Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 3 (a)/Solution 2

Let $\mathbf{𝐱}=(x_1,x_2,x_3)$ denote the point of intersection. Since $\mathbf{𝐱}\in P_1$, we must have

$\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]+s_1\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]+s_2\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right]=\left[\begin{array}{c}1+s_2\\ 1+s_1+2s_2\\ 1-s_2\end{array}\right].$

In particular, $x_1+x_3=2$. On the other hand, since $\mathbf{𝐱}\in l_1$, we must also have

$\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=t\left[\begin{array}{c}0\\ 1\\ 2\end{array}\right]=\left[\begin{array}{c}0\\ t\\ 2t\end{array}\right].$

It follows that $0+2t=2$, so $t=1$. We conclude that

$\mathbf{𝐱}=\left[\begin{array}{c}0\\ 1\\ 2\end{array}\right].$