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

MATH200 December 2012

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• April 2012 • December 2011 •

[hide] 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.

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[show] 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.

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