## Linear Algebra and Its Applications, Exercise 3.1.9

Exercise 3.1.9. For the plane in $\mathbb{R}^3$ spanned by the vectors $(1, 1, 2)$ and $(1, 2, 3)$ find the orthogonal complement (i.e., the line in $\mathbb{R}^3$ perpendicular to the plane). Note that this can be done by solving the system $Ax = 0$ where the two vectors are the rows of $A$.

$A = \begin{bmatrix} 1&1&2 \\ 1&2&3 \end{bmatrix}$

We solve the system $Ax = 0$ using Gaussian elimination, starting by subtracting 1 times row 1 from  row 2:

$\begin{bmatrix} 1&1&2 \\ 1&2&3 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&1&2 \\ 0&1&1 \end{bmatrix}$

Since the resulting echelon matrix has pivots in columns 1 and 2, we have $x_1$ and $x_2$ as basic variables and $x_3$ as a free variable. Setting $x_3 = 1$, from the second row we have $x_2 + x+3 = x_2 + 1 = 0$ or $x_2 = -1$. From the first row we have $x_1 + x_2 + 2x_3 = x_1 - 1 + 2 = 0$ or $x_1 = -1$.

So $(-1, -1, 1)$ is a solution to $Ax = 0$ and a basis for the orthogonal complement, which consists of the line through the origin and $(-1, -1, 1)$.

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.