Finding parallel unit vector

• Nov 25th 2011, 11:21 AM
benny92000
Finding parallel unit vector
Problem: A unit vector parallel to vector V = (2, -3, 6) is ?

Answer is (-.29, .43, -.85)

Thanks. I thought to find a parallel vector you have to multiply through by the same number, so I chose the answer (-2, 3, -6)
• Nov 25th 2011, 11:38 AM
Plato
Re: Finding parallel unit vector
Quote:

Originally Posted by benny92000
Problem: A unit vector parallel to vector V = (2, -3, 6) is ? Answer is (-.29, .43, -.85)

Find the length, $\displaystyle \|V\|=\sqrt{2^2+(-3)^2+6^2}=7$.

Now divide $\displaystyle \left\langle {\frac{2}{7},\frac{{ - 3}}{7},\frac{6}{7}} \right\rangle$.

There is your parallel unit vector.
• Nov 25th 2011, 12:33 PM
benny92000
Re: Finding parallel unit vector
Ok, and can the length be -7, as well? As in divide through by -7 to find the right answer? That is what my book says, so I just wanted to verify this.
• Nov 25th 2011, 12:41 PM
Plato
Re: Finding parallel unit vector
Quote:

Originally Posted by benny92000
Ok, and can the length be -7, as well? As in divide through by -7 to find the right answer? That is what my book says, so I just wanted to verify this.

Length is never negative.
However, dividing by $\displaystyle -7$ also gives a unit vector parallel to the given vector.