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

## Contents

Suggestions for hint and solution207:56, 24 March 2018

## Suggestions for hint and solution

Hi Li,

For MATH152/April 2016/Question B 01 (c), the hint appears to be a restatement of the question. Perhaps ask the student to determine whether the lines are parallel or not. Moreover, the solution appears to be too complicated. Perhaps we could just determine two things. (1) Are ${\displaystyle [-1,2,2]}$ and ${\displaystyle [-2,1,1]}$ parallel? (2) If so, pick ${\displaystyle [-1,2,2]}$ and ask if ${\displaystyle [0,2,1]-[-1,0,3]}$ is parallel to ${\displaystyle [-1,2,2]}$.

Best,

Brian

00:37, 13 March 2018
Edited by 0 users.
Last edit: 00:51, 13 March 2018

Hi Li,

Sorry, I assumed that these vectors were in two dimensions. So my suggestions about parallel lines won't work :-(

However, it still seems that the solution can be simplified like so. Determining if any point in in both lines is equivalent to finding solutions the the following system of linear equations

${\displaystyle {\begin{pmatrix}-1&2&|&-1\\2&-1&|&-2\\2&-1&|&2\end{pmatrix}}}$

This is obtained from

${\displaystyle [0,2,1]+s[-1,2,2]=[-1,0,3]+t[-2,1,1]}$

Best, Brian

00:51, 13 March 2018

I think the current solution is okay. But as you suggested, we can add another way to prove there's no solution to

${\displaystyle [0,2,1]+s[-1,2,2]=[-1,0,3]+t[-2,1,1]}$

using the matrix you mentioned above.

07:56, 24 March 2018