Science:Math Exam Resources/Courses/MATH200/December 2012/Question 01 (a)/Solution 1

The line $L$ satisfies both plane equations. We solve the second one for $x$:

$\begin{array}{rl}x& =y-2z\\ \end{array}$

and plug the result into the first equation:

$\begin{array}{rl}(y-2z)+y+z& =6\\ z& =2y-6\end{array}$

Hence

$\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}y-2z\\ y\\ 2y-6\end{array}\right)=\left(\begin{array}{c}y-2(2y-6)\\ y\\ 2y-6\end{array}\right)=\left(\begin{array}{c}12-3y\\ y\\ 2y-6\end{array}\right)=\left(\begin{array}{c}12\\ 0\\ -6\end{array}\right)+y\left(\begin{array}{c}-3\\ 1\\ 2\end{array}\right)$

Then the line is

$L=\left(\begin{array}{c}12\\ 0\\ -6\end{array}\right)+\lambda \left(\begin{array}{c}-3\\ 1\\ 2\end{array}\right)$

In the $yz$-plane, $x=0$. Hence, for the first entry of the line $L$ holds when $12-3\lambda =0\Rightarrow \lambda =4$

So, $L$ intersects with the $yz$-plane in the point

$p_{yz}=\left(\begin{array}{c}12\\ 0\\ -6\end{array}\right)+4\left(\begin{array}{c}-3\\ 1\\ 2\end{array}\right)=\left(\begin{array}{c}0\\ 4\\ 2\end{array}\right)$

In the $xz$-plane, $y=0$. Hence, for the second entry of the line $L$ holds when $\lambda =0$

So, $L$ intersects with the $xz$-plane in the point

$p_{xz}=\left(\begin{array}{c}12\\ 0\\ -6\end{array}\right)+0\left(\begin{array}{c}-3\\ 1\\ 2\end{array}\right)=\left(\begin{array}{c}12\\ 0\\ -6\end{array}\right)$

In the $xy$-plane, $z=0$. Hence, for the third entry of the line $L$ holds when $-6+2\lambda =0\Rightarrow \lambda =3$

So, $L$ intersects with the $xy$-plane in the point

$p_{xy}=\left(\begin{array}{c}12\\ 0\\ -6\end{array}\right)+3\left(\begin{array}{c}-3\\ 1\\ 2\end{array}\right)=\left(\begin{array}{c}3\\ 3\\ 0\end{array}\right)$