Finding components of lines intersecting at a point.
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I'm just working on some summer problems so that I can be more prepared when I go into my class in the fall. I found a website full of problems of the content we will be learning but it doesn't have the answers. I need a little guidance on how to do this problem. Here is the problem:
Consider the lines $L1$ and $L2$ with equations:
$L_1 : r = (11, 8, 2) + s(4, 3, -1)$
$L_2 : r = (1, 1,-7) + t(2, 1, 11)$
The lines intersect at point $P$.
a. Find the coordinates of $P$.
Would we use a formula for this? For example: $fracx_1+x_22, fracy_1+y_22, fracz_1+z_22$.
b. Show that the lines are perpendicular.
Isn’t there also a formula to find if lines are perpendicular?
coordinate-systems
asked Aug 9 at 20:49
Ella
1019
 |Â
show 2 more comments
up vote
0
down vote
favorite
I'm just working on some summer problems so that I can be more prepared when I go into my class in the fall. I found a website full of problems of the content we will be learning but it doesn't have the answers. I need a little guidance on how to do this problem. Here is the problem:
Consider the lines $L1$ and $L2$ with equations:
$L_1 : r = (11, 8, 2) + s(4, 3, -1)$
$L_2 : r = (1, 1,-7) + t(2, 1, 11)$
The lines intersect at point $P$.
a. Find the coordinates of $P$.
Would we use a formula for this? For example: $fracx_1+x_22, fracy_1+y_22, fracz_1+z_22$.
b. Show that the lines are perpendicular.
Isn’t there also a formula to find if lines are perpendicular?
coordinate-systems
asked Aug 9 at 20:49
Ella
1019
Do you mean $(11,8,2)+s(4,3,-1)?$
– mfl
Aug 9 at 20:55
@mfl The question I found says -1, not -2. But, perhaps there was an error on their part. I am not sure.
– Ella
Aug 9 at 20:58
Sorry about the $-2$. I mean the way you write the line. It is written as [11, 8, 2] + s = [4, 3, -1]
– mfl
Aug 9 at 21:00
@mfl Oh, yes! You’re right. Thank you. Just fixed it.
– Ella
Aug 9 at 21:01
I don't understand c. $R$ is the reflection of $Q$ with respect to what?
– mfl
Aug 9 at 21:06
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm just working on some summer problems so that I can be more prepared when I go into my class in the fall. I found a website full of problems of the content we will be learning but it doesn't have the answers. I need a little guidance on how to do this problem. Here is the problem:
Consider the lines $L1$ and $L2$ with equations:
$L_1 : r = (11, 8, 2) + s(4, 3, -1)$
$L_2 : r = (1, 1,-7) + t(2, 1, 11)$
The lines intersect at point $P$.
a. Find the coordinates of $P$.
Would we use a formula for this? For example: $fracx_1+x_22, fracy_1+y_22, fracz_1+z_22$.
b. Show that the lines are perpendicular.
Isn’t there also a formula to find if lines are perpendicular?
coordinate-systems
asked Aug 9 at 20:49
Ella
1019
I'm just working on some summer problems so that I can be more prepared when I go into my class in the fall. I found a website full of problems of the content we will be learning but it doesn't have the answers. I need a little guidance on how to do this problem. Here is the problem:
Consider the lines $L1$ and $L2$ with equations:
$L_1 : r = (11, 8, 2) + s(4, 3, -1)$
$L_2 : r = (1, 1,-7) + t(2, 1, 11)$
The lines intersect at point $P$.
a. Find the coordinates of $P$.
Would we use a formula for this? For example: $fracx_1+x_22, fracy_1+y_22, fracz_1+z_22$.
b. Show that the lines are perpendicular.
Isn’t there also a formula to find if lines are perpendicular?
coordinate-systems
asked Aug 9 at 20:49
Ella
1019
edited Aug 10 at 20:09
asked Aug 9 at 20:49
Ella
1019
asked Aug 9 at 20:49
Ella
1019
asked Aug 9 at 20:49
Ella
1019
1019
Do you mean $(11,8,2)+s(4,3,-1)?$
– mfl
Aug 9 at 20:55
@mfl The question I found says -1, not -2. But, perhaps there was an error on their part. I am not sure.
