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.

Answer: We have

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.

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