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

We have

${\displaystyle P_{1}:\ \ \ {\mathbf {x}}={\begin{pmatrix}1\\1\\1\end{pmatrix}}+s_{1}{\begin{pmatrix}0\\1\\0\end{pmatrix}}+s_{2}{\begin{pmatrix}1\\2\\-1\end{pmatrix}}}$

and

${\displaystyle P_{2}:\ \ \ {\mathbf {x}}=t_{1}{\begin{pmatrix}-1\\0\\-1\end{pmatrix}}+t_{2}{\begin{pmatrix}0\\1\\2\end{pmatrix}}}$

for some ${\displaystyle s_{1}}$, ${\displaystyle s_{2}}$, ${\displaystyle t_{1}}$ and ${\displaystyle t_{2}}$ from the question and part (b).

It is easy to see that the line ${\displaystyle l_{2}}$ is just the intersection of plane ${\displaystyle P_{1},P_{2}}$, i. e., any point ${\displaystyle {\mathbf {x}}}$ on the line has to satisfy both equations. Thus we have following equation for line ${\displaystyle l_{2}}$ .

${\displaystyle {\mathbf {x}}={\begin{pmatrix}1\\1\\1\end{pmatrix}}+s_{1}{\begin{pmatrix}0\\1\\0\end{pmatrix}}+s_{2}{\begin{pmatrix}1\\2\\-1\end{pmatrix}}=t_{1}{\begin{pmatrix}-1\\0\\-1\end{pmatrix}}+t_{2}{\begin{pmatrix}0\\1\\2\end{pmatrix}}}$.

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

${\displaystyle l_{2}:{\mathbf {x}}=t_{1}{\begin{pmatrix}-1\\0\\-1\end{pmatrix}}+(1+t_{1}){\begin{pmatrix}0\\1\\2\end{pmatrix}}=t_{1}{\begin{pmatrix}-1\\1\\1\end{pmatrix}}+{\begin{pmatrix}0\\1\\2\end{pmatrix}},}$

which has is equivalent equation form; for any point ${\displaystyle {\mathbf {x}}=(x_{1},x_{2},x_{3})^{T}}$ on the line ${\displaystyle l_{2}}$ satisfies

{\displaystyle \color {blue}{\begin{aligned}{\begin{cases}x_{1}+x_{2}=1\\x_{1}+x_{3}=2\end{cases}}\end{aligned}}}