# Basically solved, Last coordinate does not match?

• allanwinters
In summary, the line with equation (x, y, z) = (5, -4, 6) + u(1,4,-1) lies in the plane with equation (x, y, z) = (3, 0, 2) + s(1,1,-1) + t(2, -1, 1). Justification of the answer algebraically was given.
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?

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?

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?

allanwinters
allanwinters said:
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.

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

## 1. What does "Basically solved, Last coordinate does not match?" mean?

This means that the problem or experiment being conducted has been mostly solved or completed, but the final result or data point does not match the expected or desired outcome.

## 2. Why is the last coordinate not matching a problem?

There could be several reasons for this, such as human error, equipment malfunction, or a flaw in the experimental design. It is important to carefully analyze the data and methodology to determine the cause of the discrepancy.

## 3. How can I fix the last coordinate not matching?

Depending on the cause of the discrepancy, there are different approaches to addressing this issue. This may include recalibrating equipment, adjusting the experimental design, or repeating the experiment to gather more data. It is important to carefully evaluate the situation and choose the most appropriate solution.

## 4. Is the experiment or study still valid if the last coordinate does not match?

It depends on the significance of the discrepancy and the overall results of the experiment. If the last coordinate is the only outlier and the rest of the data is consistent, the experiment may still be valid. However, if the last coordinate significantly impacts the overall conclusion, it may be necessary to repeat the experiment or make adjustments to the methodology.

## 5. How can I prevent the last coordinate from not matching in future experiments?

To prevent this issue, it is important to carefully plan and design the experiment, properly calibrate and maintain equipment, and closely monitor the data collection process. It is also helpful to have multiple researchers independently collect and analyze the data to ensure accuracy and identify any potential errors.

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