– Ella
Aug 9 at 20:58
Sorry about the $-2$. I mean the way you write the line. It is written as [11, 8, 2] + s = [4, 3, -1]
– mfl
Aug 9 at 21:00
@mfl Oh, yes! You’re right. Thank you. Just fixed it.
– Ella
Aug 9 at 21:01
I don't understand c. $R$ is the reflection of $Q$ with respect to what?
– mfl
Aug 9 at 21:06
 |Â
show 2 more comments
Do you mean $(11,8,2)+s(4,3,-1)?$
– mfl
Aug 9 at 20:55
@mfl The question I found says -1, not -2. But, perhaps there was an error on their part. I am not sure.
– Ella
Aug 9 at 20:58
Sorry about the $-2$. I mean the way you write the line. It is written as [11, 8, 2] + s = [4, 3, -1]
– mfl
Aug 9 at 21:00
@mfl Oh, yes! You’re right. Thank you. Just fixed it.
– Ella
Aug 9 at 21:01
I don't understand c. $R$ is the reflection of $Q$ with respect to what?
– mfl
Aug 9 at 21:06
Do you mean $(11,8,2)+s(4,3,-1)?$
– mfl
Aug 9 at 20:55
Do you mean $(11,8,2)+s(4,3,-1)?$
– mfl
Aug 9 at 20:55
@mfl The question I found says -1, not -2. But, perhaps there was an error on their part. I am not sure.
– Ella
Aug 9 at 20:58
@mfl The question I found says -1, not -2. But, perhaps there was an error on their part. I am not sure.
– Ella
Aug 9 at 20:58
Sorry about the $-2$. I mean the way you write the line. It is written as [11, 8, 2] + s = [4, 3, -1]
– mfl
Aug 9 at 21:00
Sorry about the $-2$. I mean the way you write the line. It is written as [11, 8, 2] + s = [4, 3, -1]
– mfl
Aug 9 at 21:00
@mfl Oh, yes! You’re right. Thank you. Just fixed it.
– Ella
Aug 9 at 21:01
@mfl Oh, yes! You’re right. Thank you. Just fixed it.
– Ella
Aug 9 at 21:01
I don't understand c. $R$ is the reflection of $Q$ with respect to what?
– mfl
Aug 9 at 21:06
I don't understand c. $R$ is the reflection of $Q$ with respect to what?
– mfl
Aug 9 at 21:06
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Saying that
$L_1:r=(11,8,2)+s(4,3,−1)$
means that every point on the line is of the form
(x, y, z)= (11+ 4s, 8+ 3s, 2- s) for some number s.
In other words, x= 11+ 4s, y= 8+ 3s, z= 2- s.
$L_2:r=(1,1,−7)+t(2,1,11)$
means that every point on the line is of the form
(x, y, z)= (1+ 2t, 1+ t, -7+ 11t) for some number t.
In other words, x= 1+ 2t, y= 1+ t, z= -7+ 11t.
Where the two lines intercept the x, y, z coordinates must be the same: 11+ 4s= 1+ 2t, 8+ 3s= 1+ t, 2- s= -7+ 11t.
Notice that these are three equations in only two unknowns, s and t. In general, two lines in three dimensions do NOT intersect- most often they are "skew". What we can do is solve two of the equations for s and t then see if those s and t satisfy the third equation.
11+ 4s= 1+ 2t is the same as 10+ 4s= 2t or 5+ 2s= t. Setting t= 5+ 2s in the second equation, 8+ 3s= 1+ t= 6+ 2s. s= -2 and then t= 5- 4= 1.
Checking the third equation, 2- s= -7+ 11t, 2- (-2)= 4= -7+ 11. Yes, those values are the same so the lines intersect at x= 11+ 4(-2)= 11- 8= 3, y= 8+ 3s= 8+ 3(-2)= 8- 6= 2, z= 2- s= 2- (-2)= 4. The two lines intersect at P= (3, 2, 4).
answered Aug 9 at 21:27
user247327
9,7311515
add a comment |Â
up vote
0
down vote
Ideas to get the answer
Part a):
You need to solve the system
beginalign
11+4s&=1+2t\
8+3s&=1+t\
2-s&=-7+11t
endalign
Part b):
The lines are perpendicular if and only if its directions are perpendicular. That is, if and only if $$(4,3,-1)cdot (2,1,11)=0.$$
answered Aug 9 at 21:00
mfl
24.6k12040
For part a, would I combine like terms and solve for both s and t?
