Normal vectors

Vector calculus in plane and space: Distances, Angles and Inner Product

Normal vectors

The dot product in $\mathbb{R}^3$ can be used to describe a linear equation with three unknowns.

Normal vector

Let $a_1$ , $a_2$ , $a_3$ , and $d$ be real numbers. Write $\vec{a}=\rv{a_1,a_2,a_3}$ . The linear equation $a_1\cdot x_1 + a_2\cdot x_2 + a_3 \cdot x_3 = d$ can be written with the aid of the dot product as $\dotprod{\vec{a}}{ \vec{x}}=d$ where $\vec{x}=\rv{x_1,x_2,x_3}$ . If $\vec{a}$ is not the zero vector, then the set of solutions of the equation is a plane and we call $\vec{a}$ a normal vector of the plane.

If $\vec{u}=\rv{u_1, u_2, u_3}$ and $\vec{v}=\rv{v_1, v_2, v_3}$ are two vectors in the plane $V$ with equation $2x_1 -x_2 + 3x_3 = 6$ , then we have $2u_1 -u_2 + 3u_3 = 6$ and $2v_1 -v_2 + 3v_3 = 6$ . Subtracting these two equalities yields
$2(u_1 - v_1) -(u_2 -v_2)+3(u_3-v_3) = 0$
We can also read this equality as follows:
$\rv{2,-1,3}\cdot \rv{u_1-v_1, u_2-v_2, u_3-v_3}= 0$
that is, the difference vector $\vec{u}-\vec{v}$ is perpendicular to the vector $\rv{2,-1,3}$ . In particular, $\rv{2,-1,3}$ is a vector which is perpendicular to all the directional vectors of the plane: a normal vector of the plane.

Let $V$ be a plane in $\mathbb{R}^3$ , and let $\vec{a}$ be a normal vector of $V$ .

The vector $\vec{a}$ is uniquely determined, apart from a scalar multiple.
A vector $\vec{r}$ in $\mathbb{R}^3$ is then, and only then, a directional vector of $V$ if $\vec{a}\cdot\vec{r}=0$ .

If $\vec{u}$ and $\vec{v}$ are two vectors in the plane, then $\vec{a}\cdot \vec{u} = d$ and $\vec{a}\cdot \vec{v} = d$ apply. Subtracting these equalities yields $(\vec{a}\cdot \vec{u}) - (\vec{a}\cdot \vec{v}) = 0$ , which means that through the properties of the dot product (additivity and associativity with respect to scalar multiplication): $\vec{a}\cdot( \vec{u}-\vec{v})=0$ In other words, the difference vector $\vec{u}-\vec{v}$ is perpendicular to $\vec{a}$ . In particular, directional vectors of the plane $V$ are perpendicular to the normal vector $\vec{a}$ .

Something similar applies to lines in the plane: if $a_1\cdot x_1 + a_2 \cdot x_2 = d$ is the equation of a line, then $\rv{a_1, a_2}$ is a normal vector of the line. The vector is perpendicular to each directional vector of the line.

Let $V$ be the plane with parametric representation $\left[-4,0,1\right]+\lambda\cdot \rv{-5,2,2}+\mu\cdot\rv{-3,1,4}\tiny.$

Determine a normal vector of $V$ .

Give your answer in the form $\rv{a_1,a_2,a_3}$ , wherein $a_1$ , $a_2$ , $a_3$ are integers.

$\left[ -6 , -14 , -1 \right]$

To see this, we find a linear equation to which $V$ is the set of solutions. We do this by choosing three points of $V$ that do not to go through one line, and entering them in the general form of the equation: $a_1\cdot x_1+a_2\cdot x_2+a_3\cdot x_3=d\tiny.$
We choose the three points by the parametric choices $\rv{\lambda,\mu}=\rv{0,0}$ , $\rv{1,0}$ , $\rv{0,1}$ . In coordinates:
$\begin{array}{cl}& \rv{ -4,0,1 }\\ & \rv{ -9 , 2 , 3 }\\ & \rv{ -7 , 1 , 5 }\end{array}$ Entering the coordinates of each of these points in the equation above, gives the following system of linear equations in $a_1$ , $a_2$ , $a_3$ , $d$ :
$\eqs{ -4\cdot a_{1}+a_{3} &=& d\cr -9\cdot a_{1}+2\cdot a_{2}+3\cdot a_{3} &=& d\cr -7\cdot a_{1}+a_{2}+5\cdot a_{3} &=& d }$
A solution with integer values is $\rv{a_1,a_2,a_3,d}=\left[ -6 , -14 , -1 , 23 \right]$ . This means that the three points, and therefore all points of $V$ , are solutions of the equation $-6\cdot x_{1}-14\cdot x_{2}-x_{3}=23\tiny.$ According to the theory, $\left[ -6 , -14 , -1 \right]$ is a normal vector.

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