Basically solved, Last coordinate does not match?

[allanwinters]

Homework Statement

1. Does the line with equation (x, y, z) = (5, -4, 6) + u(1,4,-1) lie in the plane with equation (x, y, z) = (3, 0, 2) + s(1,1,-1) + t(2, -1, 1)? Justify your answer algebraically.

Homework Equations

(x,y,z) = (x0,y0,z0) +s(a1.a2,b3) + t(b1,b2,b3)

The Attempt at a Solution

u=-2+s+2t
s=-4 +4u +t
t=4 -u +s
t=4 -(-2 +s+2t) + s, t=-6
s=-4 +4(-2 +s +2(-6)) -6, s=22
u= -2 +22 +2(-6), u=8
(x,y,z) = (5,-4,6) + 8(1,4,-1) = (13,28,-2)
(x,y,z) = (3,0,2) +22(1,1,-1) -6(2,-1,1) =(13,28,-26)
I don't know why the z -coordinate does not match with each other? Does the point not lie in the plane?

 

[andrewkirk]

I can't follow what you were trying to do.

But to find out if the line lies in the plane, write equations for the parameters that specify point(s) that are in both the line and the plane. That is

(5, -4, 6) + u(1,4,-1) = (3, 0, 2) + s(1,1,-1) + t(2, -1, 1)

You can write this as a matrix equation
$$\mathbf A\pmatrix{s\\t\\u}=\vec b$$

where $\mathbf A$ is a 3 x 3 matrix and $\vec b$ is a 3-element vector, both obtained from the above equation.

If the equation has only one solution, or no solutions, the line is not in the plane. If it has infinitely many solutions, it is in the plane. Do you know the linear algebra necessary to solve that?

 

[allanwinters]

In my grade 12 calculus course, we're not taught matrixes, and only learn about vector and parametric equations of a plane. Is there an alternative way to solve this problem?

 

[bluenoise]

Don't worry, a matrix equation is just a convenient way to write a system of linear equations compactly. What you have is:

$$\pmatrix{5\\-4\\6}+u\pmatrix{1\\4\\-1}=\pmatrix{3\\0\\2}+s\pmatrix{1\\1\\-1}+t\pmatrix{2\\-1\\1}$$

Sorting the stuff gives you:

$$u\pmatrix{1\\4\\-1}+s\pmatrix{-1\\-1\\1}+t\pmatrix{-2\\1\\-1}=\pmatrix{-2\\4\\-4}$$

which you can rewrite into a system of linear equations, where u, s, t are the unknowns (hopefully I made no mistake):

$$\begin{align}
u-s-2t &= -2\nonumber\\
4u-s+t &= 4\nonumber\\
-u+s-t &= -4\nonumber\end{align}$$

This system can be solved e.g. by Gaussian elemination. Were you taught how to solve linear systems?

 

[Ray Vickson]

In my grade 12 calculus course, we're not taught matrixes, and only learn about vector and parametric equations of a plane. Is there an alternative way to solve this problem?

As andrewkirk pointed out in post #2, you need to have

(5, -4, 6) + u(1,4,-1) = (3, 0, 2) + s(1,1,-1) + t(2, -1, 1)

Looking at this one component at a time, you get
$$\text{first component:} \;\; 5 + 1 u = 3 + 1s + 2t \; \Longrightarrow \: s + 2t - u = 5-3 = 2$$
Looking at the other two components gives you two other equations.

So, you have three linear equations in the three unknowns u,s,t. Just solve them using high-school methods. These methods were being used hundreds of years before matrices were even invented.

Note added in edit: I see that the equations have already been written down completely---which looks very much like a violation of PF policy ----but the post did not appear on my screen until after I pressed the "enter" button.

 

[allanwinters]

Thank you for all the help! I solved the question.