– Ella
Aug 9 at 21:19
Yes. Note that if $(x,y,z)$ is the intersection point then it must be $x=11+4s$ (line 1) and $x=1+2t$ (line 2). The same for $y,z.$ Once you get a solution of the system ($s,t$ must be a solution of the $3$ equations) you substitute $s$ in the first line (or $t$ in the second line) and you'll get $P$.
– mfl
Aug 9 at 21:23
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Saying that
$L_1:r=(11,8,2)+s(4,3,−1)$
means that every point on the line is of the form
(x, y, z)= (11+ 4s, 8+ 3s, 2- s) for some number s.
In other words, x= 11+ 4s, y= 8+ 3s, z= 2- s.
$L_2:r=(1,1,−7)+t(2,1,11)$
means that every point on the line is of the form
(x, y, z)= (1+ 2t, 1+ t, -7+ 11t) for some number t.
In other words, x= 1+ 2t, y= 1+ t, z= -7+ 11t.
Where the two lines intercept the x, y, z coordinates must be the same: 11+ 4s= 1+ 2t, 8+ 3s= 1+ t, 2- s= -7+ 11t.
Notice that these are three equations in only two unknowns, s and t. In general, two lines in three dimensions do NOT intersect- most often they are "skew". What we can do is solve two of the equations for s and t then see if those s and t satisfy the third equation.
11+ 4s= 1+ 2t is the same as 10+ 4s= 2t or 5+ 2s= t. Setting t= 5+ 2s in the second equation, 8+ 3s= 1+ t= 6+ 2s. s= -2 and then t= 5- 4= 1.
Checking the third equation, 2- s= -7+ 11t, 2- (-2)= 4= -7+ 11. Yes, those values are the same so the lines intersect at x= 11+ 4(-2)= 11- 8= 3, y= 8+ 3s= 8+ 3(-2)= 8- 6= 2, z= 2- s= 2- (-2)= 4. The two lines intersect at P= (3, 2, 4).
answered Aug 9 at 21:27
user247327
9,7311515
add a comment |Â
up vote
1
down vote
accepted
Saying that
$L_1:r=(11,8,2)+s(4,3,−1)$
means that every point on the line is of the form
(x, y, z)= (11+ 4s, 8+ 3s, 2- s) for some number s.
In other words, x= 11+ 4s, y= 8+ 3s, z= 2- s.
$L_2:r=(1,1,−7)+t(2,1,11)$
means that every point on the line is of the form
(x, y, z)= (1+ 2t, 1+ t, -7+ 11t) for some number t.
In other words, x= 1+ 2t, y= 1+ t, z= -7+ 11t.
Where the two lines intercept the x, y, z coordinates must be the same: 11+ 4s= 1+ 2t, 8+ 3s= 1+ t, 2- s= -7+ 11t.
Notice that these are three equations in only two unknowns, s and t. In general, two lines in three dimensions do NOT intersect- most often they are "skew". What we can do is solve two of the equations for s and t then see if those s and t satisfy the third equation.
11+ 4s= 1+ 2t is the same as 10+ 4s= 2t or 5+ 2s= t. Setting t= 5+ 2s in the second equation, 8+ 3s= 1+ t= 6+ 2s. s= -2 and then t= 5- 4= 1.
Checking the third equation, 2- s= -7+ 11t, 2- (-2)= 4= -7+ 11. Yes, those values are the same so the lines intersect at x= 11+ 4(-2)= 11- 8= 3, y= 8+ 3s= 8+ 3(-2)= 8- 6= 2, z= 2- s= 2- (-2)= 4. The two lines intersect at P= (3, 2, 4).
answered Aug 9 at 21:27
user247327
9,7311515
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Saying that
$L_1:r=(11,8,2)+s(4,3,−1)$
means that every point on the line is of the form
(x, y, z)= (11+ 4s, 8+ 3s, 2- s) for some number s.
In other words, x= 11+ 4s, y= 8+ 3s, z= 2- s.
