Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 3 (c)/Solution 1

We have

$P_1:\mathbf{𝐱}=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)+s_1\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)+s_2\left(\begin{array}{c}1\\ 2\\ -1\end{array}\right)$

and

$P_2:\mathbf{𝐱}=t_1\left(\begin{array}{c}-1\\ 0\\ -1\end{array}\right)+t_2\left(\begin{array}{c}0\\ 1\\ 2\end{array}\right)$

for some $s_1$, $s_2$, $t_1$ and $t_2$ from the question and part (b).

It is easy to see that the line $l_2$ is just the intersection of plane $P_1,P_2$, i. e., any point $\mathbf{𝐱}$ on the line has to satisfy both equations. Thus we have following equation for line $l_2$ .

$\mathbf{𝐱}=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)+s_1\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)+s_2\left(\begin{array}{c}1\\ 2\\ -1\end{array}\right)=t_1\left(\begin{array}{c}-1\\ 0\\ -1\end{array}\right)+t_2\left(\begin{array}{c}0\\ 1\\ 2\end{array}\right)$.

Now let's find out the relation among parameters $s_1,s_2,t_1,t_2$ . The first coordinate shows $1+s_2=-t_1$, while the third shows $1-s_2=-t_1+2t_2$. Summing these equations gives $2=-2t_1+2t_2$, hence $t_2=1+t_1$. Substituting it back gives: for any point $\mathbf{𝐱}$ on line it satisfies

$l_2:\mathbf{𝐱}=t_1\left(\begin{array}{c}-1\\ 0\\ -1\end{array}\right)+(1+t_1)\left(\begin{array}{c}0\\ 1\\ 2\end{array}\right)=t_1\left(\begin{array}{c}-1\\ 1\\ 1\end{array}\right)+\left(\begin{array}{c}0\\ 1\\ 2\end{array}\right),$

which has is equivalent equation form; for any point $\mathbf{𝐱}=(x_1,x_2,x_3{)}^T$ on the line $l_2$ satisfies

$\begin{array}{r}\{\begin{array}{l}{x}_1+x_2=1\\ x_1+x_3=2\end{array}\end{array}$