Semseter One revision/questions thread

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the.watchman

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Re: Semseter One revision/questions thread

« Reply #15 on: June 15, 2010, 08:02:45 pm »

0

Quote from: cltf on June 15, 2010, 07:03:57 pm

because it is perpendicular $cos\theta =90^{\circ}=0$ so no matter the dot product, when multiplied by cos the result is 0 and hence perpendicular. however, how do you determine the value of $cos\theta$ because in this case we are only given one vector

As you said yourself, $\cos \theta = 0$
There are many answers to this question, as long as the magnitude is kept at one (unit vector)

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cltf

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Re: Semseter One revision/questions thread

« Reply #16 on: June 23, 2010, 06:39:49 pm »

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why is the reciprocal of $x^3-1$ the shape it is.

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cltf

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Re: Semseter One revision/questions thread

« Reply #17 on: July 13, 2010, 05:02:16 pm »

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If $a \sin^2 \theta + b \cos^2 \theta = c$ find $\sin \theta$ and $\cos \theta$ in terms of a,b,c

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