### 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.

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.

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:

\[\begin{array}{cl}& \rv{ -2,-6,8 }\\ & \rv{ -6 , -5 , 7 }\\ & \rv{ -7 , -3 , 8 }\end{array}\] 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#:

\[\eqs{ -2\cdot a_{1}-6\cdot a_{2}+8\cdot a_{3} &=& d\cr -6\cdot a_{1}-5 \cdot a_{2}+7\cdot a_{3} &=& d\cr -7\cdot a_{1}-3\cdot a_{2}+8\cdot a_{3} &=& d }\]

A solution with integer values is #\rv{a_1,a_2,a_3,d}=\left[ -3 , -5 , 7 , 92 \right]#. This means that the three points, and therefore all points of #V#, are solutions of the equation \[-3\cdot x_{1}-5\cdot x_{2}+7\cdot x_{3}=92\tiny.\] According to the theory, #\left[ -3 , -5 , 7 \right]# is a normal vector.

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