Mixing
How to create an equation with 3 variables, given two points in 3-dimensions?
$begingroup$
I have two points: (20.33, 16.1, 0.0150) & (20.48, 19, 0.0123),
and I would like an equation of the line that connects these two points, but I'm not entirely sure how to do it. Is there a form of Y = MX + B, but in 3-dimensions? Iv'e already got the parametric equations and vector equations, but that doesn't seem to get me what I'm looking for. I would like to be able to enter in two values, say Y & Z, and then find out what X is. The vector equation is:
(x-0.15)/20.33 = (y-2.9)/16.1 = (z+0.0027)/0.0150
I'm not sure if what I'm asking is impossible or not...
Instead of just plugging in X, and calculation Y and Z - I would like to plug both X and Y, and calculate Z.
Thanks so much for any feedback!
- Cole
calculus linear-algebra multivariable-calculus multilinear-algebra
asked Sep 26 '15 at 3:13
InductionInduction
1011
$endgroup$
add a comment |
$begingroup$
I have two points: (20.33, 16.1, 0.0150) & (20.48, 19, 0.0123),
and I would like an equation of the line that connects these two points, but I'm not entirely sure how to do it. Is there a form of Y = MX + B, but in 3-dimensions? Iv'e already got the parametric equations and vector equations, but that doesn't seem to get me what I'm looking for. I would like to be able to enter in two values, say Y & Z, and then find out what X is. The vector equation is:
(x-0.15)/20.33 = (y-2.9)/16.1 = (z+0.0027)/0.0150
I'm not sure if what I'm asking is impossible or not...
Instead of just plugging in X, and calculation Y and Z - I would like to plug both X and Y, and calculate Z.
Thanks so much for any feedback!
- Cole
calculus linear-algebra multivariable-calculus multilinear-algebra
asked Sep 26 '15 at 3:13
InductionInduction
1011
$endgroup$
$begingroup$
There is no $y=mx+b$ for a line in three dimensions. You will have to go by parametric form or as you describe, a vector equation
$endgroup$
– imranfat
Sep 26 '15 at 3:15
$begingroup$
The equation you wrote is exactly what you want. For a point on a line, one of the three coordinates ($x$, $y$ or $z$) is sufficient to determine where it is.
$endgroup$
– Asydot
Sep 26 '15 at 3:17
$begingroup$
Ah thank you, that makes more sense I suppose. Perhaps what I'm really looking for is a plane to describe my data rather than a line.
$endgroup$
– Induction
Sep 26 '15 at 3:21
$begingroup$
Maybe, but for a plane in the three dim world, you need three points, not two. And these three points can't be collinear (why?)
$endgroup$
– imranfat
Sep 26 '15 at 3:22
$begingroup$
Well I would actually have three points, I was only using two for simplification. The points above all correspond to (extraction, ratio, total dissolved solids), these are referring to brewing coffee actually - just to put things into perspective. I would like to specify both extraction and ratio, and then find out the corresponding total dissolved solids accordingly. Now that I think of it though, I couldn't imagine a line being able to represent it.
$endgroup$
– Induction
Sep 26 '15 at 3:27
add a comment |
$begingroup$
I have two points: (20.33, 16.1, 0.0150) & (20.48, 19, 0.0123),
and I would like an equation of the line that connects these two points, but I'm not entirely sure how to do it. Is there a form of Y = MX + B, but in 3-dimensions? Iv'e already got the parametric equations and vector equations, but that doesn't seem to get me what I'm looking for. I would like to be able to enter in two values, say Y & Z, and then find out what X is. The vector equation is:
(x-0.15)/20.33 = (y-2.9)/16.1 = (z+0.0027)/0.0150
I'm not sure if what I'm asking is impossible or not...
Instead of just plugging in X, and calculation Y and Z - I would like to plug both X and Y, and calculate Z.
Thanks so much for any feedback!
