Could someone explain this to me$?$

Also how would you do this in Matlab? Thank you so much.

$2\mathrm{.}1\mathrm{.}1$ Example: Computing the Traction Vector

Suppose we are given that the horizontal components of the stress tensor are

$=\left[\begin{array}{cc}{}_{}& {}_{xy}\\ {}_{xy}& {}_{yy}\end{array}\right]=\left[\begin{array}{cc}-40& -10\\ -10& -60\end{array}\right]$MPa

Assuming this is a two$-$dimensional problem, let us compute the forces acting across a fault oriented at $45($clockwise$)$ from the $x-$axis. We typically assume that the $x-$axis points east and the $y-$axis points north, so in this case the fault is trending from the northwest to the southeast. To compute the traction vector from equation $(2\mathrm{.}3),$ we need the normal vector hat$(n).$ This vector is perpendicular to the fault and thus points to the northeast, or parallel to the vector $(1,1)$ in our $(x,y)$ coordinate system ${?}^1.$ However, remember that hat$(n)$ is a unit vector and thus we must normalize its length to obtain

hat$(n)=\left[\begin{array}{c}\frac{1}{\sqrt[\phantom{2}]{2}}\\ \frac{1}{\sqrt[\phantom{2}]{2}}\end{array}\right]=\left[\begin{array}{c}0\mathrm{.}7071\\ 0\mathrm{.}7071\end{array}\right]$

Substituting into $(2\mathrm{.}3),$ we have

$t(hat(n))=hat(n)=\left[\begin{array}{cc}-40& -10\\ -10& -60\end{array}\right]\left[\begin{array}{c}\frac{1}{\sqrt[\phantom{2}]{2}}\\ \frac{1}{\sqrt[\phantom{2}]{2}}\end{array}\right]=\left[\begin{array}{c}-\frac{50}{\sqrt[\phantom{2}]{2}}\\ -\frac{70}{\sqrt[\phantom{2}]{2}}\end{array}\right]$~~$\left[\begin{array}{c}-35\mathrm{.}4\\ -49\mathrm{.}4\end{array}\right]$MPa

Note that the traction vector points approximately southwest $($see Figure $2\mathrm{.}2).$ This is the force exerted by the northeast side of the fault $($i$.$e$.,$ in the direction of our normal vector$)$ on the southwest side of the fault. Thus we see that there is fault normal compression on the fault. To resolve the normal and shear stress on the fault, we compute the dot products with unit vectors perpendicular $(hat(n))$ and parallel $(hat(f))$ to the fault

$t_N=t*hat(n)=(-\frac{50}{\sqrt[\phantom{2}]{2}},-\frac{70}{\sqrt[\phantom{2}]{2}})*(\frac{1}{\sqrt[\phantom{2}]{2}},\frac{1}{\sqrt[\phantom{2}]{2}})=-60$MPa

$t_S=t*hat(f)=(-\frac{50}{\sqrt[\phantom{2}]{2}},-\frac{70}{\sqrt[\phantom{2}]{2}})*(\frac{1}{\sqrt[\phantom{2}]{2}},-\frac{1}{\sqrt[\phantom{2}]{2}})=10$MPa

The fault normal compression is $60$MPa. The shear stress is $10$MPa.

${?}^1$ There is of course another fault$-$normal vector that points to the southwest. The choice of which vector to use as a reference for the fault$-$normal direction for this problem is arbitrary, as is our subsequent choice among two opposing vectors for the fault$-$parallel direction.