Finding point coordinates that have been reflected.
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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 $P$.
The coordinates of $P$ are (3, 2, 4).
The lines are also perpendicular.
The point $Q(7,5,3)$ lies on $L_1$. The point $R$ is the reflection of $Q$ in the line $L_2$.
Question: Find the coordinates of $R$.
So, trying to work this out, Q is on $L_1$. $L_1 : r = (11, 8, 2) + s(4, 3, -1)$. And $R$ is on $L_2$. $L_2 : r = (1, 1,-7) + t(2, 1, 11)$
I’m guessing I figure out how $Q$ is correlated to $L_1$. And then relate that to $R$. But how would that work?
coordinate-systems reflection
asked Aug 10 at 20:10
Ella
1189
add a comment |Â
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 $P$.
The coordinates of $P$ are (3, 2, 4).
The lines are also perpendicular.
The point $Q(7,5,3)$ lies on $L_1$. The point $R$ is the reflection of $Q$ in the line $L_2$.
Question: Find the coordinates of $R$.
So, trying to work this out, Q is on $L_1$. $L_1 : r = (11, 8, 2) + s(4, 3, -1)$. And $R$ is on $L_2$. $L_2 : r = (1, 1,-7) + t(2, 1, 11)$
I’m guessing I figure out how $Q$ is correlated to $L_1$. And then relate that to $R$. But how would that work?
coordinate-systems reflection
asked Aug 10 at 20:10
Ella
1189
If two points A,B(end points of position vectors) are mirror images of each other with regard to a line, the the line bisect the line segment AB.
– Lance
Aug 10 at 20:19
This question was asked before by you math.stackexchange.com/questions/2877680/…. You have deletec part c) and asked a new question. I asked for clarification of part c) without answer. Even you include a comment about a part c).
– mfl
Aug 10 at 20:24
@mfl I gave as much information as I was given and I am trying to work out the question to get to the final answer.
– Ella
Aug 10 at 20:26
I think it's reasonable to split out part (c) of the old question into a separate question, especially since you accepted an answer for parts (a) and (b) of the old question. I would also have included a link back to the previous question, since this is a follow-up question. We have that link now via a comment, so it's a moot point now; I merely suggest that in the next situation like this, you include a link in the question.
– David K
Aug 10 at 20:48
add a comment |Â
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 $P$.
The coordinates of $P$ are (3, 2, 4).
The lines are also perpendicular.
The point $Q(7,5,3)$ lies on $L_1$. The point $R$ is the reflection of $Q$ in the line $L_2$.
Question: Find the coordinates of $R$.
So, trying to work this out, Q is on $L_1$. $L_1 : r = (11, 8, 2) + s(4, 3, -1)$. And $R$ is on $L_2$. $L_2 : r = (1, 1,-7) + t(2, 1, 11)$
I’m guessing I figure out how $Q$ is correlated to $L_1$. And then relate that to $R$. But how would that work?
coordinate-systems reflection
asked Aug 10 at 20:10
Ella
1189
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 $P$.
The coordinates of $P$ are (3, 2, 4).
The lines are also perpendicular.
The point $Q(7,5,3)$ lies on $L_1$. The point $R$ is the reflection of $Q$ in the line $L_2$.
Question: Find the coordinates of $R$.
So, trying to work this out, Q is on $L_1$. $L_1 : r = (11, 8, 2) + s(4, 3, -1)$. And $R$ is on $L_2$. $L_2 : r = (1, 1,-7) + t(2, 1, 11)$
I’m guessing I figure out how $Q$ is correlated to $L_1$. And then relate that to $R$. But how would that work?
coordinate-systems reflection
asked Aug 10 at 20:10
Ella
1189
edited Aug 10 at 21:15
asked Aug 10 at 20:10
Ella
1189
asked Aug 10 at 20:10
Ella
1189
asked Aug 10 at 20:10
Ella
1189
1189
If two points A,B(end points of position vectors) are mirror images of each other with regard to a line, the the line bisect the line segment AB.