- Cole
calculus linear-algebra multivariable-calculus multilinear-algebra
asked Sep 26 '15 at 3:13
InductionInduction
1011
$endgroup$
I have two points: (20.33, 16.1, 0.0150) & (20.48, 19, 0.0123),
and I would like an equation of the line that connects these two points, but I'm not entirely sure how to do it. Is there a form of Y = MX + B, but in 3-dimensions? Iv'e already got the parametric equations and vector equations, but that doesn't seem to get me what I'm looking for. I would like to be able to enter in two values, say Y & Z, and then find out what X is. The vector equation is:
(x-0.15)/20.33 = (y-2.9)/16.1 = (z+0.0027)/0.0150
I'm not sure if what I'm asking is impossible or not...
Instead of just plugging in X, and calculation Y and Z - I would like to plug both X and Y, and calculate Z.
Thanks so much for any feedback!
- Cole
calculus linear-algebra multivariable-calculus multilinear-algebra
calculus linear-algebra multivariable-calculus multilinear-algebra
asked Sep 26 '15 at 3:13
InductionInduction
1011
asked Sep 26 '15 at 3:13
InductionInduction
1011
asked Sep 26 '15 at 3:13
InductionInduction
1011
asked Sep 26 '15 at 3:13
InductionInduction
1011
asked Sep 26 '15 at 3:13
InductionInduction
1011
1011
$begingroup$
There is no $y=mx+b$ for a line in three dimensions. You will have to go by parametric form or as you describe, a vector equation
$endgroup$
– imranfat
Sep 26 '15 at 3:15
$begingroup$
The equation you wrote is exactly what you want. For a point on a line, one of the three coordinates ($x$, $y$ or $z$) is sufficient to determine where it is.
$endgroup$
– Asydot
Sep 26 '15 at 3:17
$begingroup$
Ah thank you, that makes more sense I suppose. Perhaps what I'm really looking for is a plane to describe my data rather than a line.
$endgroup$
– Induction
Sep 26 '15 at 3:21
$begingroup$
Maybe, but for a plane in the three dim world, you need three points, not two. And these three points can't be collinear (why?)
$endgroup$
– imranfat
Sep 26 '15 at 3:22
$begingroup$
Well I would actually have three points, I was only using two for simplification. The points above all correspond to (extraction, ratio, total dissolved solids), these are referring to brewing coffee actually - just to put things into perspective. I would like to specify both extraction and ratio, and then find out the corresponding total dissolved solids accordingly. Now that I think of it though, I couldn't imagine a line being able to represent it.
$endgroup$
– Induction
Sep 26 '15 at 3:27
add a comment |
$begingroup$
There is no $y=mx+b$ for a line in three dimensions. You will have to go by parametric form or as you describe, a vector equation
$endgroup$
– imranfat
Sep 26 '15 at 3:15
$begingroup$
The equation you wrote is exactly what you want. For a point on a line, one of the three coordinates ($x$, $y$ or $z$) is sufficient to determine where it is.
$endgroup$
– Asydot
Sep 26 '15 at 3:17
$begingroup$
Ah thank you, that makes more sense I suppose. Perhaps what I'm really looking for is a plane to describe my data rather than a line.
$endgroup$
– Induction
Sep 26 '15 at 3:21
$begingroup$
Maybe, but for a plane in the three dim world, you need three points, not two. And these three points can't be collinear (why?)
$endgroup$
– imranfat
Sep 26 '15 at 3:22
$begingroup$
Well I would actually have three points, I was only using two for simplification. The points above all correspond to (extraction, ratio, total dissolved solids), these are referring to brewing coffee actually - just to put things into perspective. I would like to specify both extraction and ratio, and then find out the corresponding total dissolved solids accordingly. Now that I think of it though, I couldn't imagine a line being able to represent it.
$endgroup$
– Induction
Sep 26 '15 at 3:27
$begingroup$
There is no $y=mx+b$ for a line in three dimensions. You will have to go by parametric form or as you describe, a vector equation
$endgroup$
– imranfat
Sep 26 '15 at 3:15
$begingroup$
There is no $y=mx+b$ for a line in three dimensions. You will have to go by parametric form or as you describe, a vector equation
$endgroup$
– imranfat
Sep 26 '15 at 3:15
$begingroup$
The equation you wrote is exactly what you want. For a point on a line, one of the three coordinates ($x$, $y$ or $z$) is sufficient to determine where it is.
