## 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.