## 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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