Finding Vector Components in Different Directions

[zooxanthellae]

Homework Statement

Find the component of the force \vec{F} = 2\hat{i} - 2\hat{j} + \hat{k} in:

a) the direction (\hat{i} + \hat{j} - \hat{k})/\sqrt{3}

b) the direction of the vector 3\hat{i} + 2\hat{j} - 6\hat{k}

Homework Equations

dir \vec{A} = \vec{A}/|\vec{A}|

The Attempt at a Solution

a) I broke \vec{F} down into components <2\hat{i}, 2\hat{j}, \hat{k}>. Then I figured that F goes 2\hat{i} in direction i, goes -2\hat{j} in direction j, then goes -\hat{k} in direction k, giving an overall Force component of <2\hat{i}, -2\hat{j}, -\hat{k}>.

b) I did the same thing and since the signs in each direction were the same as in a), I ended up with the same answer: <2\hat{i}, -2\hat{j}, -\hat{k}>.

It seems to make sense, but I am pretty uncertain, so I thought I would ask here.

Thanks!

 

[HallsofIvy]

What you are looking for is the projection of the force vector, \vec{u} on the vector giving the direction, \vec{v}. Imagine a line extended in the direction of the given vector to be projected on and a line segment representing the force vector, having the same beginning point. Dropping a perpendicular to the line from the end of the segment, you have a right triangle with hypotenuse the force vector. Since cos \theta is "near side over hypotenuse", the "near side", the length of the projection, is |\vec{u}|cos(\theta).

You also need to know that \vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta) so that \vec{u}|cos(\theta)= \vec{u}\cdot\vec{v}/|\vec{v}|. That is the length of the projection. To get it as a vector, multiply by the unit vector in the direction of \vec{v}, \vec{v}/|\vec{v}| which gives
\frac{\vec{u}\cdot\vec{v}}{|\vec{v}|^2}\vec{v}.

(For the vector in (a), |\vec{v}|= 1, for (b), it is 7.)

 

[zooxanthellae]

So we're imagining a right triangle in which the given \vec{F} is the adjacent side, and the hypotenuse is the line segment stretching from the origin to the point where a perpendicular from the end of \vec{F} crosses the line formed by an infinite extension of <\hat{i} + \hat{j} - \hat{k}>/\sqrt{3}.

Then you say that the length of the projection is |\vec{u}|cos\theta. What exactly is \vec{u}?