Cross Product and Planes

Chapter 9 Cross Product and Planes

“Happiness lies in the joy of achievement and the thrill of creative effort.” - Franklin D. Roosevelt

In Problem 8.34, you were given two vectors and asked to find a vector that was perpendicular to both. Because there are infinitely many vectors perpendicular to any two given vectors, we added another condition (\(x_1+x_2+x_3=1\)) so that the answer would be unique. If, instead of this choice, we had chosen the third condtion to be that it have length that is the area of the parallelogram defined by the two vectors, we would have narrowed it down to one of two choices. The following definition is one of them and the negative of this vector is the other.

Definition 9.1.

The cross product of the vectors \(\oa u\) and \(\oa v\) is the vector

\begin{equation*} \oa u \times \oa v = (u_{2}v_{3}-u_{3}v_{2},\ u_{3}v_{1}-u_{1}v_{3},\ u_{1}v_{2}-u_{2}v_{1})\text{.} \end{equation*}

An easy way to compute the cross products requires a tool from linear algebra, determinants. If I have not done so already, ask me during class to show you how to compute the determinant of 2 by 2 and 3 by 3 matrices.

Computing Cross Products Using Determinants. If we define

\begin{equation*} \oa i = \oa{(1,0,0)}, \oa j = \oa{(0,1,0)}, \text{ and } \oa k = \oa{(0,0,1)}, \end{equation*}

then we may compute the cross product as:

\begin{equation*} \oa u \times \oa v = det \begin{pmatrix}\oa i \amp \oa j \amp \oa k \\ u_1 \amp u_2 \amp u_3 \\ v_1 \amp v_2 \amp v_3 \end{pmatrix} . \end{equation*}

Problem 9.2.

Prove or give a counter example for each statement, assuming \(\oa u = (u_1,u_2,u_3), \oa v = (v_1,v_2,v_3), \oa w = (w_1, w_2, w_3)\) and \(k \in \re.\)

\(k(\oa{u}\times \oa{v})=(k\oa{u})\times \oa{v} = \oa{u}\times (k\oa{v})\)
\(\oa{u}\times \oa{v} = - \big( \oa{v}\times \oa{u} \big)\)
\(\oa{u}\times \oa{u} = \oa{u}\times \oa 0 = \oa 0\times \oa{u} = \oa{0}\)
\((\oa{u}\times \oa{v})\cdot \oa{w} = \oa{u}\cdot (\oa{v}\times \oa{w})\)
\(k + ( \oa u \times \oa v) = (k + \oa u) \times (k + \oa v)\)
\(\oa{u}\times (\oa{v}+\oa{w}) = (\oa{u}\times \oa{v})+(\oa{u}\times \oa{w})\)
\(\oa{u}\times (\oa{v} \cdot \oa{w}) = (\oa{u}\times \oa{v})\cdot(\oa{u}\times \oa{w})\)
\(\oa{u}\times (\oa{v}\times \oa{w}) = (\oa{u}\cdot \oa{w})\oa{v}-(\oa{u}\cdot \oa{v})\oa{w}\)

One might think of planes in \({\re}^3\) as analogous to lines in \({\re}^2\text{.}\) This is because a line is a one-dimensional object in \(\re^2\) — that is it has dimension one less than the dimension of the space. In \(\re^3\) a plane is two-dimensional — it has dimension one less than the dimension of the space.

Definition 9.3.

Given \(a,b, c \in \re\) where \(a\) and \(b\) are not both zero, the line determined by \(a,b,\) and \(c\) is the collection of all points \((x,y) \in \re^2\) satisfying \(ax+by=c.\)

Given this definition we can define a plane in the same manner.

Definition 9.4.

Given \(a,b, c, d \in \re\) where \(a,b\) and \(c\) are not all zero, the plane determined by \(a,b,c\) and \(d\) is the collection of all points \((x,y,z) \in \re^3\) satisfying \(ax+by +cz=d.\)

If algebraically we think of a plane as all \((x,y,z)\in {\re}^3\) satisfying \(ax+by+cz=d\) where not all of \(a,b\) and \(c\) are zero, then geometrically we can think of a plane as uniquely determined by a vector and a point where the plane is perpendicular to the vector and contains the point. We need both because there are infinitely many planes perpendicular to a given vector, but knowing one point in the plane uniquely determines the plane.

Problem 9.5.

Show that \((3,-2,5)\) is perpendicular to the plane \(3x -2y +5z = 7\) by choosing two points, \(x=(x_1,x_2,x_3)\) and \(y=(y_1,y_2,y_3),\) in the plane and showing that \((3, -2, 5) \cdot \oa{x-y} = 0.\)

Problem 9.6.

Show that \((a,b,c)\in {\re}^3\) is orthogonal to the plane \(ax+by+cz=d\) by showing that if \(x=(x_1,x_2,x_3)\) and \(y=(y_1,y_2,y_3)\) are two points on the plane, then \((a, b, c) \cdot \oa{x-y} = 0.\)

Problem 9.7.

Determine whether these two planes are parallel.

\(2x-3y+\frac{5}{2}z=9\)
\(x-\frac{3}{2}y+\frac{5}{4}z=12\)

Problem 9.8.

Write in standard form (ax+by+cz=d) the equation of a plane \(\perp\) to the first plane from previous problem and containing the point \((9,2,3)\text{.}\)

Problem 9.9.

Find all the planes parallel to the plane \(x+y-z=4\) and at a distance of one unit away from the plane. When does the “distance between two planes” make sense?

Problem 9.10.

Find both angles between these two planes:

\(2x-3y+4z=10\)
\(4x+3y-6z=-4\)

Problem 9.11.

Find the equation of the plane containing \((2,3,4)\text{,}\) \((1,2,3)\) and \((6,-2,5)\text{.}\)

For the next two problems, there is a formula on the web or in some book, but it's probably wrong because of a typographical error. Find a vector \(\perp\) to both planes, determine the equation of a parametric line, \(\oa l\text{,}\) passing through both planes. Find the points where \(\oa l\) intersects each plane. Find the distance between these points.

Problem 9.12.

Find the distance between the two planes, \(x+y+z=1\) and \(x+y+z=2\text{.}\)

Problem 9.13.

Find the distance between the two planes, \(3x-4y+5z=9\) and \(3x-4y+5z=4\text{.}\)

Problem 9.14.

Find an equation for the distance between two planes, \(ax + by + cz = e\) and \(ax + by + cz = f.\)

Problem 9.15.

Find the equation of the plane containing the line \(\oa l(t)=(1+2t,\ -1+3t,\ 4+t)\) and the point \((1,-1,5)\text{.}\)

The next problem asks for the intersection of two planes. What are all the possibilities for the intersection of any two planes? One possibility is a line. To find the equation of the line, there are a couple of ways we could think about this. First, we could find two points in the intersection and then find the equation of that line. Or we could observe that the intersection of two planes must be contained in each plane and thus is parallel to both planes. How can we find a vector that is parallel to both planes?

Problem 9.16.

Find the intersection of the two planes \(3x-2y+6z=1\) and \(3x-4y+5z=1.\)