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!
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.
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 + vt
= (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
represents the equation of a line where is a point on the line and 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) = where is a point on the line, you mean to start at the point (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) = , you replace the point with a point not on the line, then you get the equation of another line that is parallel to L(t) = .
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.
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 ?