Science:Math Exam Resources/Courses/MATH152/April 2016/Question B 01 (c)

MATH152 April 2016

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Question B 01 (c)
Consider the lines ${\begin{aligned}L_{1}:&&[0,2,1]+s[-1,2,2]\\L_{2}:&&[-1,0,3]+t[-2,1,1]\end{aligned}}$ (c) Do the lines $L_{1}$ and $L_{2}$ intersect? If so, find the intersection point. If not, explain.

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
A point $[x,y,z]$ lies on $L_{1}$ if and only if $[x,y,z]=[0,2,1]+s_{0}[-1,2,2]$ for some $s_{0}$. Similarly, $[x,y,z]\in L_{2}$ iff $[x,y,z]=[-1,0,3]+t_{0}[-2,1,1]$ for some $t_{0}$.

Hint 2
Translate this problem into a system of linear equations problem.

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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. If a point $[x,y,z]$ is a intersection point of the lines $L_{1}$ and $L_{2}$, then $[x,y,z]\in L_{1}$ and $[x,y,z]\in L_{2}$. In other words, $[x,y,z]=[0,2,1]+s_{0}[-1,2,2]=[-1,0,3]+t_{0}[-2,1,1]$ for some $s_{0}$ and $t_{0}$. By expanding the second equality, we have $[-s_{0},2s_{0}+2,2s_{0}+1]=[-2t_{0}-1,t_{0},t_{0}+3].$ The first component implies that $s_{0}=2t_{0}+1$ and we plug this into the equation from the second component; $t_{0}=2s_{0}+2=2(2t_{0}+1)+2=4t_{0}+4$ So, we have $t_{0}=-{\frac {4}{3}}$ and $s_{0}=2\left(-{\frac {4}{3}}\right)+1=-{\frac {5}{3}}$. However, these values of $t_{0}$ and $s_{0}$ doesn't satisfy the equation from the third component; $-{\frac {7}{3}}=2\left(-{\frac {5}{3}}\right)+1=2s_{0}+1\neq t_{0}+3=-{\frac {4}{3}}+3={\frac {5}{3}}.$ This means there's no $t_{0}$ and $s_{0}$ to make $[0,2,1]+s_{0}[-1,2,2]=[-1,0,3]+t_{0}[-2,1,1]$ hold. In other words, there's no intersection point.

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. Translate this problem into a problem involving a system of linear equations. To do this, note that $L_{1}$ and $L_{2}$ intersect exactly when there are numbers $s$ and $t$ that satisfy $[0,2,1]+s[-1,2,2]=[-1,0,3]+t[-2,1,1]$ rearranging this gives $s[-1,2,2]-t[-2,1,1]=[-1,0,3]-[0,2,1]$ and $s[-1,2,2]+t[2,-1,-1]=[-1-0,0-2,3-1]=[-1,-2,2]$ Writing as a system of linear equations gives, $[-s+2t,2s-t,2s-t]=[-1,-2,2]$ In matrix form, ${\begin{pmatrix}-1&2&|&-1\\2&-1&|&-2\\2&-1&|&2\end{pmatrix}}$ In other words, $L_{1}$ and $L_{2}$ intersect exactly when the above system of linear equations has at least one solution. Using Gaussian Elimination, we obtain ${\begin{pmatrix}-1&2&|&-1\\2&-1&|&-2\\2&-1&|&2\end{pmatrix}}\to {\begin{pmatrix}-2&4&|&-2\\2&-1&|&-2\\2&-1&|&2\end{pmatrix}}\to {\begin{pmatrix}-2&4&|&-2\\0&3&|&-4\\0&3&|&0\end{pmatrix}}\to {\begin{pmatrix}-2&4&|&-2\\0&0&|&-4\\0&3&|&0\end{pmatrix}}$ The second row of the last matrix reads $0s+0t=-4$ which is impossible. Hence, the system of linear equations has no solution and the lines $L_{1}$ and $L_{2}$ do not intersect.