Parametric and canonical equation of the line

[says]

Homework Statement

Find the parametric and canonical equation of the line L passing through the points A = [1, 0, 2] and B = [3, 1, −2]; check whether the point M = [7, 3, 1] lies on L.

Homework Equations

Canonical equation of a line in space
x-x₀ / l = y-y₀ / m = z-z₀ / n

Parametric equation of a line
x=lt+x₀
y=mt+y₀
z=nt+z₀

where x₀ y₀ z₀ are coordinates of a point sitting on the line
l,m,n are coordinates of the direction of the line

The Attempt at a Solution

The vector AB = B - A = [3, 1, −2] - [1, 0, 2] = [2,1,-4]

To see if M sits on the line we sub in it's coordinates into both equations:

Canonical =

7-1 / 2 = 3-0/1 = 1-2/4

6/2=3/1≠-1/-4

Parametric

7 = 2t+1
3 = 1t+0
1 = -4t+2

6=2t
3=1t
-1=-4t

Because we have an inequaity / inconsistency in the z component we can conclude that the point M = [7,3,1] does not lie on the line L. For M to lie on the line L all three equations would have to equal each other in canonical form, and in parametric form all values of t would have to be the same.

Just wondering if my conclusion is correct? Thanks :)

 

[HallsofIvy]

Yes, your conclusion, that M is not on the line, is correct. However, I notice that you did NOT actually show the equations of this. Did you find them and just not show them here?

 

[Mark44]

Canonical equation of a line in space
x-x₀ / l = y-y₀ / m = z-z₀ / n

7-1 / 2 = 3-0/1 = 1-2/4

When you write fractions like the ones above on one line, you need parentheses.
Otherwise, because of the precedence of division over subtraction, the line just above means $7 - \frac 1 2 = 3 - \frac 0 1 = 1 - \frac 2 4$, which is certainly not what you meant.
Instead of 7 - 1/2, write (7 - 1)/2 when you write this on one line.