$L_2:r=(1,1,−7)+t(2,1,11)$
means that every point on the line is of the form
(x, y, z)= (1+ 2t, 1+ t, -7+ 11t) for some number t.
In other words, x= 1+ 2t, y= 1+ t, z= -7+ 11t.
Where the two lines intercept the x, y, z coordinates must be the same: 11+ 4s= 1+ 2t, 8+ 3s= 1+ t, 2- s= -7+ 11t.
Notice that these are three equations in only two unknowns, s and t. In general, two lines in three dimensions do NOT intersect- most often they are "skew". What we can do is solve two of the equations for s and t then see if those s and t satisfy the third equation.
11+ 4s= 1+ 2t is the same as 10+ 4s= 2t or 5+ 2s= t. Setting t= 5+ 2s in the second equation, 8+ 3s= 1+ t= 6+ 2s. s= -2 and then t= 5- 4= 1.
Checking the third equation, 2- s= -7+ 11t, 2- (-2)= 4= -7+ 11. Yes, those values are the same so the lines intersect at x= 11+ 4(-2)= 11- 8= 3, y= 8+ 3s= 8+ 3(-2)= 8- 6= 2, z= 2- s= 2- (-2)= 4. The two lines intersect at P= (3, 2, 4).
answered Aug 9 at 21:27
user247327
9,7311515
Saying that
$L_1:r=(11,8,2)+s(4,3,−1)$
means that every point on the line is of the form
(x, y, z)= (11+ 4s, 8+ 3s, 2- s) for some number s.
In other words, x= 11+ 4s, y= 8+ 3s, z= 2- s.
$L_2:r=(1,1,−7)+t(2,1,11)$
means that every point on the line is of the form
(x, y, z)= (1+ 2t, 1+ t, -7+ 11t) for some number t.
In other words, x= 1+ 2t, y= 1+ t, z= -7+ 11t.
Where the two lines intercept the x, y, z coordinates must be the same: 11+ 4s= 1+ 2t, 8+ 3s= 1+ t, 2- s= -7+ 11t.
Notice that these are three equations in only two unknowns, s and t. In general, two lines in three dimensions do NOT intersect- most often they are "skew". What we can do is solve two of the equations for s and t then see if those s and t satisfy the third equation.
11+ 4s= 1+ 2t is the same as 10+ 4s= 2t or 5+ 2s= t. Setting t= 5+ 2s in the second equation, 8+ 3s= 1+ t= 6+ 2s. s= -2 and then t= 5- 4= 1.
Checking the third equation, 2- s= -7+ 11t, 2- (-2)= 4= -7+ 11. Yes, those values are the same so the lines intersect at x= 11+ 4(-2)= 11- 8= 3, y= 8+ 3s= 8+ 3(-2)= 8- 6= 2, z= 2- s= 2- (-2)= 4. The two lines intersect at P= (3, 2, 4).
answered Aug 9 at 21:27
user247327
9,7311515
answered Aug 9 at 21:27
user247327
9,7311515
answered Aug 9 at 21:27
user247327
9,7311515
answered Aug 9 at 21:27
user247327
9,7311515
9,7311515
add a comment |Â
add a comment |Â
up vote
0
down vote
Ideas to get the answer
Part a):
You need to solve the system
beginalign
11+4s&=1+2t\
8+3s&=1+t\
2-s&=-7+11t
endalign
Part b):
The lines are perpendicular if and only if its directions are perpendicular. That is, if and only if $$(4,3,-1)cdot (2,1,11)=0.$$
answered Aug 9 at 21:00
mfl
24.6k12040
For part a, would I combine like terms and solve for both s and t?
– Ella
Aug 9 at 21:19
Yes. Note that if $(x,y,z)$ is the intersection point then it must be $x=11+4s$ (line 1) and $x=1+2t$ (line 2). The same for $y,z.$ Once you get a solution of the system ($s,t$ must be a solution of the $3$ equations) you substitute $s$ in the first line (or $t$ in the second line) and you'll get $P$.
– mfl
Aug 9 at 21:23
add a comment |Â
up vote
0
down vote
Ideas to get the answer
Part a):
You need to solve the system
beginalign
11+4s&=1+2t\
8+3s&=1+t\
2-s&=-7+11t
endalign
Part b):
The lines are perpendicular if and only if its directions are perpendicular. That is, if and only if $$(4,3,-1)cdot (2,1,11)=0.$$
answered Aug 9 at 21:00
mfl
24.6k12040
For part a, would I combine like terms and solve for both s and t?
