Vector equation problem

chengbin · #1 $Old$ July 6th, 2009, 08:53 AM

I don't get number 1. I don't get how they got the first and second steps, how they jumped from this to that.

Plato · #2 $Old$ July 6th, 2009, 09:43 AM

Quote:

Originally Posted by chengbin $View Post$

I don't get number 1. I don't get how they got the first and second steps, how they jumped from this to that.

The answer is quite simple: Ordinary vector addition.
$\overrightarrow {OA} + \overrightarrow {AP} = \overrightarrow {OP}$

chengbin · #3 $Old$ July 6th, 2009, 10:28 AM

Do you mind spelling it out for me? I really suck at geometry.

HallsofIvy · #4 $Old$ July 6th, 2009, 10:34 AM

Plato answered how they got (1)- simply vector addition. You go from O to P by going first from O to A and then from A to P. $\vec{p}$ is the vector from O to P, $\vec{OA}$ , and $\vec{AP}$ is the vector from A to P. "Going from O to A and then from A to P" is, by the definition of vector addition is $\vec{p}= \vec{OA}+ \vec{AP}$ .

As for (2), if two vectors point in the same direction (are parallel), then one is a multiple of the other: $\vec{AP}= t\vec{b}$ where t is simply that "multiple".

Finally, they get (3) just by replacing $\vec{AP}$ with $t\vec{b}$ which you can do because, by (2), they are the same.

chengbin · #5 $Old$ July 6th, 2009, 11:27 AM

We have a misunderstanding.

I didn't mean the example on the first half of the page. I meant the question, number 1, on the second half of the page. I get the example. I don't get question 1.

HallsofIvy · #6 $Old$ July 6th, 2009, 04:23 PM

So that is: given two point M and P, find the equation of the line through M and P. And it is given in a rather general format.

Again, as in the example, you can write the equation as $\vec{OP}= \vec{OM}+ t\vec{MN}$ where "O" is some fixed point- typically the origin of a coordinate system but not necessarily. Here they are writing $\vec{OP}$ , the vector from O to P, as $\vec{p}$ , $\vec{ON}$ , the vector from O to N, as $\vec{n}$ , and [math]\vec{OM}, the vector from O to M, as $\vec{m}$ .
The vector form N to M is $\vec{n}- \vec{m}$

Using that notation, $\vec{OP}= \vec{OM}+ t\vec{MN}$ becomes $\vec{p}= \vec{m}+ t(\vec{n}- \vec{n})$ . Notice that if t= 0, this gives $\vec{p}= \vec{m}$ , the "position vector" for the point M, and if t= 1 it gives $\vec{p}= \vec{m}+ (\vec{n}- \vec{m})= \vec{n}$ , the "position vector" of the point N. This line goes through M and N and two points determine a line.

Once you have that, $\vec{p}= \vec{m}+ t\vec{n}- t\vec{m}= (1-t)\vec{m}+ t\vec{n}$ as given.

For a more specific problem, suppose, in some coordinate system, M= (a, b, c) and N= (d, e, f). Then, writing P as (x, y, z), and taking O= (0, 0, 0), $\vec{p}= x\vec{i}+ y\vec{j}+ z\vec{k}$ , $\vec{m}= a\vec{i}+ b\vec{j}+ c\vec{k}$ , and $\vec{n}= d\vec{i}+ e\vec{j}+ f\vec{k}$ . The formula above becomes $\vec{p}= x\vec{i}+ y\vec{j}+ z\vec{k}$ $= (1-t)\vec{m}+ t\vec{n}= (1-t)(a\vec{i}+ b\vec{j}+ c\vec{k})+ t(d\vec{i}+ e\vec{j}+ f\vec{k})$ .

Again, notice that when t= 0 that gives $(1-0)(a\vec{i}+ b\vec{j}+ c\vec{k})+ 0(d\vec{i}+ e\vec{j}+ f\vec{k})$ $= a\vec{i}+ b\vec{j}+ c\vec{k}$ , giving the point M= (a, b, c) and if t= 1, $(1-1)(a\vec{i}+ b\vec{j}+ c\vec{k})+ 1(d\vec{i}+ e\vec{j}+ f\vec{k})$ $d\vec{i}+ e\vec{j}+ f\vec{k}$ giving the point N= (d, e, f).

chengbin · #7 $Old$ July 6th, 2009, 06:12 PM

OK, I get it. There is still one thing I don't get. How do you get the starting equation $\overrightarrow {MP}=t\overrightarrow {MN}$