– Lance
Aug 10 at 20:19
This question was asked before by you math.stackexchange.com/questions/2877680/…. You have deletec part c) and asked a new question. I asked for clarification of part c) without answer. Even you include a comment about a part c).
– mfl
Aug 10 at 20:24
@mfl I gave as much information as I was given and I am trying to work out the question to get to the final answer.
– Ella
Aug 10 at 20:26
I think it's reasonable to split out part (c) of the old question into a separate question, especially since you accepted an answer for parts (a) and (b) of the old question. I would also have included a link back to the previous question, since this is a follow-up question. We have that link now via a comment, so it's a moot point now; I merely suggest that in the next situation like this, you include a link in the question.
– David K
Aug 10 at 20:48
add a comment |Â
If two points A,B(end points of position vectors) are mirror images of each other with regard to a line, the the line bisect the line segment AB.
– Lance
Aug 10 at 20:19
This question was asked before by you math.stackexchange.com/questions/2877680/…. You have deletec part c) and asked a new question. I asked for clarification of part c) without answer. Even you include a comment about a part c).
– mfl
Aug 10 at 20:24
@mfl I gave as much information as I was given and I am trying to work out the question to get to the final answer.
– Ella
Aug 10 at 20:26
I think it's reasonable to split out part (c) of the old question into a separate question, especially since you accepted an answer for parts (a) and (b) of the old question. I would also have included a link back to the previous question, since this is a follow-up question. We have that link now via a comment, so it's a moot point now; I merely suggest that in the next situation like this, you include a link in the question.
– David K
Aug 10 at 20:48
If two points A,B(end points of position vectors) are mirror images of each other with regard to a line, the the line bisect the line segment AB.
– Lance
Aug 10 at 20:19
If two points A,B(end points of position vectors) are mirror images of each other with regard to a line, the the line bisect the line segment AB.
– Lance
Aug 10 at 20:19
This question was asked before by you math.stackexchange.com/questions/2877680/…. You have deletec part c) and asked a new question. I asked for clarification of part c) without answer. Even you include a comment about a part c).
– mfl
Aug 10 at 20:24
This question was asked before by you math.stackexchange.com/questions/2877680/…. You have deletec part c) and asked a new question. I asked for clarification of part c) without answer. Even you include a comment about a part c).
– mfl
Aug 10 at 20:24
@mfl I gave as much information as I was given and I am trying to work out the question to get to the final answer.
– Ella
Aug 10 at 20:26
@mfl I gave as much information as I was given and I am trying to work out the question to get to the final answer.
– Ella
Aug 10 at 20:26
I think it's reasonable to split out part (c) of the old question into a separate question, especially since you accepted an answer for parts (a) and (b) of the old question. I would also have included a link back to the previous question, since this is a follow-up question. We have that link now via a comment, so it's a moot point now; I merely suggest that in the next situation like this, you include a link in the question.
– David K
Aug 10 at 20:48
I think it's reasonable to split out part (c) of the old question into a separate question, especially since you accepted an answer for parts (a) and (b) of the old question. I would also have included a link back to the previous question, since this is a follow-up question. We have that link now via a comment, so it's a moot point now; I merely suggest that in the next situation like this, you include a link in the question.
– David K
Aug 10 at 20:48
add a comment |Â
1 Answer
1
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oldest
votes
up vote
1
down vote
accepted
The reflection of a point $Q$ in a line $L$ can be described as follows:
Drop a perpendicular from $Q$ to the line $L.$
Suppose we say $M$ is the name of the point at the foot of the perpendicular.
That is, $M$ is on $L$ and the segment $QM$ is perpendicular to $L.$
The reflection of $Q$ in the line $L$ is the point (call it $Q'$) that is the same distance from $M$ as $Q$ is, but in the exact opposite direction.
That way, the segment $QQ'$ is perpendicular to $L,$ and $L$ cuts through $QQ'$ at $M,$ exactly midway between $Q$ and $Q'$.