– Ella
Aug 9 at 21:19
Yes. Note that if $(x,y,z)$ is the intersection point then it must be $x=11+4s$ (line 1) and $x=1+2t$ (line 2). The same for $y,z.$ Once you get a solution of the system ($s,t$ must be a solution of the $3$ equations) you substitute $s$ in the first line (or $t$ in the second line) and you'll get $P$.
– mfl
Aug 9 at 21:23
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Ideas to get the answer
Part a):
You need to solve the system
beginalign
11+4s&=1+2t\
8+3s&=1+t\
2-s&=-7+11t
endalign
Part b):
The lines are perpendicular if and only if its directions are perpendicular. That is, if and only if $$(4,3,-1)cdot (2,1,11)=0.$$
answered Aug 9 at 21:00
mfl
24.6k12040
Ideas to get the answer
Part a):
You need to solve the system
beginalign
11+4s&=1+2t\
8+3s&=1+t\
2-s&=-7+11t
endalign
Part b):
The lines are perpendicular if and only if its directions are perpendicular. That is, if and only if $$(4,3,-1)cdot (2,1,11)=0.$$
answered Aug 9 at 21:00
mfl
24.6k12040
edited Aug 9 at 21:07
answered Aug 9 at 21:00
mfl
24.6k12040
answered Aug 9 at 21:00
mfl
24.6k12040
answered Aug 9 at 21:00
mfl
24.6k12040
24.6k12040
For part a, would I combine like terms and solve for both s and t?
– Ella
Aug 9 at 21:19
Yes. Note that if $(x,y,z)$ is the intersection point then it must be $x=11+4s$ (line 1) and $x=1+2t$ (line 2). The same for $y,z.$ Once you get a solution of the system ($s,t$ must be a solution of the $3$ equations) you substitute $s$ in the first line (or $t$ in the second line) and you'll get $P$.
– mfl
Aug 9 at 21:23
add a comment |Â
For part a, would I combine like terms and solve for both s and t?
– Ella
Aug 9 at 21:19
Yes. Note that if $(x,y,z)$ is the intersection point then it must be $x=11+4s$ (line 1) and $x=1+2t$ (line 2). The same for $y,z.$ Once you get a solution of the system ($s,t$ must be a solution of the $3$ equations) you substitute $s$ in the first line (or $t$ in the second line) and you'll get $P$.
– mfl
Aug 9 at 21:23
For part a, would I combine like terms and solve for both s and t?
– Ella
Aug 9 at 21:19
For part a, would I combine like terms and solve for both s and t?
– Ella
Aug 9 at 21:19
Yes. Note that if $(x,y,z)$ is the intersection point then it must be $x=11+4s$ (line 1) and $x=1+2t$ (line 2). The same for $y,z.$ Once you get a solution of the system ($s,t$ must be a solution of the $3$ equations) you substitute $s$ in the first line (or $t$ in the second line) and you'll get $P$.
– mfl
Aug 9 at 21:23
Yes. Note that if $(x,y,z)$ is the intersection point then it must be $x=11+4s$ (line 1) and $x=1+2t$ (line 2). The same for $y,z.$ Once you get a solution of the system ($s,t$ must be a solution of the $3$ equations) you substitute $s$ in the first line (or $t$ in the second line) and you'll get $P$.
– mfl
Aug 9 at 21:23
add a comment |Â
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Do you mean $(11,8,2)+s(4,3,-1)?$
– mfl
Aug 9 at 20:55
@mfl The question I found says -1, not -2. But, perhaps there was an error on their part. I am not sure.
– Ella
Aug 9 at 20:58
Sorry about the $-2$. I mean the way you write the line. It is written as [11, 8, 2] + s = [4, 3, -1]
– mfl
Aug 9 at 21:00
@mfl Oh, yes! You’re right. Thank you. Just fixed it.
– Ella
Aug 9 at 21:01
I don't understand c. $R$ is the reflection of $Q$ with respect to what?
– mfl
Aug 9 at 21:06