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Find $t$ if the given vectors are parallel:

$$OR = (t , t + 2)$$

$$OS = (3 , -4)$$

I easily understand the question if the vectors are perpendicular, because the dot product of perpendicular vectors is $0$, which means you can equate the vectors to $0$ and work it out that way, but I don't understand how to answer this question with parallel vectors! And it's very frustrating!

Is there a similar rule with parallel vectors that I am unaware of?

Also, my apologies for the rubbish formatting. I haven't got the hang of mathjax yet.

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  • $\begingroup$ Math formatting has been taken care of. $\endgroup$ – mlc May 1 '17 at 22:31
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Two vectors are parallel if they have the same slope. The slope of $(x,y)$ is $\frac{y}{x}$; so it suffices to equate $$\frac{t+2}{t} = \frac{-4}{3}$$ and find $t=-\frac{6}{7}$.

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  • $\begingroup$ Thank-you so much. That makes a lot of sense. Would a similar method work with vectors in three or four dimensions? I've never worked out the gradient of a three-dimensional linear graph. $\endgroup$ – ben8615 May 1 '17 at 22:49
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The dot product of two parallel vectors is equal to the product of their magnitudes. Thus you have $3t -4(t+2) = 5\sqrt{t^2 + (t+2)^2}$. The second solution comes from the fact that the dot product of two antiparallel vectors is also equal to the product of their magnitudes.

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  • $\begingroup$ I'm not sure what you did, because that doesn't have solutions, apparently. $\endgroup$ – John Lou May 1 '17 at 22:39
  • $\begingroup$ $v\cdot w = \|v\|\|w\|\cos \theta$ $\endgroup$ – JMJ May 1 '17 at 22:42
  • $\begingroup$ perhaps some calculation error... $\endgroup$ – JMJ May 1 '17 at 22:44
  • $\begingroup$ No, it needs to be the negative square root. See graph $\endgroup$ – John Lou May 1 '17 at 22:47
  • $\begingroup$ that's fair enough. $\endgroup$ – JMJ May 1 '17 at 22:53

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