$endgroup$
– Asydot
Sep 26 '15 at 3:17
$begingroup$
The equation you wrote is exactly what you want. For a point on a line, one of the three coordinates ($x$, $y$ or $z$) is sufficient to determine where it is.
$endgroup$
– Asydot
Sep 26 '15 at 3:17
$begingroup$
Ah thank you, that makes more sense I suppose. Perhaps what I'm really looking for is a plane to describe my data rather than a line.
$endgroup$
– Induction
Sep 26 '15 at 3:21
$begingroup$
Ah thank you, that makes more sense I suppose. Perhaps what I'm really looking for is a plane to describe my data rather than a line.
$endgroup$
– Induction
Sep 26 '15 at 3:21
$begingroup$
Maybe, but for a plane in the three dim world, you need three points, not two. And these three points can't be collinear (why?)
$endgroup$
– imranfat
Sep 26 '15 at 3:22
$begingroup$
Maybe, but for a plane in the three dim world, you need three points, not two. And these three points can't be collinear (why?)
$endgroup$
– imranfat
Sep 26 '15 at 3:22
$begingroup$
Well I would actually have three points, I was only using two for simplification. The points above all correspond to (extraction, ratio, total dissolved solids), these are referring to brewing coffee actually - just to put things into perspective. I would like to specify both extraction and ratio, and then find out the corresponding total dissolved solids accordingly. Now that I think of it though, I couldn't imagine a line being able to represent it.
$endgroup$
– Induction
Sep 26 '15 at 3:27
$begingroup$
Well I would actually have three points, I was only using two for simplification. The points above all correspond to (extraction, ratio, total dissolved solids), these are referring to brewing coffee actually - just to put things into perspective. I would like to specify both extraction and ratio, and then find out the corresponding total dissolved solids accordingly. Now that I think of it though, I couldn't imagine a line being able to represent it.
$endgroup$
– Induction
Sep 26 '15 at 3:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Given two points $P$ and $Q$, an equation for the line connecting them is $P+(1-t)Q$, where $t$ is the parametric variable for the line.
answered Sep 26 '15 at 3:18
pre-kidneypre-kidney
12.9k1849
$endgroup$
$begingroup$
Thanks for that, although I don't quite understand what t is. Or at least how that would apply to my situation. Also if there is no equality, wouldn't that not be an equation?
$endgroup$
– Induction
Sep 26 '15 at 3:23
$begingroup$
To get an equation, you could write out the three coordinates: $x=20.33+(1-t)20.48$, $y=16.1+(1-t)19$, and lastly $z=0.0150+(1-t)0.0123$.
$endgroup$
– pre-kidney
Sep 26 '15 at 3:25
add a comment |
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1 Answer
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$begingroup$
Given two points $P$ and $Q$, an equation for the line connecting them is $P+(1-t)Q$, where $t$ is the parametric variable for the line.
answered Sep 26 '15 at 3:18
pre-kidneypre-kidney
12.9k1849
$endgroup$
$begingroup$
Thanks for that, although I don't quite understand what t is. Or at least how that would apply to my situation. Also if there is no equality, wouldn't that not be an equation?
$endgroup$
– Induction
Sep 26 '15 at 3:23
$begingroup$
To get an equation, you could write out the three coordinates: $x=20.33+(1-t)20.48$, $y=16.1+(1-t)19$, and lastly $z=0.0150+(1-t)0.0123$.
$endgroup$
– pre-kidney
Sep 26 '15 at 3:25
add a comment |
$begingroup$
Given two points $P$ and $Q$, an equation for the line connecting them is $P+(1-t)Q$, where $t$ is the parametric variable for the line.
answered Sep 26 '15 at 3:18
pre-kidneypre-kidney
12.9k1849
$endgroup$
$begingroup$
Thanks for that, although I don't quite understand what t is. Or at least how that would apply to my situation. Also if there is no equality, wouldn't that not be an equation?
$endgroup$
– Induction
Sep 26 '15 at 3:23
$begingroup$
To get an equation, you could write out the three coordinates: $x=20.33+(1-t)20.48$, $y=16.1+(1-t)19$, and lastly $z=0.0150+(1-t)0.0123$.
