Finding vectors parallel to a given vector

AI Thread Summary
The discussion revolves around confusion regarding the representation of parallel vectors in a problem, specifically why the answers provided are unit vectors, ##(\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}})## and ##(-\frac{2}{\sqrt{13}}, -\frac{3}{\sqrt{13}})##, instead of the expected vectors ##(2,3)## and ##(-2,-3)##. It is clarified that the problem did not ask for unit vectors, leading to the conclusion that the published answers are incorrect. Participants also express confusion over the notation used, particularly the combination of an arrow and a caret, which is considered unusual. The conversation concludes with confirmation that the lecturer's solution was indeed the only one provided, with no missing information. The issue highlights the importance of clarity in mathematical notation and problem requirements.
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Homework Statement
Please see below
Relevant Equations
Coordinate form of vector
For (b) of this problem,
1681363361588.png

The solution is,
1681363414521.png

However, I am confused why the two parallel vectors are ##(\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}})## and ## (-\frac{2}{\sqrt{13}}, -\frac{3}{\sqrt{13}}) ## should it not be ##(2,3)## and ##(-2,-3)##. Do somebody please know why they wrote that?

Also I am very confused with this notation
1681363571599.png


Many thanks!
 
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You are correct. The length of ##\overrightarrow {PQ}## is ##\sqrt {13}##, so they gave a unit vector as the answer. Their notation appears to mean the unit vector in that direction. That is not what the problem asked for.
 
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ChiralSuperfields said:
However, I am confused why the two parallel vectors are ##(\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}})## and ## (-\frac{2}{\sqrt{13}}, -\frac{3}{\sqrt{13}}) ## should it not be ##(2,3)## and ##(-2,-3)##. Do somebody please know why they wrote that?
Did the published answer to the question give ##(\frac{2}{\sqrt{13}}## and ## \frac{3}{\sqrt{13}})## as the answers? If so, these answers are wrong as they did not ask for the vectors to be unit vectors.

The other thing you asked about, PQ with both an arrow above it and a caret (or hat, for a unit vector), is unusual notation, in my experience. Usually, one or the other is used, but not both.
 
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The only vectors which are parallel to a given vector and which are of the same length as that vector are that vector itself and its negative.
 
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I wonder if the answer was cut off and the unit vectors were multiplied in the final step that is not shown?
 
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Mark44 said:
Did the published answer to the question give ##(\frac{2}{\sqrt{13}}## and ## \frac{3}{\sqrt{13}})## as the answers? If so, these answers are wrong as they did not ask for the vectors to be unit vectors.

The other thing you asked about, PQ with both an arrow above it and a caret (or hat, for a unit vector), is unusual notation, in my experience. Usually, one or the other is used, but not both.
pasmith said:
The only vectors which are parallel to a given vector and which are of the same length as that vector are that vector itself and its negative.
FactChecker said:
I wonder if the answer was cut off and the unit vectors were multiplied in the final step that is not shown?
Thank you for your replies @Mark44, @pasmith and @FactChecker !

No @Mark44 that is was the only the solution published by the lecturer.

I checked the solution again @FactChecker, and nothing was cut-off, it was just the answer to part (c) of the question.

Many thanks!
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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