anonymous
  • anonymous
One point on the surface of a sphere centred at the origin has the coordinates (8, 3, 5). a. Represent the radius that connects this point to the origin as a position vector. b. Determine the angle that this radius makes with each of the coordinate axes. c. Suggest the position vector of a radius of this sphere that lies below the xy-plane.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Loser66
  • Loser66
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Loser66
  • Loser66
I think to define the angle between this vector with the axis, we should consider pair of coordinate. I mean, to z axix, we consider (8,3,5) with (0,0,1) by formula of cos theta = dot product / the magnitude of the 2
Loser66
  • Loser66
@FutureMathProfessor hehe, rescue me, friend,

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More answers

anonymous
  • anonymous
A: This one should be easy, since it's coming from the origin, the vector is just <8,3,5>
anonymous
  • anonymous
b & c?
anonymous
  • anonymous
B: This one might be a little bit more challenging. Remember the angle of the product between two vectors theta is described by the following equation: \[\theta = \cos^{-1} \frac{ v . w }{ |v||w| }\]First we will need to find the actual radius of the vector, we can do this by finding the magnitude \[|v|\] from the origin to our point in three-space. \[\sqrt{8^2 + 3^2 + 5^2 }=\sqrt{64+9+25}=\sqrt{100}=10=R\] Since 10 is our radius, this will make it easier. We know that the following axes will have the following coordinates for the radius of the circle: \[Xaxis=<10,0,0>\]\[Yaxis=<0,10,0>\]\[Zaxis=<0,0,10>\] Now if you find the dot products and divide them by the product of the magnitude using the formula above, you'll be able to find your three desired angles. Think you can do that on your own? :)
anonymous
  • anonymous
uhh
anonymous
  • anonymous
C. This one is very easy. Since it has to be below the XY-plane, that means Z<0. So just change <8,3,5> to <8,3,-5> :)

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