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