Normal Vectors from Two Directions

Why Two Directions Determine a Normal

A flat surface has directions that lie along it. If two of those directions are not parallel, they determine the orientation of the surface: a vector perpendicular to both points straight across the plane. Such a vector is called a normal vector.

The cross product provides a direct way to calculate one. The result is useful for describing the orientation of a plane or a flat geometric surface, such as a panel or a face of a solid. The direction of the normal matters, but its length is usually not important: any nonzero scalar multiple points along the same normal direction.

Key idea: If \(\mathbf{a}\) and \(\mathbf{b}\) are nonparallel direction vectors lying in a plane, then \(\mathbf{a}\times\mathbf{b}\) is a nonzero vector perpendicular to both. It is therefore a normal vector to that plane.

Definition and Calculation

For vectors \(\mathbf{a}=(a_1,a_2,a_3)\) and \(\mathbf{b}=(b_1,b_2,b_3)\) in three-dimensional space, their cross product is the vector

\mathbf{a}\times\mathbf{b} = \left( a_2b_3-a_3b_2,\; a_3b_1-a_1b_3,\; a_1b_2-a_2b_1 \right).

The order of the vectors matters: reversing the order changes the sign, so \(\mathbf{b}\times\mathbf{a}=-(\mathbf{a}\times\mathbf{b})\). These two results point in opposite directions but are both perpendicular to the same plane.

A vector is perpendicular to another vector when their dot product is zero. Thus, a proposed normal \(\mathbf{n}\) to a plane can be checked by confirming that its dot product with each of two nonparallel direction vectors in the plane is zero.

Definition: A normal vector to a plane is a nonzero vector perpendicular to every direction in the plane. It is enough to check perpendicularity to two nonparallel directions in the plane: those directions span all directions parallel to the plane.

Why the Cross Product Is Perpendicular

The cross product formula is designed so that its dot product with either input vector cancels to zero. This cancellation gives a short proof of the key fact.

Write the cross product.
Let \(\mathbf{c}=\mathbf{a}\times\mathbf{b}\), so its components are the three differences in the definition.

Take the dot product with \(\mathbf{a}\).
Multiply corresponding components and group the terms that cancel.

Repeat with \(\mathbf{b}\).
The same cancellation shows that \(\mathbf{c}\) is perpendicular to the other input direction as well.

In full, using the dot product rule,

\begin{aligned} \mathbf{a}\cdot\mathbf{c} &=a_1(a_2b_3-a_3b_2) +a_2(a_3b_1-a_1b_3) +a_3(a_1b_2-a_2b_1)\\ &=a_1a_2b_3-a_1a_3b_2 +a_2a_3b_1-a_1a_2b_3 +a_1a_3b_2-a_2a_3b_1\\ &=0. \end{aligned}

For the other input,

\begin{aligned} \mathbf{b}\cdot\mathbf{c} &=b_1(a_2b_3-a_3b_2) +b_2(a_3b_1-a_1b_3) +b_3(a_1b_2-a_2b_1)\\ &=a_2b_1b_3-a_3b_1b_2 +a_3b_1b_2-a_1b_2b_3 +a_1b_2b_3-a_2b_1b_3\\ &=0. \end{aligned}

Therefore \(\mathbf{a}\times\mathbf{b}\) is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\). If these two vectors are nonparallel, their cross product is nonzero, so it can serve as a normal vector. If the vectors are parallel, the cross product is zero and cannot be a normal vector.

Worked Examples

Worked Example: Finding a Normal from Two Directions

Find a normal vector to a plane containing the nonparallel directions \(\mathbf{a}=(2,-1,1)\) and \(\mathbf{b}=(1,3,2)\).

Take the cross product in the stated order:

\begin{aligned} \mathbf{a}\times\mathbf{b} &=\left( (-1)(2)-(1)(3),\; (1)(1)-(2)(2),\; (2)(3)-(-1)(1) \right)\\ &=(-2-3,\;1-4,\;6+1)\\ &=(-5,-3,7). \end{aligned}

Thus \(\mathbf{n}=(-5,-3,7)\) is a candidate normal. Check it against both input directions:

\begin{aligned} \mathbf{n}\cdot\mathbf{a} &=(-5)(2)+(-3)(-1)+(7)(1)\\ &=-10+3+7=0,\\[4pt] \mathbf{n}\cdot\mathbf{b} &=(-5)(1)+(-3)(3)+(7)(2)\\ &=-5-9+14=0. \end{aligned}

Both dot products are zero, so \(\mathbf{n}\) is perpendicular to both directions. Since the directions are nonparallel, they lie in and determine a plane, and this nonzero vector is normal to that plane.

Worked Example: A Normal to a Flat Panel

Two nonparallel directions along a flat panel are \(\mathbf{u}=(1,2,0)\) and \(\mathbf{v}=(0,1,2)\). Find and verify a normal to the panel.