One way to look at this is, if you have $M,$ you can take the vector from $M$ to $Q,$
and then reverse the direction of the vector to get a vector from $M$ to $Q'.$
Now, knowing $M$ and the vector from $M$ to $Q',$ you find $Q'.$
Now review the information in the question statement.
You have the coordinates of $Q.$ You also have the coordinates of $M$ (not called by that name, but if you look at the facts already given in the question, you should find the coordinates there).
So you can find the vector $MQ,$ reverse it, and find $Q'.$
Alternatively, instead of using vectors, use the fact that $M$ is the midpoint of $Q$ and $Q'.$ Therefore
$$ x_M = fracx_Q + x_Q'2, quad
y_M = fracy_Q + y_Q'2, quad textandquad
z_M = fracz_Q + z_Q'2.$$
Put the coordinates of $Q$ and $M$ in these equations, then solve for $x_Q',$ $y_Q',$
and $z_Q'.$
answered Aug 10 at 21:00
David K
48.5k340108
Point $Q$ has $x, y, and z$. So wouldn’t $R$ also have those 3 parts?
– Ella
Aug 10 at 21:17
1
Right, $z$ is dealt with similarly to $x$ and $y.$ I have added that detail to the answer.
– David K
Aug 10 at 21:35
I got (-1,-1, 5) as my answer. Does this seem correct?
– Ella
Aug 10 at 23:34
1
Yes, I get the same.
– David K
Aug 10 at 23:51
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The reflection of a point $Q$ in a line $L$ can be described as follows:
Drop a perpendicular from $Q$ to the line $L.$
Suppose we say $M$ is the name of the point at the foot of the perpendicular.
That is, $M$ is on $L$ and the segment $QM$ is perpendicular to $L.$
The reflection of $Q$ in the line $L$ is the point (call it $Q'$) that is the same distance from $M$ as $Q$ is, but in the exact opposite direction.
That way, the segment $QQ'$ is perpendicular to $L,$ and $L$ cuts through $QQ'$ at $M,$ exactly midway between $Q$ and $Q'$.
One way to look at this is, if you have $M,$ you can take the vector from $M$ to $Q,$
and then reverse the direction of the vector to get a vector from $M$ to $Q'.$
Now, knowing $M$ and the vector from $M$ to $Q',$ you find $Q'.$
Now review the information in the question statement.
You have the coordinates of $Q.$ You also have the coordinates of $M$ (not called by that name, but if you look at the facts already given in the question, you should find the coordinates there).
So you can find the vector $MQ,$ reverse it, and find $Q'.$
Alternatively, instead of using vectors, use the fact that $M$ is the midpoint of $Q$ and $Q'.$ Therefore
$$ x_M = fracx_Q + x_Q'2, quad
y_M = fracy_Q + y_Q'2, quad textandquad
z_M = fracz_Q + z_Q'2.$$
Put the coordinates of $Q$ and $M$ in these equations, then solve for $x_Q',$ $y_Q',$
and $z_Q'.$
answered Aug 10 at 21:00
David K
48.5k340108
Point $Q$ has $x, y, and z$. So wouldn’t $R$ also have those 3 parts?
– Ella
Aug 10 at 21:17
1
Right, $z$ is dealt with similarly to $x$ and $y.$ I have added that detail to the answer.
– David K
Aug 10 at 21:35
I got (-1,-1, 5) as my answer. Does this seem correct?
– Ella
Aug 10 at 23:34
1
Yes, I get the same.
– David K
Aug 10 at 23:51
add a comment |Â
up vote
1
down vote
accepted
The reflection of a point $Q$ in a line $L$ can be described as follows:
Drop a perpendicular from $Q$ to the line $L.$
Suppose we say $M$ is the name of the point at the foot of the perpendicular.
