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 \(70.5288^\circ\).

E 13.4 Exercises