# Thornton & Marion (Fifth Edition), Chapter 01, Exercise Problem 01 Solution

## Chapter 01. Matrices, Vectors, and Vector Calculus

### Problem 01. The transformation matrix

##### Solution
The direction cosines $\lambda_{ij}$$\lambda_{ij}$ can be determined using the definition of Equation (1.3).

$\lambda_{11} = \cos(x'_1,x_1) = \frac{\sqrt{2}}{2}$$\lambda_{11} = \cos(x'_1,x_1) = \frac{\sqrt{2}}{2}$
$\lambda_{12} = \cos(x'_1,x_2) = 0$$\lambda_{12} = \cos(x'_1,x_2) = 0$
$\lambda_{13} = \cos(x'_1,x_3) = \cos 135^{\circ} = - \frac{\sqrt{2}}{2}$$\lambda_{13} = \cos(x'_1,x_3) = \cos 135^{\circ} = - \frac{\sqrt{2}}{2}$
$\lambda_{21} = \cos(x'_2,x_1) = 0$$\lambda_{21} = \cos(x'_2,x_1) = 0$
$\lambda_{22} = \cos(x'_2,x_2) = 1$$\lambda_{22} = \cos(x'_2,x_2) = 1$
$\lambda_{23} = \cos(x'_2,x_3) = 0$$\lambda_{23} = \cos(x'_2,x_3) = 0$
$\lambda_{31} = \cos(x'_3,x_1) = \frac{\sqrt{2}}{2}$$\lambda_{31} = \cos(x'_3,x_1) = \frac{\sqrt{2}}{2}$
$\lambda_{32} = \cos(x'_3,x_2) = 0$$\lambda_{32} = \cos(x'_3,x_2) = 0$
$\lambda_{33} = \cos(x'_3,x_3) = \frac{\sqrt{2}}{2}$$\lambda_{33} = \cos(x'_3,x_3) = \frac{\sqrt{2}}{2}$

Therefore, the transformation matrix is

$\lambda = \begin{pmatrix} \frac{\sqrt{2}}{2} & 0 & -\frac{\sqrt{2}}{2} \\ 0 & 1 & 0 \\ \frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2} \end{pmatrix}$$\lambda = \begin{pmatrix} \frac{\sqrt{2}}{2} & 0 & -\frac{\sqrt{2}}{2} \\ 0 & 1 & 0 \\ \frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2} \end{pmatrix}$