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

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

$\displaystyle {\begin{aligned}x&=y-2z\\\end{aligned}}$

and plug the result into the first equation:

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

Hence

$\displaystyle {\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}y-2z\\y\\2y-6\end{pmatrix}}={\begin{pmatrix}y-2(2y-6)\\y\\2y-6\end{pmatrix}}={\begin{pmatrix}12-3y\\y\\2y-6\end{pmatrix}}={\begin{pmatrix}12\\0\\-6\end{pmatrix}}+y{\begin{pmatrix}-3\\1\\2\end{pmatrix}}$

Then the line is

$\displaystyle L={\begin{pmatrix}12\\0\\-6\end{pmatrix}}+\lambda {\begin{pmatrix}-3\\1\\2\end{pmatrix}}$

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

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

$\displaystyle p_{yz}={\begin{pmatrix}12\\0\\-6\end{pmatrix}}+4{\begin{pmatrix}-3\\1\\2\end{pmatrix}}={\begin{pmatrix}0\\4\\2\end{pmatrix}}$

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

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

$\displaystyle p_{xz}={\begin{pmatrix}12\\0\\-6\end{pmatrix}}+0{\begin{pmatrix}-3\\1\\2\end{pmatrix}}={\begin{pmatrix}12\\0\\-6\end{pmatrix}}$

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

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

$\displaystyle p_{xy}={\begin{pmatrix}12\\0\\-6\end{pmatrix}}+3{\begin{pmatrix}-3\\1\\2\end{pmatrix}}={\begin{pmatrix}3\\3\\0\end{pmatrix}}$