# Math Help - What does llvll mean?

1. ## What does llvll mean?

I have to figure out a question for my linear algebra course and ever since the beginning of the course, we've been using llvll or llull (v and u being vectors with a line over top of them) and I've been wrongly (I now know) assuming these are weird-looking absolute value signs. I've looked in my textbook but can't find what it means. (I've also tried Google with no luck). Could somebody please explain? Thank you so much!

2. Originally Posted by sberxa
I have to figure out a question for my linear algebra course and ever since the beginning of the course, we've been using llvll or llull (v and u being vectors with a line over top of them) and I've been wrongly (I now know) assuming these are weird-looking absolute value signs. I've looked in my textbook but can't find what it means. (I've also tried Google with no luck). Could somebody please explain? Thank you so much!
There are two words that can summarize what the double bars represent:

1. Magnitude
2. Norm

Its just asking you to find the length of the vector; and it can be done by applying this definition:

$||\vec{v}||=\sqrt{v_x^2+v_y^2}$ ----- [Two Dimensions]
$||\vec{v}||=\sqrt{v_x^2+v_y^2+v_z^2}$ ----- [Three Dimensions]
$||\vec{v}||=\sqrt{v_1^2+v_2^2+\dots+v_n^2}$ ----- [ $n$ Dimensions]

You can prove the two and three dimensional space by applying the Pythagorean Theorem. This then, can be generalized for a vector that has $n$ components.

Does this help?

--Chris

3. so for finding the magnitude of a 3D vector, let's say (x,y,z), do I just plug it into . or i guess i need two point, or to take one as the origin?
thanks!

4. Originally Posted by sberxa
so for finding the magnitude of a 3D vector, let's say (x,y,z), do I just plug it into . or i guess i need two point, or to take one as the origin?
thanks!
If the vector is given as a point, then we assume that the other point is the origin. So you would just plug in the x, y, and z components into the equation for finding the norm.

So $\vec{v}=\left$

Thus, $||\vec{v}||=\sqrt{x^2+y^2+z^2}$

Does this make sense?

--Chris