Line-Line Intersection
Given these two lines, where the first one comes as the intersection between two planes, i struggle to find a solution.
L1:
$$x + y − z − 1 = 0$$
$$2x − y + 2z − 3 = 0 $$
L2 :
$$X = (1 − k, k + 1, 3k), k in mathbb{R}$$
So far, i've got the equation for the first line, but i really don't know where to start with the second one.
algebra-precalculus
edited Nov 29 '18 at 1:01
David Diaz
954420
asked Nov 29 '18 at 0:38
Tomas Vortali
33
add a comment |
Given these two lines, where the first one comes as the intersection between two planes, i struggle to find a solution.
L1:
$$x + y − z − 1 = 0$$
$$2x − y + 2z − 3 = 0 $$
L2 :
$$X = (1 − k, k + 1, 3k), k in mathbb{R}$$
So far, i've got the equation for the first line, but i really don't know where to start with the second one.
algebra-precalculus
edited Nov 29 '18 at 1:01
David Diaz
954420
asked Nov 29 '18 at 0:38
Tomas Vortali
33
1
If the lines do intersect, then that intersection point is also the intersection of $L_2$ with either plane.
– amd
Nov 29 '18 at 1:07
add a comment |
Given these two lines, where the first one comes as the intersection between two planes, i struggle to find a solution.
L1:
$$x + y − z − 1 = 0$$
$$2x − y + 2z − 3 = 0 $$
L2 :
$$X = (1 − k, k + 1, 3k), k in mathbb{R}$$
So far, i've got the equation for the first line, but i really don't know where to start with the second one.
algebra-precalculus
edited Nov 29 '18 at 1:01
David Diaz
954420
asked Nov 29 '18 at 0:38
Tomas Vortali
33
Given these two lines, where the first one comes as the intersection between two planes, i struggle to find a solution.
L1:
$$x + y − z − 1 = 0$$
$$2x − y + 2z − 3 = 0 $$
L2 :
$$X = (1 − k, k + 1, 3k), k in mathbb{R}$$
So far, i've got the equation for the first line, but i really don't know where to start with the second one.
algebra-precalculus
algebra-precalculus
edited Nov 29 '18 at 1:01
David Diaz
954420
asked Nov 29 '18 at 0:38
Tomas Vortali
33
edited Nov 29 '18 at 1:01
David Diaz
954420
asked Nov 29 '18 at 0:38
Tomas Vortali
33
edited Nov 29 '18 at 1:01
David Diaz
954420
edited Nov 29 '18 at 1:01
David Diaz
954420
edited Nov 29 '18 at 1:01
David Diaz
954420
954420
asked Nov 29 '18 at 0:38
Tomas Vortali
33
asked Nov 29 '18 at 0:38
Tomas Vortali
33
asked Nov 29 '18 at 0:38
Tomas Vortali
33
33
1
If the lines do intersect, then that intersection point is also the intersection of $L_2$ with either plane.
– amd
Nov 29 '18 at 1:07
add a comment |
1
If the lines do intersect, then that intersection point is also the intersection of $L_2$ with either plane.
– amd
Nov 29 '18 at 1:07
1
1
If the lines do intersect, then that intersection point is also the intersection of $L_2$ with either plane.
– amd
Nov 29 '18 at 1:07
If the lines do intersect, then that intersection point is also the intersection of $L_2$ with either plane.
– amd
Nov 29 '18 at 1:07
add a comment |
1 Answer
1
active
oldest
votes
There is no solution.
Let's take an arbitrary point on $L_2$ and plug it into the first equation:
begin{align}
underbrace{1-k}_{x} + underbrace{k+1}_{y} - underbrace{3k}_{z} - 1 &= 0\
-3k+1 &= 0\
k&=frac{1}{3}
end{align}
$$therefore X = bigg(frac{2}{3}, frac{4}{3}, 1bigg)$$
$L_2$ and the plane represented by the first equation intersect at exactly one point. We can plug this point in to the second plane to check if it is a solution:
begin{align}
2cdotunderbrace{frac{2}{3}}_{x}-underbrace{frac{4}{3}}_{y}+2cdotunderbrace{1}_{z}-3 &= 0\
-1&=0
end{align}
It is not. Therefore, there are no solutions.
answered Nov 29 '18 at 2:00
David Diaz
954420
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
There is no solution.
