Question about introductory vectors.

[icesalmon]

Homework Statement

Given ||v|| = 6 and u = < 0,3 >
find the vector v with the given magnitude and the same direction as u

Homework Equations

u = v / || v ||

The Attempt at a Solution

It seems simple enough, substitute the appropriate numbers into the equation and solve for v so < 0,3 > = v / 6 so v = 6 * < 0,3 > or v = < 0,18 >. The answer is that v = < 0,6 >. I'm really not sure what is going on.

 

[SteamKing]

Answer these questions: if u = <0, 3>, what is its direction?
If a vector of magnitude 6 is in the same direction as vector u, then what are the components of this new vector?

 

[icesalmon]

if u = < 0,3 > it's direction, magnitude, or size is 3
< 0,6 > are the components of a vector with magnitude 6. Oh, and because it's direction is the same all you have to change is it's magnitude, okay I understand.

That doesn't answer why I didn't get v = < 0,6 > using the above equation.

What you're saying makes sense, but I don't know if it's that easy with the following problems.

 

[Mark44]

if u = < 0,3 > it's direction, magnitude, or size is 3

Its magnitude or size is 3, but its direction is <0, 1>. To get this, multiply the vector by the reciprocal of the magnitude. The direction of a vector v is the unit vector (1/|v|) v.

< 0,6 > are the components of a vector with magnitude 6. Oh, and because it's direction is the same all you have to change is it's magnitude, okay I understand.

That doesn't answer why I didn't get v = < 0,6 > using the above equation.

What you're saying makes sense, but I don't know if it's that easy with the following problems.

 

[icesalmon]

Okay, so I need to re-read this section before I put a pencil to paper. I clearly don't have any sort of grasp for these things. Thanks for your help, i'll post back if I have any further questions.

 

[HallsofIvy]

To find a vector of magnitude "x" in the direction of vector $\vec{v}$, you have to multiply a unit vector in that direction by x. A unit vector in the same direction as v is given by
$$\frac{\vec{v}}{|\vec{v}|}$$

In this particular problem, $\vec{v}= <0, 3>$, which has length 3, so a unit vector in that direction is
$$\frac{<0, 3>}{3}= <0, 1>$$

A vector of length 6 in the direction of [itex]\vec{v}\itex] is<br /> $$6\frac{<0, 3>}{3}= 6<0, 1>= <0, 6>$$[/itex]

 

[icesalmon]

okay, I understand the problem now. A lot more than I did before, it's not difficult, and maybe I should have posted it in the pre-calculus section. But this is where we started off in my Calculus II course. Thanks HallsofIvy.