Thread: Finding a Vector

Find a vector of length 3 which has:
a) the same direction as 5i + 2j
b) the opposite direction to 4i + 5j

$\displaystyle a) |5-2j| = (5^2 + (-2)^2)^.5 = (29)^.5$
then
$\displaystyle 3[(1/(29)^.5)(5i + 2j)] = 15/(29)^.5i - 6/(29)^.5j$
the $\displaystyle ^.5$ is square root...but I don't know how to show that on here. is my math correct?

and is b) done the same way except I would multiply the whole thing by -1?

Thanks

$\displaystyle \left\| {5i - 2j} \right\| = \sqrt {29} $ to need to learn this.
So the vector you want is $\displaystyle \frac{{15}}{{\sqrt {29} }}i - \frac{6}{{\sqrt {29} }}j$

Originally Posted by Plato

$\displaystyle \left\| {5i - 2j} \right\| = \sqrt {29} $ to need to learn this.
So the vector you want is $\displaystyle \frac{{15}}{{\sqrt {29} }}i - \frac{6}{{\sqrt {29} }}j$

ah, okay, so my math was correct.
I'll go ahead and assume the same format is used for b), but it's negative.

Thanks