# 13.4 | The Cross Product

If $$\mathbf{a}$$ and $$\mathbf{b}$$ are nonzero three-dimensional vectors, the cross product of $$\mathbf{a}$$ and $$\mathbf{b}$$ is the vector
$$\mathbf{a} \times \mathbf{b} = \left(|\mathbf{a}|| \mathbf{b}| \sin \theta \right) \mathbf{n}$$
where $$\theta$$ is the angle between $$\mathbf{a}$$ and $$\mathbf{b}$$, $$0 \leq \theta \leq \pi$$, and $$\mathbf{n}$$ is a unit vector perpendicular to both $$\mathbf{a}$$ and $$\mathbf{b}$$ and whose direction is given by the right-hand rule: If the fingers of your right hand curl through the angle $$\theta$$ from $$\mathbf{a}$$ to $$\mathbf{b}$$, then your thumb points in the direction of $$\mathbf{n}$$.
Two vectors nonzero vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ are parallel if and only if $$\mathbf{a} \times \mathbf{b} = 0$$
If $$\mathbf{a}$$,$$\mathbf{b}$$, and $$\mathbf{c}$$ are vectors and $$c$$ is a scalar, then

1. $$\mathbf{a} \times \mathbf{b}= -\mathbf{b} \times \mathbf{a}$$
2. $$(c\mathbf{a}) \times \mathbf{b}=c(\mathbf{a} \times \mathbf{b}) = \mathbf{a} \times (c\mathbf{b})$$
3. $$\mathbf{a} \times ( \mathbf{b} + \mathbf{c})= \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}$$
4. $$( \mathbf{a} + \mathbf{b}) \times \mathbf{c} = \mathbf{a} \times \mathbf{c} + \mathbf{b} \times \mathbf{c}$$
The length of the cross product $$\mathbf{a} \times \mathbf{b}$$ is equal to the area of the parallelogram determined by $$\mathbf{a}$$ and $$\mathbf{b}$$.
If $$\mathbf{a}= \langle a_1, a_2, a_3 \rangle$$ and $$\mathbf{b} = \langle b_1, b_2, b_3 \rangle$$, then
\begin{aligned} \mathbf{a} \times \mathbf{b} & = \langle a_2b_3 – a_3b_2, a_3b_1 -a_1 b_3, a_1b_2-a_2b_1 \rangle \\ & = \left| \begin{array}{c c} a_2 & a_3 \\ b_2 & b_3 \end{array}\right| \mathbf{i} – \left| \begin{array}{c c} a_1 & a_3 \\ b_1 & b_3\end{array}\right| \mathbf{j} + \left| \begin{array}{c c} a_1 & a_2 \\ b_1 & b_2 \end{array} \right| \mathbf{k} \\ & = \left| \begin{array}{c c c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{array} \right| \end{aligned}
The volume of the parallelepiped determined by the vectors $$\mathbf{a}$$, $$\mathbf{b}$$, and $$\mathbf{c}$$ is the magnitude of their scalar triple product:
$$V = \left| \mathbf{a \cdot} (\mathbf{b} \times \mathbf{c}) \right|$$

Find the surface area of the tetrahedron defined by the points
$$A(3,0,0), \; B(0,2,0), \; C(-2,-2,0), \; D(1,1,1)$$

