# Math Help - Equaitons of Lines and planes

1. ## Equaitons of Lines and planes

Hi, just wanted to double check this question off with you people.

Find a vector equation through the point (0, 14, -10) and parallel to the line x= -1 + 2t, y= 6-3t, z=3+9t

r = r0 + tv
r0 = 0i + 14j - 10k

the vector we need can be derived from the line x= -1 + 2t, y= 6-3t, z=3+9t
it is <2, -3, 9> [i was unsure at this part, am I correct here?]

so r0 + tv = (2t)i + (14-3t)j + (9t-10)k

hence vector equation = (2t)i + (14-3t)j + (9t-10)k

parametric equation = x =2t, y=14-3t, z=9t-10

another question i wanted to ask is why is it helpful to know how to find the parametric equation? How does it make life more convenient, or how is it useful?

also anyone have resources on where I can learn more with lines and planes. I feel my conceptual skills could use some tightening up.

furthermore, do any programs exist that would allow me to plot planes and lines in 3-d spaces? I tried wolfram but either it can't do it (tried plotting x=1, y=0, z=0) or i just don't know how to use it.

Thanks!

2. ## Re: Equaitons of Lines and planes

Originally Posted by 99.95
Find a vector equation through the point (0, 14, -10) and parallel to the line x= -1 + 2t, y= 6-3t, z=3+9t

Pick any vector that is perpendicular to $<2,-3,9>.$

$N=<-3,1,1>$ will do. Now write the equation of the plane with normal $N$ containing $(0,14,-10)$.

3. ## Re: Equaitons of Lines and planes

The line can be written
r = (-1,6,3) + (2,-3,9)t
which is the equation of the line through the point (-1,6,3) in the direction (2,-3,9), by vector addition.

The equation of the line in the same direction through the point (0,14,10) is:
r = (0,14,10) + (2,-3,9)t

t enters naturally as the parameter time, for example, r = r(t) is the position of a particle in space as a function of time with velocity v = dr/dt.

Any calculus text with a section on vectors discusses lines and planes. The geometry without vectors gets bogged down in a lot of x’s and y’s.

4. ## Re: Equaitons of Lines and planes

thanks for the help.

5. ## Re: Equaitons of Lines and planes

Another question:

Find parametric equations and symmetric equations for the line passing through the origin and the point (1,2,3).

So, the vector equation of a line can be expressed in the form r = r0 + vt where v is a vector parallel to L and t is a scalar.

I thought I'd use the position vector OP (1,2,3) and then find a vector parallel to this line by crossing it with an unknown vector a,b,c and letting it equal zero, then solving.

so i did:
(1,2,3) x (a,b,c) = 0

ended up with 3b - 2c - 3a + c + 2a - b = 0
-> 2b-c-a=0
let b=1, c = 2 and a =0

we have v= <0,1,2>

r = r0 + v
t
= (1i + 2j + 3k) + <0i + j + 2k>t
= 1t + (2+t)j + (3+2t)k

the parametric equations are therefore:
x = t, y = 2+t, z = 3+2t

and symmetric equations:
x = y-2/b = z-3/2c
is this correct?

does anyone have a LaTeX guide

6. ## Re: Equaitons of Lines and planes

Originally Posted by 99.95
Another question:

Find parametric equations and symmetric equations for the line passing through the origin and the point (1,2,3).

So, the vector equation of a line can be expressed in the form r = r0 + vt where v is a vector parallel to L and t is a scalar.

I thought I'd use the position vector OP (1,2,3) and then find a vector parallel to this line by crossing it with an unknown vector a,b,c and letting it equal zero, then solving.

am i on the correct track?
$r = r_0 + vt$ represents the equation of a line where $r_0$ is a point on the line and $v$ is a vector parallel to the line.

If you know 2 points on your line A,B, then B-A represents a vector parallel to the line. So since you have a line passing through the origin (0,0,0) and (1,2,3)
(1-0,2-0,3-0) = (1,2,3) is a vector parallel to your line.

To understand the format of the equation, and what it is doing you must first try to understand the difference between vectors and points. Points and vectors are sometimes written in the same format (as i careless also did above) but they are not the same things. A point represents a fixed place in space. A vector only represents magnitude and direction (how much and which direction to move). If i have a point and add it to a vector i get another point, if i add 2 vectors i get a vector. if you have two points A and B, then B-A is a vector (which when added to A will get you to point B).

So in writing the equation of a line L(t) = $p_0 + tv$ where $p_0$ is a point on the line, you mean to start at the point $p_0$ (fixed in space), and move in the direction of the vector v or the opposite direction of v and by any amount (magnitude) by changing your parameter t.

Now if for the same equation as above L(t) = $p_0 + tv$, you replace the point $p_0$ with a point not on the line, then you get the equation $L(t) = p_1 + vt$ of another line that is parallel to L(t) = $p_0 + tv$.

Hope this gives you a bit more intuition on this stuff.

So since you are given 2 points, (0,0,0) and (1,2,3) . (1-0,2-0,3-0) is a vector paralell to your line. now you can use any point on the line, so i use (0,0,0).
so L(t) = (0,0,0) + (1,2,3)t or L(t) = (1,2,3)t gives the equation of the line.

7. ## Re: Equaitons of Lines and planes

thanks dave, it certainly does, i think i understand more how a vector equation of a line actually maps the line out, as the scalar t changes.

What's my take on the problem, would you say it was correct in finding another vector parallel to the position vector OP ?

8. ## Re: Equaitons of Lines and planes

no, you only need one parallel vector. and you already have one. (1,2,3).

9. ## Re: Equaitons of Lines and planes

thanks, book solutions were incorrect, which caused some confusion.

10. ## Re: Equaitons of Lines and planes

Originally Posted by 99.95
Another question:

Find parametric equations and symmetric equations for the line passing through the origin and the point (1,2,3).
r = t(1,2,3)
x=t, y=2t, z=3t

11. ## Re: Equaitons of Lines and planes

Given a parametric equation x = p + t*a, if a is a unit vector, t represents the distance from the point p in the direction of a...