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


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 $$\bigg<\frac{-3}{\sqrt{11}},\frac{-1}{\sqrt{11}},\frac1{\sqrt{11}}\bigg>$$

  • $\begingroup$ 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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