The area of a parallelogram is given by the length of the cross product of the two vectors which define it. Consequently, the area of a triangle is given by half the length of the cross product of the two vectors which define it.
A vector equation of a line is given by
$$\mathbf{r} = \mathbf{r_0} + t \mathbf{v}$$
where $$\mathbf{r_0}$$ is a point on the line and $$\mathbf{v}$$ is a vector pointing in the direction of the line. Suppose $$\mathbf{r} =\langle x, y, z \rangle$$, $$\mathbf{r_0} = \langle x_0, y_0, z_0 \rangle$$, and $$\mathbf{v} = \langle a, b, c \rangle$$, then the vector equation becomes
$$\langle x,y,z \rangle = \langle x_0,y_0,z_0 \rangle + t\langle a,b,c \rangle$$
which is equivalent to the parametric equations
$$x = x_0 + at \; \; y = y_0 + bt \; \; z = z_0 +ct$$
which is equivalent to the symmetric equations
$$\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$$
provided $$a,b,$$ and $$c$$ are nonzero.
The line segment from $$\mathbf{r_0}$$ to $$\mathbf{r_1}$$ is given by the vector equation
$$\mathbf{r}(t) = (1-t) \mathbf{r_0} + t \mathbf{r_1}, \; \; \; 0 \leq t \leq 1$$
where $$\mathbf{r_0}$$ is the initial point and $$\mathbf{r_1}$$ is the terminal point.
A vector equation of a plane is given by
$$\mathbf{n \cdot } (\mathbf{r} – \mathbf{r_0}) = 0$$
where $$\mathbf{r_0}$$ is a point in the plane and $$\mathbf{n}$$ is a vector perpendicular to the plane. Suppose $$\mathbf{r} =\langle x, y, z \rangle$$, $$\mathbf{r_0} = \langle x_0, y_0, z_0 \rangle$$, and $$\mathbf{n} = \langle a, b, c \rangle$$, then the vector equation becomes
$$\langle a, b, c \rangle \mathbf{ \cdot } (\langle x, y, z \rangle – \langle x_0, y_0, z_0 \rangle) =0$$
which is equivalent to
$$a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$$
which is equivalent to
$$ax + by + cz + d = 0$$
where $$d = -(ax_0 + by_0 + cz_0)$$.
The angle between two planes is defined to be the angle between their normal vectors. That is, if
\begin{aligned} P_1: \; \; & \; \; \mathbf{n_1}(\mathbf{r}-\mathbf{r_{1}}) = 0\\ P_2: \; \; & \; \; \mathbf{n_2}(\mathbf{r}-\mathbf{r_{2}}) = 0 \end{aligned}
then the angle $$\theta$$ between the planes is the value of $$\theta$$ between $$0 \leq \theta \leq \pi/2$$ that satisfies
$$\mathbf{n_1 \cdot n_2} = |\mathbf{n_1}||\mathbf{n_2}| \cos \theta$$

Determine whether the lines $$L_1$$ and $$L_2$$ are parallel, skew, or intersecting. If they intersect, find the point of intersection.
\begin{aligned} L_1: \; \; & x = \frac{y-1}{2} = \frac{z-2}{3} \\ L_2: \; \; & \frac{x-3}{-4} = \frac{y-2}{-3} = \frac{z-1}{2} \end{aligned}

Rewrite the symmetric equations for $$L_1$$ and $$L_2$$ in vector form:
\begin{aligned} L_1: \; \; \mathbf{r}_1(t) & = \langle 0, 1, 2 \rangle + t \langle 1, 2, 3 \rangle \\ L_2: \; \; \mathbf{r}_2(s) & = \langle 3, 2, 1 \rangle + s \langle -4,-3, 2\rangle \end{aligned}

Let $$\mathbf{v}_1$$ and $$\mathbf{v}_2$$ be the directional vectors of $$L_1$$ and $$L_2$$, respectively. So, $$\mathbf{v}_1 = \langle 1, 2, 3 \rangle$$ and $$\mathbf{v}_2 = \langle -4,-3, 2\rangle$$. The two lines are parallel if and only if $$\mathbf{v}_1 \times \mathbf{v}_2 = \mathbf{0}$$. But $$\mathbf{v}_1 \times \mathbf{v}_2 = \langle 13, -14, 5 \rangle$$ and so the lines are not parallel.
The lines intersect if and only if there exists a value of $$t$$ and $$s$$ such that $$\mathbf{r}_1(t) = \mathbf{r}_2(s)$$. That is,
\left \{\begin{aligned} t & = -4s+3 \\ 2t & = -3s +2 \\ 3t & = 2s +1 \end{aligned} \right.
This system has no solution and consequently, the two lines do not intersect. Hence the two lines are skew.

Find parametric equations and symmetric equations for the line through $$(1,-1,1)$$ and parallel to the line $$x+2 = \frac{1}{2}y = z-3$$.

If we have the vector equation of the line, then from there we can get the parametric equations and symmetric equations. The form of the vector equation of the line will be
$$\mathbf{r}(t) = \mathbf{r_0} + t \mathbf{v}$$
where $$\mathbf{r_0}$$ is a point on the line and $$\mathbf{v}$$ is the directional vector of the line. So we may take $$\mathbf{r_0} = \langle 1,-1,1\rangle$$. To find $$\mathbf{v}$$, note that two lines are parallel if and only if their directional vectors are nonzero scalar multiples of one another. The vector equation for the given line is:
$$\mathbf{r}_{//}(t) = \langle -2,0,3 \rangle + t \langle 1, 2, 1 \rangle$$
and so has directional vector $$\mathbf{v}_{//} = \langle 1,2,1 \rangle$$. Consequently, the directional vector for our line must be a nonzero scalar multiple of $$\mathbf{v}_{//}$$ and so taking that scalar to be one suffices.
Therefore, the vector equation of our line is
$$\mathbf{r}(t) = \langle 1, -1, 1 \rangle + t \langle 1,2,1 \rangle$$
The parametric equations are obtained by simplifying the vector equation and setting the respective components equal to $$x,y,$$ and $$z$$:
$$x = t+1, y=2t-1, z = t+1$$
The symmetric equations are obtained by solving for $$t$$ in each of the parametric equations and then setting the results equal to one another:
$$x-1=\frac{y+1}{2} = z-1$$
Find an equation of the plane through the points $$(0,1,1)$$, $$(1,0,1)$$, and $$(1,1,0)$$