That is, $M$ is on $L$ and the segment $QM$ is perpendicular to $L.$
The reflection of $Q$ in the line $L$ is the point (call it $Q'$) that is the same distance from $M$ as $Q$ is, but in the exact opposite direction.
That way, the segment $QQ'$ is perpendicular to $L,$ and $L$ cuts through $QQ'$ at $M,$ exactly midway between $Q$ and $Q'$.
One way to look at this is, if you have $M,$ you can take the vector from $M$ to $Q,$
and then reverse the direction of the vector to get a vector from $M$ to $Q'.$
Now, knowing $M$ and the vector from $M$ to $Q',$ you find $Q'.$
Now review the information in the question statement.
You have the coordinates of $Q.$ You also have the coordinates of $M$ (not called by that name, but if you look at the facts already given in the question, you should find the coordinates there).
So you can find the vector $MQ,$ reverse it, and find $Q'.$
Alternatively, instead of using vectors, use the fact that $M$ is the midpoint of $Q$ and $Q'.$ Therefore
$$ x_M = fracx_Q + x_Q'2, quad
y_M = fracy_Q + y_Q'2, quad textandquad
z_M = fracz_Q + z_Q'2.$$
Put the coordinates of $Q$ and $M$ in these equations, then solve for $x_Q',$ $y_Q',$
and $z_Q'.$
answered Aug 10 at 21:00
David K
48.5k340108
Point $Q$ has $x, y, and z$. So wouldn’t $R$ also have those 3 parts?
– Ella
Aug 10 at 21:17
1
Right, $z$ is dealt with similarly to $x$ and $y.$ I have added that detail to the answer.
– David K
Aug 10 at 21:35
I got (-1,-1, 5) as my answer. Does this seem correct?
– Ella
Aug 10 at 23:34
1
Yes, I get the same.
– David K
Aug 10 at 23:51
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The reflection of a point $Q$ in a line $L$ can be described as follows:
Drop a perpendicular from $Q$ to the line $L.$
Suppose we say $M$ is the name of the point at the foot of the perpendicular.
That is, $M$ is on $L$ and the segment $QM$ is perpendicular to $L.$
The reflection of $Q$ in the line $L$ is the point (call it $Q'$) that is the same distance from $M$ as $Q$ is, but in the exact opposite direction.
That way, the segment $QQ'$ is perpendicular to $L,$ and $L$ cuts through $QQ'$ at $M,$ exactly midway between $Q$ and $Q'$.
One way to look at this is, if you have $M,$ you can take the vector from $M$ to $Q,$
and then reverse the direction of the vector to get a vector from $M$ to $Q'.$
Now, knowing $M$ and the vector from $M$ to $Q',$ you find $Q'.$
Now review the information in the question statement.
You have the coordinates of $Q.$ You also have the coordinates of $M$ (not called by that name, but if you look at the facts already given in the question, you should find the coordinates there).
So you can find the vector $MQ,$ reverse it, and find $Q'.$
Alternatively, instead of using vectors, use the fact that $M$ is the midpoint of $Q$ and $Q'.$ Therefore
$$ x_M = fracx_Q + x_Q'2, quad
y_M = fracy_Q + y_Q'2, quad textandquad
z_M = fracz_Q + z_Q'2.$$
Put the coordinates of $Q$ and $M$ in these equations, then solve for $x_Q',$ $y_Q',$
and $z_Q'.$
answered Aug 10 at 21:00
David K
48.5k340108
The reflection of a point $Q$ in a line $L$ can be described as follows:
Drop a perpendicular from $Q$ to the line $L.$
Suppose we say $M$ is the name of the point at the foot of the perpendicular.
That is, $M$ is on $L$ and the segment $QM$ is perpendicular to $L.$
The reflection of $Q$ in the line $L$ is the point (call it $Q'$) that is the same distance from $M$ as $Q$ is, but in the exact opposite direction.
That way, the segment $QQ'$ is perpendicular to $L,$ and $L$ cuts through $QQ'$ at $M,$ exactly midway between $Q$ and $Q'$.
