>[!summary] Scalar equations are an equation to verify whether lines are in a plane. We often use this type of equation in validating lines in a plane or creating equations to describe the behaviour of a plane in reference to a point. > **Key equations:** > **Scalar equations:** $\vec{n} \cdot \vec{AB} = 0$ > or >$\begin{array}{cc} \vec{n} \cdot \vec{b} = \vec{n} \cdot \vec{a} \\ n_1 b_1 + n_2b_2 + n_2b_3 = d \\ \text{Where $d = \vec{n} \cdot \vec{a} $} \end{array}$ >[!info]+ Read Time **⏱ 4 mins** # Defining Scalar Equations In [[Vectors Equations of a Line in 2D]] & [[Vectors Equations of a Line in 3D]] we define vector equations to describe the behaviour of vectors. Sometimes we just want to know whether a vector would be in a plane, or the behaviour of a vector in a plane. So well make a definition of a normal vector that is perpendicular to a plane or line. Our solution from this equation is a scalar number rather than a vector from vector equations. >[!warning] Assumptions >- Let $\vec{n} = \begin{bmatrix} n_1 \\ n_2 \\ n_3 \end{bmatrix}$ be our normal vector perpendicular to a plane >- Let $\vec{AB}$ be a line in our plane >- We will derive this equation using [[Dots Product & Angles]] ![[se_1.png|400]] In our definition we are trying to make a definition to describe whether a point is a plane or not returning a scalar $\begin{array}{c} \text{Let $\vec{n} = \begin{bmatrix} n_1 \\ n_2 \\ n_3 \end{bmatrix}$} \\ \vec{n} \cdot \vec{AB} = 0 \\\\ \text{If $\vec{AB} = (\vec{b} - \vec{a})$ we can make another equal defination from the above:}\\ \vec{n} \cdot \vec{AB} = 0 \\ \vec{n} \cdot (\vec{b} - \vec{a}) = 0 \\ \vec{n} \cdot \vec{b} - \vec{n} \cdot \vec{a} = 0 \\ \vec{n} \cdot \vec{b} = \vec{n} \cdot \vec{a} \\ n_1 b_1 + n_2b_2 + n_2b_3 = d \\ \text{Where $d = \vec{n} \cdot \vec{a} $} \end{array}$ >[!bug] Note If we know our normal vectors to a plane we can determine whether two planes are parallel or orthogonal ([[Orthogonality]]) > If the dot product between two normal vectors is zero then the each plane is orthogonal to one another. ([[Dots Product & Angles]]) > If one normal vector is a non-zero scalar multiple of another then they are parallel to one another. >The example below is adapted from an example in An Introduction To Linear Algebra For Science and Engineering by Norman, D., & Wolczuk, D. The solution is adapted and is my original interpretation of steps. >[!example] Example of a finding a scalar equation from a point and plane Find a scalar equation of a plane that contains the point P(3,-1,3) and is orthogonal to the plane $x_1 -2x_2 + 4x_3 = 2$ > Because we are looking for a plane that is orthogonal to the plane, there is many possible solutions but we will suppose a plane such as $2x_1 + 3x_2 + x_3 = d$ or a normal vector of $\vec{n_2} = \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}$. We can verify the two plane are orthogonal to one another because > >$\begin{array}{c} \text{If $\vec{n_1} = \begin{bmatrix} 1 \\ -2 \\4\end{bmatrix}$ and $\vec{n_2 } = \begin{bmatrix} 2 \\ 3 \\ 1\end{bmatrix}$} \\ \\ \vec{n_1} \cdot \vec{n_2} = 1(2) + -2(3) + 4(1) = 0 \\ \text{So they are orthogonal to one another} \\ \\ \end{array}$ > To find our $d$ scalar to conclude our scalar equation from our derivation we need to find >$\begin{array}{c} d = \vec{n_2} \cdot \vec{a} \\ \\ \text{Where $\vec{a} = P(3,-1,3)$} \\ \\ \vec{n_2} \cdot \vec{a} = 2(3) + 3(-1) + 3(1) = 6 \\ d = 6 \end{array}$ > So our scalar equation is therefore: $2x_1 + 3x_2 + x_3 = 6$ --- 📂 Want to see more structured notes like these? Help grow the project by [starring Math & Matter on GitHub](https://github.com/rajeevphysics/Obsidan-MathMatter). ---