Let's take an arbitrary point on $L_2$ and plug it into the first equation:
begin{align}
underbrace{1-k}_{x} + underbrace{k+1}_{y} - underbrace{3k}_{z} - 1 &= 0\
-3k+1 &= 0\
k&=frac{1}{3}
end{align}
$$therefore X = bigg(frac{2}{3}, frac{4}{3}, 1bigg)$$
$L_2$ and the plane represented by the first equation intersect at exactly one point. We can plug this point in to the second plane to check if it is a solution:
begin{align}
2cdotunderbrace{frac{2}{3}}_{x}-underbrace{frac{4}{3}}_{y}+2cdotunderbrace{1}_{z}-3 &= 0\
-1&=0
end{align}
It is not. Therefore, there are no solutions.
answered Nov 29 '18 at 2:00
David Diaz
954420
add a comment |
There is no solution.
Let's take an arbitrary point on $L_2$ and plug it into the first equation:
begin{align}
underbrace{1-k}_{x} + underbrace{k+1}_{y} - underbrace{3k}_{z} - 1 &= 0\
-3k+1 &= 0\
k&=frac{1}{3}
end{align}
$$therefore X = bigg(frac{2}{3}, frac{4}{3}, 1bigg)$$
$L_2$ and the plane represented by the first equation intersect at exactly one point. We can plug this point in to the second plane to check if it is a solution:
begin{align}
2cdotunderbrace{frac{2}{3}}_{x}-underbrace{frac{4}{3}}_{y}+2cdotunderbrace{1}_{z}-3 &= 0\
-1&=0
end{align}
It is not. Therefore, there are no solutions.
answered Nov 29 '18 at 2:00
David Diaz
954420
add a comment |
There is no solution.
Let's take an arbitrary point on $L_2$ and plug it into the first equation:
begin{align}
underbrace{1-k}_{x} + underbrace{k+1}_{y} - underbrace{3k}_{z} - 1 &= 0\
-3k+1 &= 0\
k&=frac{1}{3}
end{align}
$$therefore X = bigg(frac{2}{3}, frac{4}{3}, 1bigg)$$
$L_2$ and the plane represented by the first equation intersect at exactly one point. We can plug this point in to the second plane to check if it is a solution:
begin{align}
2cdotunderbrace{frac{2}{3}}_{x}-underbrace{frac{4}{3}}_{y}+2cdotunderbrace{1}_{z}-3 &= 0\
-1&=0
end{align}
It is not. Therefore, there are no solutions.
answered Nov 29 '18 at 2:00
David Diaz
954420
There is no solution.
Let's take an arbitrary point on $L_2$ and plug it into the first equation:
begin{align}
underbrace{1-k}_{x} + underbrace{k+1}_{y} - underbrace{3k}_{z} - 1 &= 0\
-3k+1 &= 0\
k&=frac{1}{3}
end{align}
$$therefore X = bigg(frac{2}{3}, frac{4}{3}, 1bigg)$$
$L_2$ and the plane represented by the first equation intersect at exactly one point. We can plug this point in to the second plane to check if it is a solution:
begin{align}
2cdotunderbrace{frac{2}{3}}_{x}-underbrace{frac{4}{3}}_{y}+2cdotunderbrace{1}_{z}-3 &= 0\
-1&=0
end{align}
It is not. Therefore, there are no solutions.
answered Nov 29 '18 at 2:00
David Diaz
954420
answered Nov 29 '18 at 2:00
David Diaz
954420
answered Nov 29 '18 at 2:00
David Diaz
954420
answered Nov 29 '18 at 2:00
David Diaz
954420
954420
add a comment |
add a comment |
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If the lines do intersect, then that intersection point is also the intersection of $L_2$ with either plane.
– amd
Nov 29 '18 at 1:07