Please help

Can you find the length of this vector?
What can you do to make this vector get a length of 1 unit?

That new vector that you get is an answer to your question.

A vector parallel to that vector will have components proportional to those of the original vector. In other words, since the original vector is (3,0,7), some examples of parallel vectors would be (6,0,14), (300,0,700), and (0.3,0,0.7). All of those vectors are in the same direction as the original, but they each have different magnitudes. In the first example, all of the components are twice as big as in the original vector, so the magnitude is therefore also twice as big (you could verify this by finding the magnitude of each using the distance formula if you felt like it). Similarly, the magnitudes of the second and third examples are 100 and 1/10 times the original. However, none of those examples fit the criteria of a unit vector. A unit vector has a magnitude of 1.

So how do we find a parallel vector with a magnitude of 1? Well, first we find the magnitude of the original vector. Then we simply divide all of the components by that magnitude so that the new magnitude is 1.

Original magnitude:

|v|=sqrt{x^2+y^2+z^2}=sqrt(3^2+0^2+7^2)=\sqrt{58}

Now to find the new vector, we just divide everything by \sqrt{58}:

v' = \frac{v}{|v|}=\frac{(3,0,7)}{\sqrt{58}}=\(\frac{3 \sqrt{58}}{58},0,\frac{7 \sqrt{58}}{58}\)

If you find the magnitude of this new vector, you'll see that it is, indeed, 1.

Sorry Jerry. Once again, the Go skin failed to show me that you had already answered hours earlier. Didn't mean to hijack your teaching session.

It's okay, don't worry :)