Two Ways To Describe a Line in R3

[rtareen]

Hello all. I’m studying Calc 3 by myself and one of the topics I’m studying is how to determine a line in R3 given a point and direction vector.

I’m using 2 different textbooks, and they both have different ways of doing it. One uses vector parametrization and the other uses vector equivalency. I have included a photo of me working out an example both ways and getting the same answer.

I think I understand the first one pretty well. But I am puzzled on how a completely different method also gives the same answer. It’s mind boggling and I want to understand it more. Any and all insight will be greatly appreciated.

 

[Dr.Peterson]

Did you observe that r(t) in the first method is OP, and that P₀P = OP - OP₀?

That makes the two methods equivalent: the first starts effectively with OP - OP₀ = tV, which is the same as the second.

 

[pka]

Suppose that \(\displaystyle P: (p,q,r)\) is a point and direction vector \(\displaystyle \vec{v}: <d_1,d_2,d_3>\).
There are three forms of the line that contains \(\displaystyle P\) with direction vector \(\displaystyle \vec{v}\).

The vector form: \(\displaystyle P+t\cdot\vec{v}=<p,q,r>+t<d_1,d_2,d_3>=<p+td_1,q+td_2,r+td_3>\)

The parametric form: \(\displaystyle \ell (t) = \left\{ \begin{array}{l}x(t) = p + d_1t\\y(t) = q + d_2t\\z(t) = r + d_3t\end{array} \right.\)

The symmetric form: \(\displaystyle \frac{x-p}{d_1}=\frac{y-q}{d_2}=\frac{z-r}{d_3}\)

 

[rtareen]

Did you observe that r(t) in the first method is OP, and that P₀P = OP - OP₀?

That makes the two methods equivalent: the first starts effectively with OP - OP₀ = tV, which is the same as the second.

So in general a vector ab - ac = cb ?

Included is the work I’ve been able to do since you showed me that
P₀P = OP - OP₀



Is there an easy way to remember this? Or do you have to be a person of understanding?

 

[Dr.Peterson]

So in general a vector ab - ac = cb ?

Included is the work I’ve been able to do since you showed me that
P₀P = OP - OP₀

Is there an easy way to remember this? Or do you have to be a person of understanding?

Well, I'd say that understanding is a good thing to have in math ...

You should have learned how position vectors are related to the vector joining two points; yes, in general AB = OB - OA, and more generally, AC - AB = BC. You've demonstrated that several times in what you wrote, and you can see it by sketching the points and vectors. Also note that AB + BC = AC.

 

[pka]

So in general a vector ab - ac = cb ?
Included is the work I’ve been able to do since you showed me that
P₀P = OP - OP₀
Is there an easy way to remember this? Or do you have to be a person of understanding?

Would you tell us what textbooks you are using?
I ask because what you have posted is confused beyond all belief.
You are self studding. Right? If so I will suggest the text MULTIVARIABLE CALCULUS by Smith & Minton.
Buy a used edition in good condition. It is a slim volume which I consider the most useful for self-study.

 

[rtareen]

Would you tell us what textbooks you are using?
I ask because what you have posted is confused beyond all belief.
You are self studding. Right? If so I will suggest the text MULTIVARIABLE CALCULUS by Smith & Minton.
Buy a used edition in good condition. It is a slim volume which I consider the most useful for self-study.

I’m actually using Smith & Minton, as well as Rogawski to supplement it.

 

[LCKurtz]

You are self studding. Right?

Off to the corner with you!

 

[rtareen]

Would you tell us what textbooks you are using?
I ask because what you have posted is confused beyond all belief.
You are self studding. Right? If so I will suggest the text MULTIVARIABLE CALCULUS by Smith & Minton.
Buy a used edition in good condition. It is a slim volume which I consider the most useful for self-study.

What is exactly is confused, and for what reason is it confused ?

 

[firemath]

Off to the corner with you!

Yes, your spelling is awful.