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

MATH200 December 2012

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  •  April 2012  •  December 2011  •  

Question 01 (a)
Let $\displaystyle L$ be the line of intersection of the planes $\displaystyle x+y+z=6$ and $\displaystyle x-y+2z=0$ (a) Find the points in which the line $\displaystyle L$ intersects the coordinate planes.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
First, find the parametric equation of the line $\displaystyle L$. Therefore, express the variables $\displaystyle x$ and $\displaystyle z$ in terms of $\displaystyle y$, using the two plane equations.

Hint 2
Once you have the parametric equation for $\displaystyle L$ with parameter $\displaystyle \lambda$, set the first entry of $\displaystyle L$ equal to $\displaystyle 0$. Solve for $\displaystyle \lambda$ and plug in your result as $\displaystyle \lambda$ in $\displaystyle L$. Then you find the point, where $\displaystyle L$ intersects the $\displaystyle yz$-plane.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $\displaystyle p_{xy}={\begin{pmatrix}12\\0\\-6\end{pmatrix}}+3{\begin{pmatrix}-3\\1\\2\end{pmatrix}}={\begin{pmatrix}3\\3\\0\end{pmatrix}}$

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