One way to look at this is, if you have $M,$ you can take the vector from $M$ to $Q,$
and then reverse the direction of the vector to get a vector from $M$ to $Q'.$
Now, knowing $M$ and the vector from $M$ to $Q',$ you find $Q'.$
Now review the information in the question statement.
You have the coordinates of $Q.$ You also have the coordinates of $M$ (not called by that name, but if you look at the facts already given in the question, you should find the coordinates there).
So you can find the vector $MQ,$ reverse it, and find $Q'.$
Alternatively, instead of using vectors, use the fact that $M$ is the midpoint of $Q$ and $Q'.$ Therefore
$$ x_M = fracx_Q + x_Q'2, quad
y_M = fracy_Q + y_Q'2, quad textandquad
z_M = fracz_Q + z_Q'2.$$
Put the coordinates of $Q$ and $M$ in these equations, then solve for $x_Q',$ $y_Q',$
and $z_Q'.$
answered Aug 10 at 21:00
David K
48.5k340108
edited Aug 10 at 21:34
answered Aug 10 at 21:00
David K
48.5k340108
answered Aug 10 at 21:00
David K
48.5k340108
answered Aug 10 at 21:00
David K
48.5k340108
48.5k340108
Point $Q$ has $x, y, and z$. So wouldn’t $R$ also have those 3 parts?
– Ella
Aug 10 at 21:17
1
Right, $z$ is dealt with similarly to $x$ and $y.$ I have added that detail to the answer.
– David K
Aug 10 at 21:35
I got (-1,-1, 5) as my answer. Does this seem correct?
– Ella
Aug 10 at 23:34
1
Yes, I get the same.
– David K
Aug 10 at 23:51
add a comment |Â
Point $Q$ has $x, y, and z$. So wouldn’t $R$ also have those 3 parts?
– Ella
Aug 10 at 21:17
1
Right, $z$ is dealt with similarly to $x$ and $y.$ I have added that detail to the answer.
– David K
Aug 10 at 21:35
I got (-1,-1, 5) as my answer. Does this seem correct?
– Ella
Aug 10 at 23:34
1
Yes, I get the same.
– David K
Aug 10 at 23:51
Point $Q$ has $x, y, and z$. So wouldn’t $R$ also have those 3 parts?
– Ella
Aug 10 at 21:17
Point $Q$ has $x, y, and z$. So wouldn’t $R$ also have those 3 parts?
– Ella
Aug 10 at 21:17
1
1
Right, $z$ is dealt with similarly to $x$ and $y.$ I have added that detail to the answer.
– David K
Aug 10 at 21:35
Right, $z$ is dealt with similarly to $x$ and $y.$ I have added that detail to the answer.
– David K
Aug 10 at 21:35
I got (-1,-1, 5) as my answer. Does this seem correct?
– Ella
Aug 10 at 23:34
I got (-1,-1, 5) as my answer. Does this seem correct?
– Ella
Aug 10 at 23:34
1
1
Yes, I get the same.
– David K
Aug 10 at 23:51
Yes, I get the same.
– David K
Aug 10 at 23:51
add a comment |Â
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If two points A,B(end points of position vectors) are mirror images of each other with regard to a line, the the line bisect the line segment AB.
– Lance
Aug 10 at 20:19
This question was asked before by you math.stackexchange.com/questions/2877680/…. You have deletec part c) and asked a new question. I asked for clarification of part c) without answer. Even you include a comment about a part c).
– mfl
Aug 10 at 20:24
@mfl I gave as much information as I was given and I am trying to work out the question to get to the final answer.
– Ella
Aug 10 at 20:26
I think it's reasonable to split out part (c) of the old question into a separate question, especially since you accepted an answer for parts (a) and (b) of the old question. I would also have included a link back to the previous question, since this is a follow-up question. We have that link now via a comment, so it's a moot point now; I merely suggest that in the next situation like this, you include a link in the question.
– David K
Aug 10 at 20:48