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

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  •  

   Other MATH200 Exams

  •  April 2012  •  December 2011  •  

Question 01 (a)
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 a vector parallel to the line L.

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
For any vector $\mathbf {u}$, The line $\mathbf {u} +t\mathbf {v}$, $t\in \mathbb {R}$ is parallel to the vector $\mathbf {v}$.

Hint 2
If a vector, v, is parallel to a plane P, then v is perpendicular to the normal vector of P. If two vectors are perpendicular, what is the dot product between them equal to?

Hint 3
(Alternative solution) The vector that we're looking for is perpendicular to two vectors. Which? What is a quick way to calculate such a vector, given the other two?

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let the line L be written in vector form as $\displaystyle [a_{0},b_{0},c_{0}]+[a_{1},b_{1},c_{1}]t.$ If L is parallel to the plane 2x + y - z = 5, then L must be perpendicular to the normal vector of said plane, n = [2,1,-1]. i.e. The dot product between n and the direction of L must be equal to zero: $[2,1,-1]\cdot [a_{1},b_{1},c_{1}]=0.$ Evaluating the dot product gives: $\mathbf {(1)} \quad [2,1,-1]\cdot [a_{1},b_{1},c_{1}]=2a_{1}+b_{1}-c_{1}=0$ Remember that we also need the vector is perpendicular to the line,$[3-t,1-2t,3t]=[3,1,0]+[-1,-2,3]t.$ Hence, the following equation must also be satisfied: $\displaystyle \mathbf {(2)} \quad [-1,-2,3]\cdot [a_{1},b_{1},c_{1}]=-a_{1}-2b_{1}+3c_{1}=0$ So we have two equations and three unknowns. This means there are infinitely many vectors that will satisfy the given conditions. We only need to find one. (Note: Clearly $a_{1}=b_{1}=c_{1}=0$ is a solution to the above equations (1), (2), but it is the trivial solution and is perpendicular to every vector, including the direction of the line L.) Solving (1) for $c_{1}$ gives $c_{1}=2a_{1}+b_{1}$. Subbing this into (2) gives $\displaystyle -a_{1}-2b_{1}+3(2a_{1}+b_{1})=0\quad \rightarrow \quad b_{1}=-5a_{1}.$ From this we get that $c_{1}=2a_{1}-5a_{1}=-3a_{1}$. Thus, any vector of the form k[1,-5,-3] where k is a constant (not equal to zero) will be parallel to L. For example, the vectors [1,-5,-3] and [-2,10,6] are acceptable solutions.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. This quicker alternative solution relies on the cross product: Given two vectors $v_{1},v_{2}$, their cross product $v_{3}=v_{1}\times v_{2}$ is perpendicular to both, $v_{1}$ and $v_{2}$. We are asked to find a vector that is parallel to a plane, and perpendicular to a line. By the definition of the normal vector of a plane, all vector parallel to a plane are perpendicular to the normal vector. Further, vectors are perpendicular to a line if they are perpendicular to the direction vector of that line. Hence, choose $v_{1}$ as the normal vector of the plane, $v_{1}=[2,1,-1]$, and choose $v_{2}$ as the direction vector of the line, $v_{2}=[-1,-2,3]$. Then $\displaystyle v_{3}=v_{1}\times v_{2}=[2,1,-1]\times [-1,-2,3]=[1,-5,-3]$ is parallel to the plane, and perpendicular to the line.

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