Linear Algebra and Its Applications, Exercise 3.1.21

Exercise 3.1.21. If $P$ is the plane in $\mathbb{R}^3$ described by $x+2y-z = 6$ what is the equation for the plane $P'$ parallel to $P$ through the origin? What is a vector perpendicular to $P'$? Find a matrix $A$ for which $P'$ is the nullspace, and a matrix $B$ for which $P'$ is the row space.

Answer:  One way to approach this problem is to find a general solution to the equation $x+2y-z = 6$ and express it as the sum of a homogeneous solution and a particular solution. The corresponding homogeneous system is $x+2y-z = 0$, which is represented by the matrix

$A = \begin{bmatrix} 1&2&-1 \end{bmatrix}$

This system has $x$ as a basic variable and $y$ and $y$ as free variables. Setting $y = 1$ and $z = 0$ we have $x+2y-z = x+2-0 = 0$ or $x = -2$. So $(-2, 1, 0)$ is one solution to the homogeneous system. Setting $y = 0$ and $z = 1$ we have $x+2y-z = x+0-1 = 0$ or $x = 1$. So $(1, 0, 1)$ is a second solution to the homogeneous system. These vectors are in the nullspace of $A$ and serve as a basis for it.

To find the particular solution we set $y = z = 0$ for the general system so that we have $x+2y-z = x+0-0 = 6$ or $x = 6$. So $(6, 0, 0)$ is a particular solution to the general system, and the general solution to the equation is then the sum of the particular solution and the homogeneous solution:

$\begin{bmatrix} 6 \\ 0 \\ 0 \end{bmatrix} + y \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} + z \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$

The plane $P$ is defined by the general solution to $x+2y-z=6$. The plane $P'$ going through the origin corresponds to the homogeneous system $x+2y-z=0$ and is spanned by the vectors $(-2, 1, 0)$ and $(1, 0, 1)$; it is the nullspace of the matrix $A$ above. The plane $P$ is parallel to $P'$ and is offset from it by the vector $(6, 0, 0)$.

To find a vector perpendicular to $P'$ (and $P$),  from the above solution to the homogeneous system $x+2y-z = 0$ we know that the vector $(1, 2, -1)$ (the first and only row of the matrix $A$ above) is orthogonal to the vectors $(-2, 1, 0)$ and $(1, 0, 1)$ (the basis vectors for the nullspace of $A$). Therefore the vector $(1, 2, -1)$ is perpendicular to the plane $P'$, the nullspace of $A$ spanned by $(-2, 1, 0)$ and $(1, 0, 1)$.

Using the vectors $(-2, 1, 0)$ and $(1, 0, 1)$ that serve as a basis for the plane $P'$ we can construct a matrix $B$ for which $P'$ is the row space:

$B = \begin{bmatrix} -2&1&0 \\ 1&0&1 \end{bmatrix}$

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

This entry was posted in linear algebra and tagged , . Bookmark the permalink.