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This is a calculus 3 problem and I need to find a vector in the third dimension but I need to do this without using the dot or cross product. So I am asked to find:

A unit vector u parallel to the vector with initial point (6,6,7) and terminal point (-3,3,10)

So to do this I think I need to combine the initial point and terminal point:

So that means v=<-3-6,3-6,10-7) which means that v=<-9,-3,3>

But after that, I do not know how to find this answer. I think I need to find the magnitude which I get to equal √(99). But this is the last thing I have any clue how to do.

Thanks you for the help

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We must first translate the given vector to be origin centered. This can be done by subtracting the initial point from the terminal point.

So, our new vector is $<-9,-3,3>$ as you've noted. Next, we have to compute this vector's magnitude, since we want the magnitude of the unit vector to be 1. The magnitude, as you correctly note, is $\sqrt{99}$. So, we divide each coefficient of the vector to get a vector whose magnitude is 1 and points in the correct direction.

So the vector you desire is $$\bigg<\frac{-3}{\sqrt{11}},\frac{-1}{\sqrt{11}},\frac1{\sqrt{11}}\bigg>$$

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  • $\begingroup$ Thank you so much $\endgroup$ Sep 4 '18 at 21:01
  • $\begingroup$ I have to wait 3 more minutes before it will let me accept this answer. As soon as I can I will hit the green check $\endgroup$ Sep 4 '18 at 21:03

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