The vector equation for a plane is given by
$$\mathbf{n} \cdot (\mathbf{r}-\mathbf{r}_0) = 0$$
where $$\mathbf{n}$$ is a vector normal to the plane, $$\mathbf{r} = \langle x, y, z\rangle$$, and $$\mathbf{r}_0$$ is a point in the plane. Consequently, we may take $$\mathbf{r}_0 = \langle 0,1,1\rangle$$ (though any of the three points will suffice). To find $$\mathbf{n}$$, note that given two vectors that lie in the plane their cross product provides us with a vector normal to the plane. Let $$A = \langle 0,1,1 \rangle$$, $$B = \langle 1,0,1 \rangle$$, and $$C = \langle 1,1,0 \rangle$$. So, $$\mathbf{n} = AB \times AC = \langle 1,1,1 \rangle$$.

Thus, the vector equation of the plane is
$$\langle 1,1,1 \rangle \cdot \langle x, y-1, z-1 \rangle = 0$$
which is equivalent to
$$x + y + z = 2$$

Find an equation of the plane that passes through the point $$(1,5,1)$$ and is perpendicular to the planes $$2x+y-2z = 2$$ and $$x + 3z = 4$$.

The vector equation for a plane is given by
$$\mathbf{n} \cdot (\mathbf{r}-\mathbf{r}_0) = 0$$
where $$\mathbf{n}$$ is a vector normal to the plane, $$\mathbf{r} = \langle x, y, z\rangle$$, and $$\mathbf{r}_0$$ is a point in the plane. Consequently, we may take $$\mathbf{r}_0 = \langle 1,5,1 \rangle$$. Note that the cross product of the normals of the given planes provides us with a normal vector to a plane that is perpendicular to both planes. Let $$\mathbf{n}_1$$ and $$\mathbf{n}_2$$ be the normal vectors of the first plane and second plane, respectively. So that, $$\mathbf{n}_1 = \langle 2,1,-2 \rangle$$ and $$\mathbf{n}_2 = \langle 1, 0, 3 \rangle$$. So,
$$\mathbf{n} = \mathbf{n}_1 \times \mathbf{n}_2 = \langle 3,-8,-1 \rangle$$

Thus, the vector equation for the plane is
$$\langle 3, -8,-1 \rangle \cdot \langle x -1, y-5, z-1 \rangle = 0$$
which is equivalent to
$$3x -8y -z = -36$$

Determine whether the planes are parallel, perpendicular, or neither. If neither, find the angle between them.
\begin{aligned} P_1: \; \; & x + y + z = 1\\ P_2: \; \; & x – y + z = 1 \end{aligned}

Let $$\mathbf{n}_1$$ and $$\mathbf{n}_2$$ be the normal vectors of $$P_1$$ and $$P_2$$, respectively. Then $$\mathbf{n}_1 = \langle 1,1,1 \rangle$$ and $$\mathbf{n}_2 = \langle 1,-1,1 \rangle$$. Note that the two planes are

• parallel if and only if $$\mathbf{n}_1 \times \mathbf{n}_2 = \mathbf{0}$$
• perpendicular if and only if $$\mathbf{n}_1 \cdot \mathbf{n}_2 =0$$

But $$\mathbf{n}_1 \times \mathbf{n}_2 = \langle 2,0,-2 \rangle$$ and $$\mathbf{n}_1 \cdot \mathbf{n}_2 =1$$ and so the planes are not parallel and not perpendicular. That is, the planes are neither. The angle $$\theta$$ between the planes satisfies
$$\cos \theta = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{\left|\mathbf{n}_1 \right| \left|\mathbf{n}_2 \right|}$$
where $$0 \leq \theta \leq \pi/2$$. But then
$$\theta = \cos^{-1}\left(\frac{1}{3} \right) \approx 1.23096$$
or approximately $$70.5288^\circ$$.