Vector and Force Components

1. Jun 11, 2010

zooxanthellae

1. The problem statement, all variables and given/known data

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 $$3\hat{i} + 2\hat{j} - 6\hat{k}$$

2. Relevant equations

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

3. 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!

Last edited: Jun 11, 2010
2. Jun 11, 2010

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.)

3. Jun 14, 2010

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 $$\vec{u}$$?