## Linear Algebra and Its Applications, Exercise 3.4.2

Exercise 3.4.2. Given two orthonormal vectors $a_1 = (\frac{2}{3}, \frac{2}{3}, -\frac{1}{3})$ and $a_2 = (-\frac{1}{3}, \frac{2}{3}, \frac{2}{3})$ and the vector $b = (0, 3, 0)$, project $b$ onto $a_1$ and $a_2$. Also find the projection $p$ of $b$ onto the plane formed by $a_1$ and $a_2$.

Answer: Since $a_1$ is orthonormal, the projection of $b$ onto $a_1$ is given by

$a_1^Tba_1 = \begin{bmatrix} \frac{2}{3}&\frac{2}{3}&-\frac{1}{3} \end{bmatrix} \begin{bmatrix} 0 \\ 3 \\ 0 \end{bmatrix} \begin{bmatrix} \frac{2}{3} \\ \frac{2}{3} \\ -\frac{1}{3} \end{bmatrix} = 2 \cdot \begin{bmatrix} \frac{2}{3} \\ \frac{2}{3} \\ -\frac{1}{3} \end{bmatrix} = \begin{bmatrix} \frac{4}{3} \\ \frac{4}{3} \\ -\frac{2}{3} \end{bmatrix}$

Similarly the projection of $b$ onto $a_2$ is given by

$a_2^Tba_2 = \begin{bmatrix} -\frac{1}{3}&\frac{2}{3}&\frac{2}{3} \end{bmatrix} \begin{bmatrix} 0 \\ 3 \\ 0 \end{bmatrix} \begin{bmatrix} -\frac{1}{3} \\ \frac{2}{3} \\ \frac{2}{3} \end{bmatrix} = 2 \cdot \begin{bmatrix} -\frac{1}{3} \\ \frac{2}{3} \\ \frac{2}{3} \end{bmatrix} = \begin{bmatrix} -\frac{2}{3} \\ \frac{4}{3} \\ \frac{4}{3} \end{bmatrix}$

The projection $p$ of $b$ onto the plane formed by $a_1$ and $a_2$ is simply the sum of the projections above:

$p = a_1^Tba_1 + a_2^Tba_2 = \begin{bmatrix} \frac{4}{3} \\ \frac{4}{3} \\ -\frac{2}{3} \end{bmatrix} + \begin{bmatrix} -\frac{2}{3} \\ \frac{4}{3} \\ \frac{4}{3} \end{bmatrix} = \begin{bmatrix} \frac{2}{3} \\ \frac{8}{3} \\ \frac{2}{3} \end{bmatrix}$

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, Fifth Edition and the accompanying free online course, and Dr Strang’s other books.

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