Science:Math Exam Resources/Courses/MATH152/April 2011/Question A 16

MATH152 April 2011

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Question A 16
Find a vector of length $1$ that is parallel to the line given by the equations $\left\{{\begin{aligned}y-z&=1\\2x+y+z&=2\end{aligned}}\right.$

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
The line is an intersection of two planes and so by definition it is on both planes. What is the relationship to any point on a plane and its normal?

Hint 2
As an alternative method, how can we re-write this problem as a matrix problem?

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. The first plane $\displaystyle {}y-z=1$ has normal n$_{1}$=[0,1,-1] (just the coefficients on x,y,z). The second plane $\displaystyle {}2x+y+z=2$ has normal n$_{2}$=[2,1,1]. Let's call the direction our line has v. Notice that since v is an intersection of the two planes then by definition it is on plane 1 and plane 2. Therefore $\mathbf {n} _{1}\times \mathbf {v} =\mathbf {0} \quad {\text{and}}\quad \mathbf {n} _{2}\times \mathbf {v} =\mathbf {0} .$ Therefore we seek v so that it is orthogonal to both normal vectors. The cross product, by definition, provides this direction. Therefore, $\mathbf {v} =\mathbf {n} _{1}\times \mathbf {n} _{2}=[0,1,-1]\times [2,1,1]=[2,-2,-2].$ We want this to have length 1 and so we see that $|\mathbf {v} |={\sqrt {12}}=2{\sqrt {3}}.$ Therefore we get that the unit norm vector, $\mathbf {\hat {v}}$ is $\mathbf {\hat {v}} ={\frac {1}{\sqrt {3}}}[1,-1,-1].$

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. Since a vector on the line has to be on each plane, it has to be a solution of a linear system. We have the following equations $\left\{{\begin{aligned}y-z&=1\\2x+y+z&=2\end{aligned}}\right.$ which is a linear system with 2 equations in 3 unknowns. We can write it in an augmented matrix as $\left[{\begin{array}{ccc|c}0&1&-1&1\\2&1&1&2\end{array}}\right]$ We will swap the first two rows and then row reduce to get $\left[{\begin{array}{ccc|c}1&0&1&1/2\\0&1&-1&1\end{array}}\right].$ We notice there is a free variable and we let x = t. We then get ${\begin{aligned}z&={\frac {1}{2}}-t\\y&=1+z={\frac {3}{2}}-t\end{aligned}}.$ Therefore we get that the solution, s, is $\mathbf {s} =\mathbf {s} _{0}+t\mathbf {v} =\left[0,{\frac {3}{2}},{\frac {1}{2}}\right]+t[1,-1,-1].$ The direction part of the line, v is [1,-1,-1]. However to make it length 1 we must divide by its magnitude, $|\mathbf {v} |={\sqrt {3}}$ to get, $\mathbf {\hat {v}} ={\frac {1}{\sqrt {3}}}[1,-1,-1]$ just like in solution 1.