to be same if they have the same magnitude and direction. This means that if we take a vector and translate it to a new position (without rotating it), then the vector we obtain after the translation is the same vector.

Given two vectors $u$ and $v$, we find their addition by translating the vector $v$ until its tail coincides with the head of $u$. Then, the directed line segment from the tail of $u$ to the head of $v$ is the vector $u + v$.

Subtraction of Vectors:

The vector $-u$, which is the opposite of $u$. The vector $u$ is the vector with the same magnitude as $u$ but that is pointed in the opposite direction.

We define subtraction as addition with the opposite of a vector:

$v - u$ = $v + (-u)$

## Formula

$v + u$

Subtraction of Vectors:

$v - u$ = $v + (-u)$

Dot Product of Vectors:

$|u \cdot v|$= $|u||v|cos \theta$

$|u|$ is the magnitude of the vector $u,\ |v|$ is the magnitude of the vector $v$. $\theta$ is the angle between the vectors.

$\theta$ = $cos^{-1}$ $(\frac{|u \cdot v|}{|u||v|})$

Cross Product of Vectors:

$u\ x\ v$ = $\begin{vmatrix}i & j & k\\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}$

$u\ x\ v$ = $\begin{vmatrix}u_2 & u_3\\ v_2 & v_3\end{vmatrix}$ $i -$ $\begin{vmatrix}u_1 & u_3\\ v_1 & v_3 \end{vmatrix}$ $j +$ $\begin{vmatrix}u_1 & u_2\\ v_1 & v_3\end{vmatrix}$ $k$

$|u\ x\ v|$ = $|u||v|\ sin\ \theta$

$|u|$ is the magnitude of the vector $u,\ |v|$ is the magnitude of the vector $v$. $\theta$ is the angle between the vectors.

$\theta$ = $sin^{-1}$ $(\frac{|u x v|}{|u||v|})$

## Orthogonal Vectors

Two vectors $u$ and $v$ are said to be orthogonal vectors if they are perpendicular to each other and whose dot product is zero.

## Orthogonal Vectors Cross Product

$u\ x\ v$ = $\begin{vmatrix}i & j & k\\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}$

$u\ x\ v$ = $\begin{vmatrix}u_2 & u_3\\ v_2 & v_3\end{vmatrix}$ $i -$ $\begin{vmatrix}u_1 & u_3\\ v_1 & v_3 \end{vmatrix}$ $j +$ $\begin{vmatrix}u_1 & u_2\\ v_1 & v_3\end{vmatrix}$ $k$

## Orthogonal Vectors Dot Product

$u \cdot (v + w)$ = $u \cdot v\ +\ u \cdot w$

$u \cdot v$ = $v \cdot u$

$v \cdot v$ = $|v|^2$

$(Cu) \cdot v$ = $u \cdot (Cv)$

$u \cdot 0$ = $0$

$u \cdot v$ = $|u||v|cos \theta$

$|u|$ is the magnitude of the vector $u$, $|v|$ is the magnitude of the vector $v \cdot \theta$ is the angle between the vectors.

$\theta$ = $cos^{-1}$ $(\frac{u \cdot v}{|u||v|})$

## Examples

Example 1:

Prove that the vectors $i - j + 2k$ and $i - j - k$ are orthogonal.

Solution:

Let $u$ = $i - j + 2k$ and $v$ = $i - j - k$

$u \cdot v$ = $1 \times 1 + (-1) \times (-1) + 2 \times (-1 )$

= $1 + 1 - 2$

= $2 - 2$

= $0$

Since the dot product of given vectors is zero. So, the given vectors are orthogonal vectors.
Example 2:

Find if the vectors

i) $(3, 1, -1)$ and $(3, 1, 10)$

ii) $(3, 4)$ and $(1, 2)$

are orthogonal to each other.

Solution:

i) Let $u$ = $(3, 1, -1)$ and $v$ = $(3, 1, 10)$

Dot product

$u \cdot v$ = $3 \times 3 + 1 \times 1 + (-1) \times 10$

= $9 + 1 - 10$

= $10 - 10$

= $0$

Since, the dot product of given vectors is zero. So, the given vectors  are orthogonal vectors.

ii) Let $u$ = $(3, 4)$ and $v$ = $(1, 2)$

Dot product

= $3 \times 1 + 4 \times 2$

= $3 + 8$

= $11$

Since, the dot product of given vectors is not equal to zero. So, the given vectors are not orthogonal.
Example 3:

If the vectors $i - j + 2k$ and $2i + 2yj − 4k$ are orthogonal, then find the value of $y$.

Solution:

Let $u$ = $i - j + 2k$ and $v$ = $2i + 2yj − 4k$

If the given vectors are orthogonal vectors,  then their dot product must be zero. That is

$u \cdot v$ = $0$

$1 \times 2 + (-1) \times 2y + 2 \times (- 4)$ = $0$

$2 - 2y - 8$ = $0$

$- 6 - 2 y$ = $0$

adding $6$ on both sides, we get

$- 6 - 2y + 6$ = $6$

$- 2y$ = $6$

Dividing by $-2$ on both sides, we get

$\frac{-2y}{-2}$ = $\frac{6}{-2}$

$Y$ = $-3$.
Example 4:

Determine the angle between the vectors $u$ = $(1, 3, 5)$ and $v$ = $(-2, 4, 1)$?

Solution:

$u \cdot v$ = $1 \times -2 + 3 \times 4 + 5 \times 1$

= $-2 + 12 + 5$

= $15$

$|u|$ = $\sqrt{1^2 + 3^2 + 5^2}$

= $\sqrt{1 + 9 + 25}$

= $\sqrt{35}$

$|v|$ = $\sqrt{- 2^2 + 4^2 + 1^2}$

= $\sqrt{4 + 16 + 1}$

= $\sqrt{21}$

$cos \theta$ = $(\frac{u \cdot v}{|u||v|})$

$cos \theta$ = $\frac{15}{\sqrt{35} x \sqrt{21}}$

= $0.553$
Example 5:

If $u$ = $(1, -1, 2)$ and $v$ = $(2, 3, -1)$ and compute $u x v$?

Solution:

$u\ x\ v$ = $\begin{vmatrix}i & j & k \\ 1 & -1 & 2\\ 2 & 3 & -1\end{vmatrix}$

= $i(1 - 6) - j(-1 -4) + k (3 + 2)$

= $-5i + 5j + 5k$