**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 \(\mathbf{n}\).

- \(\mathbf{a} \times \mathbf{b}= -\mathbf{b} \times \mathbf{a}\)
- \((c\mathbf{a}) \times \mathbf{b}=c(\mathbf{a} \times \mathbf{b}) = \mathbf{a} \times (c\mathbf{b})\)
- \(\mathbf{a} \times ( \mathbf{b} + \mathbf{c})= \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}\)
- \(( \mathbf{a} + \mathbf{b}) \times \mathbf{c} = \mathbf{a} \times \mathbf{c} + \mathbf{b} \times \mathbf{c}\)

$$

\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}

$$

$$

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.

$$

\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.

$$

\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.

$$

\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 \(d = -(ax_0 + by_0 + cz_0)\).

$$

\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.

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

$$

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

$$

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

$$

$$

\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 \(70.5288^\circ\).