Flaw in 3D Vector Definition: Solving for t and x Step by Step

[soandos]

where is the flaw in the following:
it is possible to define a 3d vector using the form:
(x,y,z) = (x_1,y_1,z_1)+t(a,b,c)
this can then be parameterized to become

x = x_1 + a*t
y = y_1 + b*t
z = z_1 +c*t

so, for example, given the vector (x,y,z) = (1,2,3) + t(5,6,7)
one can parametrize it to become:

x = 1 + 5t
y = 2 + 6t
z = 3 + 7t

solving the first equation for t gives
t = (x-1)/5

plugging that into the other two equations gives:

y = 2 + 6/5 * (x-1)
z = 3 + 7/5 * (x-1)

solving the second one for x gives:

x = (5*z-8)/7

changing y = 2 + 6/5 * (x-1) to y = 2 + 6/5 * (2x-x-1)
and in the -x spot putting -(5*z-8)/7 yeilds:

y = 2 + 6/5 * (2x-(5*z-8)/7-1)

simplifying gets:
76 = 35 y + 30 z - 84 x

which is the equation of a plane.
this cannot possibly be right as i started out with a 2-d object (the vector)

where did i mess up?
(p.s. i tried to do all of this algebraically but it got too ugly for me)

thanks.

 

[djeitnstine]

First of all, that is not the equation of a vector, it is a vector equation of a line.

In that equation there exists 2 vectors, the position vector r and direction vector v

so in your example

r = <1,2,3>
v = <5,6,7>

so your line starts at (1,2,3) and points in the direction <5,6,7>

 

[soandos]

sorry if i used the wrong words. the fact remains that i began with a line/line segment/vector and ended up with an equivalent plane. this is not possible. where is the mistake?

 

[adriank]

You think you have made a mistake where you haven't (except possibly a conceptual one).

You get the equation of a plane containing the line.

Your calculations are correct. All you have done is introduced extra solutions. What you have essentially proven is that if (x, y, z) = (1, 2, 3) + t(5, 6, 7) for some t, then 76 = 35y + 30z - 84x. It, of course, is not true that if 76 = 35y + 30z - 84x, then (x, y, z) = (1, 2, 3) + t(5, 6, 7) for some t, but you haven't proven this (which is good, since it's false).

I imagine with some slight variations, you will find an equation for a different plane. However, it will still be a plane containing the original line; the two planes you found would intersect at this line.

 

[djeitnstine]

There are only 3 forms of a line in 3d. The vector, parametric and symmetric forms.

Taking your example gives:

vector form: \vec{r}=&lt;1,2,3&gt; + t&lt;5,6,7&gt;

symmetric form: \frac{x-1}{5}=\frac{y-2}{6}=\frac{z-3}{7}

parametric form: x = 1+5t , y = 2+6t , z = 3+7t

Of course you can figure out how the line behaves in each plane (which I presume you are searching for) by solving the symmetric form gives you the trivial equation y=mx+c on a particular plane. For example z=my+c (zy plane) or z=mx+c (zx plane) etc...

Edit I left out the 2 point form which looks like: \frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}} so there are 4 (viewed at http://www.cs.fit.edu/~wds/classes/cse5255/thesis/lineEqn/lineEqn#2ptEqn)

 

[soandos]

let me try to explain:
i got three parametric equations.
i simplified them into one equation
i got the equation of a plane.
therefore, i know something went wrong.
read the PDF to see exactly what i did.

 

[adriank]

Read my post. Then read it again.

 

[djeitnstine]

let me try to explain:
i got three parametric equations.
i simplified them into one equation
i got the equation of a plane.
therefore, i know something went wrong.
read the PDF to see exactly what i did.

adriank is correct. All you have shown is a plane that the line lies upon. There are infinitely many.

 

[soandos]

that is not the case, as i have locked it down to one plane. what happened to all of the others?

 

[djeitnstine]

that is not the case, as i have locked it down to one plane. what happened to all of the others?

I will answer you question with another, how many equations did you solve for t? I see 3 different solutions for t

Also, how many lines of the same slope and direction lie upon a single plane ?

 

[vladibo]

... and by the way if you want to visualize your vectors in 3D you could use this program http://www.bodurov.com/VectorVisualizer/"

 

[soandos]

In response to djeitnstine, i only solved one.
doe that mean anything?

 

[juliaroberts]

show that the line containing the points (0,0,5) and (1,-1,4) is perpendicular to the line with equations:

x/7=y-3/4=z+9/3

i need help with this question..

 

[adriank]

Find vectors parallel to each line, and show that their dot product is zero.

 

[juliaroberts]

thank you adriank:)
so;

x/7=y-3/4=z+9/3=t

solving the first equation for t gives;

x/7=t so x=7t

and other two:

y-3/4=t so y=4t+3.
z+9/3=t so z=3t-9.

u=(7,4,3) and P= (0,3,-9) is on l.
u is parallel to l.

then we find another point on l.
x=2 so y=29/7 and z= -57/7.

P1(2,29/7,-57/7)

as we have Q (0,0,5) and Q1 (1,-1,4) ; Q1-Q= (1,-1,-1).

P(0,0,-9) and P1( 2,29/7,-57/7) ; P1-P= (2,8/7,6/7)

P1P*Q1Q=2-8/7-6/7=0 so we can say Q1Q is perpendiclar to l.

is it true? could you pls explain? I'm not sure..