Science:Math Exam Resources/Courses/MATH200/April 2012/Question 01 (b)

MATH200 April 2012

  •  Q1 (a)  •  Q1 (b)  •  Q2 (a)  •  Q2 (b)  •  Q3  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q5 (a)  •  Q5 (b)  •  Q6 (a)  •  Q6 (b)  •  Q7  •  Q8  •  Q9  •  Q10  •  

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  •  April 2012  •  December 2011  •  

Question 01 (b)
Let L be the line which is parallel to the plane $\displaystyle {}2x+y-z=5$ and perpendicular to the line $x=3-t\qquad {}y=1-2t\qquad {}z=3t.$ Find parametric equations for the line L if L passes through a point Q(a,b,c) where a<0, b>0, c>0 and the distances from Q to the xy-plane, xz-plane, and the yz-plane are 2, 3, and 4 respectively.

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
From part (a) we know that the direction vector of the line $\displaystyle L$ is $\displaystyle {\begin{pmatrix}1\\-5\\-3\end{pmatrix}}$ $\displaystyle L=Q+t{\begin{pmatrix}1\\-5\\-3\end{pmatrix}}={\begin{pmatrix}a\\b\\c\end{pmatrix}}+t{\begin{pmatrix}1\\-5\\-3\end{pmatrix}}$

Hint 2

Recall that the distance of a point $\displaystyle P=(p_{1},p_{2},p_{3})$ to the $\displaystyle xy$-plane is the absolute value of the $\displaystyle z$-coordinate $\displaystyle |p_{3}|$, to the $\displaystyle xz$-plane is the absolute value of the $\displaystyle y$-coordinate $\displaystyle |p_{2}|$, to the $\displaystyle yz$-plane is the absolute value of the $\displaystyle x$-coordinate $\displaystyle |p_{1}|$.

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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. To find the line to the line $\displaystyle L$, we must find $\displaystyle a,b,c$ such that the point $\displaystyle Q={\begin{pmatrix}a\\b\\c\end{pmatrix}}$ has distance $\displaystyle 2$ to the $\displaystyle xy$-plane. This means $\displaystyle |c|=2$. Since $\displaystyle c>0$, we have $\displaystyle c=2$. The distance from $\displaystyle Q$ to the $\displaystyle xz$-plane is $\displaystyle 3$, hence $\displaystyle |b|=3$. With $\displaystyle b>0$, we have $\displaystyle b=3$. The distance from $\displaystyle Q$ to the $\displaystyle yz$-plane is $\displaystyle 4$, hence $\displaystyle |a|=4$. With $\displaystyle a<0$, we have $\displaystyle a=-4$. Hence $\displaystyle Q={\begin{pmatrix}-4\\3\\2\end{pmatrix}}$ and $\displaystyle L={\begin{pmatrix}-4\\3\\2\end{pmatrix}}+t{\begin{pmatrix}1\\-5\\-3\end{pmatrix}}$.

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