$endgroup$
– pre-kidney
Sep 26 '15 at 3:25
add a comment |
$begingroup$
Given two points $P$ and $Q$, an equation for the line connecting them is $P+(1-t)Q$, where $t$ is the parametric variable for the line.
answered Sep 26 '15 at 3:18
pre-kidneypre-kidney
12.9k1849
$endgroup$
Given two points $P$ and $Q$, an equation for the line connecting them is $P+(1-t)Q$, where $t$ is the parametric variable for the line.
answered Sep 26 '15 at 3:18
pre-kidneypre-kidney
12.9k1849
answered Sep 26 '15 at 3:18
pre-kidneypre-kidney
12.9k1849
answered Sep 26 '15 at 3:18
pre-kidneypre-kidney
12.9k1849
answered Sep 26 '15 at 3:18
pre-kidneypre-kidney
12.9k1849
12.9k1849
$begingroup$
Thanks for that, although I don't quite understand what t is. Or at least how that would apply to my situation. Also if there is no equality, wouldn't that not be an equation?
$endgroup$
– Induction
Sep 26 '15 at 3:23
$begingroup$
To get an equation, you could write out the three coordinates: $x=20.33+(1-t)20.48$, $y=16.1+(1-t)19$, and lastly $z=0.0150+(1-t)0.0123$.
$endgroup$
– pre-kidney
Sep 26 '15 at 3:25
add a comment |
$begingroup$
Thanks for that, although I don't quite understand what t is. Or at least how that would apply to my situation. Also if there is no equality, wouldn't that not be an equation?
$endgroup$
– Induction
Sep 26 '15 at 3:23
$begingroup$
To get an equation, you could write out the three coordinates: $x=20.33+(1-t)20.48$, $y=16.1+(1-t)19$, and lastly $z=0.0150+(1-t)0.0123$.
$endgroup$
– pre-kidney
Sep 26 '15 at 3:25
$begingroup$
Thanks for that, although I don't quite understand what t is. Or at least how that would apply to my situation. Also if there is no equality, wouldn't that not be an equation?
$endgroup$
– Induction
Sep 26 '15 at 3:23
$begingroup$
Thanks for that, although I don't quite understand what t is. Or at least how that would apply to my situation. Also if there is no equality, wouldn't that not be an equation?
$endgroup$
– Induction
Sep 26 '15 at 3:23
$begingroup$
To get an equation, you could write out the three coordinates: $x=20.33+(1-t)20.48$, $y=16.1+(1-t)19$, and lastly $z=0.0150+(1-t)0.0123$.
$endgroup$
– pre-kidney
Sep 26 '15 at 3:25
$begingroup$
To get an equation, you could write out the three coordinates: $x=20.33+(1-t)20.48$, $y=16.1+(1-t)19$, and lastly $z=0.0150+(1-t)0.0123$.
$endgroup$
– pre-kidney
Sep 26 '15 at 3:25
add a comment |
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$begingroup$
There is no $y=mx+b$ for a line in three dimensions. You will have to go by parametric form or as you describe, a vector equation
$endgroup$
– imranfat
Sep 26 '15 at 3:15
$begingroup$
The equation you wrote is exactly what you want. For a point on a line, one of the three coordinates ($x$, $y$ or $z$) is sufficient to determine where it is.
$endgroup$
– Asydot
Sep 26 '15 at 3:17
$begingroup$
Ah thank you, that makes more sense I suppose. Perhaps what I'm really looking for is a plane to describe my data rather than a line.
$endgroup$
– Induction
Sep 26 '15 at 3:21
$begingroup$
Maybe, but for a plane in the three dim world, you need three points, not two. And these three points can't be collinear (why?)
$endgroup$
– imranfat
Sep 26 '15 at 3:22
$begingroup$
Well I would actually have three points, I was only using two for simplification. The points above all correspond to (extraction, ratio, total dissolved solids), these are referring to brewing coffee actually - just to put things into perspective. I would like to specify both extraction and ratio, and then find out the corresponding total dissolved solids accordingly. Now that I think of it though, I couldn't imagine a line being able to represent it.
$endgroup$
– Induction
Sep 26 '15 at 3:27