# Orthogonal/Perpendicular Vectors

Definition: Two vector $\vec{u}, \vec{v} \in \mathbb{R}^n$ are said to be Orthogonal or Perpendicular if their dot product is zero, that is $\vec{u} \cdot \vec{v} = 0$. If $\vec{u}$ and $\vec{v}$ are orthogonal, then we write $\vec{u} \perp \vec{v}$. |

# Vectors Perpendicular to a Line

Let us first defined a line $L_1$ by the two points $P(x_1, y_1)$ and $Q(x_2, y_2)$. A vector $\vec{PQ} = (x_2 – x_1, y_2 – y_1$ represents this line. Now let’s construct another vector $\vec{n} = (a, b)$ to be perpendicular to this line. Since $\vec{PQ} \perp \vec{n}$, it follows that their dot product is equal to zero, that is $\vec{n} \cdot \vec{PQ} = 0$. Expanding this formula we obtain the general form of a line, that is:

(1)

Alternatively, let $P$ and $Q$ be distinct points on a line. We obtain the following equations:

(2)

Subtraction of these two equations results in $a(x_{2} – x_{1}) + b(y_{2} – y_{1}) = 0$ as well.

## Example 1

**Given the points $P(1, 2)$ and $Q(4, 6)$ that define a line $L1$, what are one set of values of $a$ and $b$ such that the vector $n = (a, b)$ is perpendicular to $L1$?**

We first note that $\vec{PQ} = (3, 4)$, and thus it follows we need to construct a dot product $\vec{n} \cdot \vec{PQ} = 0$. It thus follows that we need a linear combination of $a$ and $b$ that satisfies the equation $3a + 4b = 0$. Let $a = t$. Therefore $b = -\frac{3t}{4}$. Thus it follows that $n = (t, -\frac{3t}{4})$ where $t \in \mathbb{R}$. One set of values occurs when $t = 4$ and thus $a = 4$ and $b = -3$.