Calculate each component of the cross product:

\begin{aligned} \mathbf{u}\times\mathbf{v} &=\left( (2)(2)-(0)(1),\; (0)(0)-(1)(2),\; (1)(1)-(2)(0) \right)\\ &=(4,-2,1). \end{aligned}

Check the result using dot products:

\begin{aligned} (4,-2,1)\cdot(1,2,0)&=4-4+0=0,\\ (4,-2,1)\cdot(0,1,2)&=0-2+2=0. \end{aligned}

The vector \((4,-2,1)\) is nonzero and perpendicular to both panel directions, so it is a normal to the panel's plane. The vector \((-4,2,-1)\) would also be a valid normal, since it is the negative of this one and remains perpendicular to both.

Worked Example: Normal from Two Edge Directions

A flat triangular surface has two edge directions from one of its corners given by \(\mathbf{p}=(3,0,1)\) and \(\mathbf{q}=(1,2,-1)\). Find a normal vector to the surface.

The cross product is

\begin{aligned} \mathbf{p}\times\mathbf{q} &=\left( (0)(-1)-(1)(2),\; (1)(1)-(3)(-1),\; (3)(2)-(0)(1) \right)\\ &=(-2,4,6)\\ &=2(-1,2,3). \end{aligned}

Because nonzero scalar multiples have the same direction, the simpler vector \(\mathbf{n}=(-1,2,3)\) can be used. Verify it directly:

\begin{aligned} \mathbf{n}\cdot\mathbf{p} &=(-1)(3)+(2)(0)+(3)(1)\\ &=-3+0+3=0,\\ \mathbf{n}\cdot\mathbf{q} &=(-1)(1)+(2)(2)+(3)(-1)\\ &=-1+4-3=0. \end{aligned}

The vector \(\mathbf{n}\) is nonzero and perpendicular to both edge directions. Since those directions are nonparallel, it is a normal to the plane of the triangular surface. The factor of \(2\) in the unsimplified cross product does not affect its normal direction.

Conditions and Interpretation

The method depends on having two nonparallel directions that lie in the plane or flat surface of interest. The cross product is always perpendicular to its two input vectors, but that alone does not establish that it is normal to a particular plane. The input vectors must actually be directions in that plane.

There are two useful limitations to keep in mind:

Parallel directions: If the inputs are parallel, their cross product is the zero vector. The zero vector is not a normal vector, because a normal must be nonzero. Choose another direction in the plane that is not parallel to the first.
Curved surfaces: The method here identifies a normal to a plane or flat surface from two directions lying in it. It does not by itself define a normal to a curved surface, where the relevant directions and notion of a normal require additional ideas.

The cross product's magnitude relates to the area of the parallelogram formed by its input vectors, a fact from the earlier material on parallelogram and triangle areas. For finding a normal direction, however, the key facts are that the cross product is perpendicular to both inputs and is nonzero when the inputs are nonparallel.

A Common Sign Error

The second component of the cross product is especially easy to compute with the wrong sign. For the vectors in the first example, an incorrect calculation might use \(a_1b_3-a_3b_1\) for that component. This would give the proposed vector \((-5,3,7)\), rather than the correct \((-5,-3,7)\).

Test the incorrect result against \(\mathbf{a}=(2,-1,1)\):

(-5,3,7)\cdot(2,-1,1) =-10-3+7=-6\ne 0.

It is not perpendicular to \(\mathbf{a}\), so it cannot be the required normal. The correct second component is \(a_3b_1-a_1b_3=(1)(1)-(2)(2)=1-4=-3\). Checking dot products is a reliable way to catch this sign error before using the result.

Why Normal Vectors Are Useful

A normal vector records the orientation of a flat surface in a compact form. It can help distinguish which way a plane faces, compare the orientations of two planes, and support later calculations involving planes. At this stage, the essential task is to find a correct normal from directions already known to lie in the plane.

When checking your work, confirm all three points: the two input directions are nonparallel, they lie in the plane, and the calculated nonzero vector has dot product zero with each. These checks establish that the result is genuinely a normal, rather than merely a vector produced by a calculation.

Check Your Understanding

For each question, use the cross-product definition and, where appropriate, verify the result with dot products.

Find a normal vector to a plane containing directions \((1,0,1)\) and \((0,2,1)\).
Why does taking the cross product of two nonparallel directions in a plane give a normal to the plane?
If \(\mathbf{a}\times\mathbf{b}=\mathbf{n}\), what is the relationship between \(\mathbf{b}\times\mathbf{a}\) and \(\mathbf{n}\)? Is it also a normal to the same plane?
What happens when the two input directions are parallel, and why can the result not be used as a normal?
A computed vector has dot product zero with one direction in a plane but not with a second nonparallel direction in that plane. What does this tell you about the proposed normal?