Shortest distance between two lines and common perpendicular
$begingroup$
Line 1 has equation: $dfrac{x-8}{3}=dfrac{y+9}{-16}=dfrac{z+1}{-2}$
Line 2 has equation:
$left(begin{matrix}x\y\z
end{matrix}right)=left(begin{matrix}15\29\5
end{matrix}right) + left(begin{matrix}3\8\-5
end{matrix}right)t$
How do you find the shortest distance between lines 1 and 2?
Also how would I find the coordinates of the points where the common perpendicular meets the line 1 and 2? Would I use cross product of direction vectors from both lines? But how would I find the coordinate that starts from?
vectors
edited Dec 1 '15 at 14:30
Alex M.
28.1k103158
asked Dec 1 '15 at 13:43
IvyIvy
95112
$endgroup$
|
show 2 more comments
$begingroup$
Line 1 has equation: $dfrac{x-8}{3}=dfrac{y+9}{-16}=dfrac{z+1}{-2}$
Line 2 has equation:
$left(begin{matrix}x\y\z
end{matrix}right)=left(begin{matrix}15\29\5
end{matrix}right) + left(begin{matrix}3\8\-5
end{matrix}right)t$
How do you find the shortest distance between lines 1 and 2?
Also how would I find the coordinates of the points where the common perpendicular meets the line 1 and 2? Would I use cross product of direction vectors from both lines? But how would I find the coordinate that starts from?
vectors
edited Dec 1 '15 at 14:30
Alex M.
28.1k103158
asked Dec 1 '15 at 13:43
IvyIvy
95112
$endgroup$
$begingroup$
math.stackexchange.com/questions/210848/…
$endgroup$
– user137731
Dec 1 '15 at 14:05
$begingroup$
Hint: to find two points on Line 1, you can look for values for (x;y;z) which result in 0 or 1. This leads to points (8; -9; -1) and (11; -25; -3). From there, you directly get an equation of the second form, including both direction vectors.
$endgroup$
– Axel Kemper
Dec 1 '15 at 14:09
$begingroup$
What do you mean by second form??
$endgroup$
– Ivy
Dec 1 '15 at 14:17
$begingroup$
The shortest distance between two lines is the length of projection of any vector between the two lines onto the cross product of the direction of the two lines.
$endgroup$
– Yiyuan Lee
Dec 1 '15 at 14:20
$begingroup$
The "second form" is the form used for Line 2: point plus multiple of direction vector. Once you have two points, you can use their difference as direction vector and one of the points as starting point.
$endgroup$
– Axel Kemper
Dec 1 '15 at 14:28
|
show 2 more comments
$begingroup$
Line 1 has equation: $dfrac{x-8}{3}=dfrac{y+9}{-16}=dfrac{z+1}{-2}$
Line 2 has equation:
$left(begin{matrix}x\y\z
end{matrix}right)=left(begin{matrix}15\29\5
end{matrix}right) + left(begin{matrix}3\8\-5
end{matrix}right)t$
How do you find the shortest distance between lines 1 and 2?
Also how would I find the coordinates of the points where the common perpendicular meets the line 1 and 2? Would I use cross product of direction vectors from both lines? But how would I find the coordinate that starts from?
vectors
edited Dec 1 '15 at 14:30
Alex M.
28.1k103158
asked Dec 1 '15 at 13:43
IvyIvy
95112
$endgroup$
Line 1 has equation: $dfrac{x-8}{3}=dfrac{y+9}{-16}=dfrac{z+1}{-2}$
Line 2 has equation:
$left(begin{matrix}x\y\z
end{matrix}right)=left(begin{matrix}15\29\5
end{matrix}right) + left(begin{matrix}3\8\-5
end{matrix}right)t$
How do you find the shortest distance between lines 1 and 2?
Also how would I find the coordinates of the points where the common perpendicular meets the line 1 and 2? Would I use cross product of direction vectors from both lines? But how would I find the coordinate that starts from?
vectors
vectors
edited Dec 1 '15 at 14:30
Alex M.
28.1k103158
asked Dec 1 '15 at 13:43
IvyIvy
95112
edited Dec 1 '15 at 14:30
Alex M.
28.1k103158
asked Dec 1 '15 at 13:43
IvyIvy
95112
edited Dec 1 '15 at 14:30
Alex M.
28.1k103158
edited Dec 1 '15 at 14:30
Alex M.
28.1k103158
edited Dec 1 '15 at 14:30
Alex M.
28.1k103158
28.1k103158
asked Dec 1 '15 at 13:43
IvyIvy
95112
asked Dec 1 '15 at 13:43
IvyIvy
95112
asked Dec 1 '15 at 13:43
IvyIvy
95112
95112
$begingroup$
math.stackexchange.com/questions/210848/…
$endgroup$
– user137731
Dec 1 '15 at 14:05
$begingroup$
Hint: to find two points on Line 1, you can look for values for (x;y;z) which result in 0 or 1. This leads to points (8; -9; -1) and (11; -25; -3). From there, you directly get an equation of the second form, including both direction vectors.
$endgroup$
– Axel Kemper
Dec 1 '15 at 14:09
$begingroup$
What do you mean by second form??
$endgroup$
– Ivy
Dec 1 '15 at 14:17
$begingroup$
The shortest distance between two lines is the length of projection of any vector between the two lines onto the cross product of the direction of the two lines.
$endgroup$
– Yiyuan Lee
Dec 1 '15 at 14:20
$begingroup$
The "second form" is the form used for Line 2: point plus multiple of direction vector. Once you have two points, you can use their difference as direction vector and one of the points as starting point.
$endgroup$
– Axel Kemper
Dec 1 '15 at 14:28
|
show 2 more comments
$begingroup$
math.stackexchange.com/questions/210848/…
$endgroup$
– user137731
Dec 1 '15 at 14:05
$begingroup$
Hint: to find two points on Line 1, you can look for values for (x;y;z) which result in 0 or 1. This leads to points (8; -9; -1) and (11; -25; -3). From there, you directly get an equation of the second form, including both direction vectors.
$endgroup$
– Axel Kemper
Dec 1 '15 at 14:09
$begingroup$
What do you mean by second form??
$endgroup$
– Ivy
Dec 1 '15 at 14:17
$begingroup$
The shortest distance between two lines is the length of projection of any vector between the two lines onto the cross product of the direction of the two lines.
$endgroup$
– Yiyuan Lee
Dec 1 '15 at 14:20
$begingroup$
The "second form" is the form used for Line 2: point plus multiple of direction vector. Once you have two points, you can use their difference as direction vector and one of the points as starting point.
$endgroup$
– Axel Kemper
Dec 1 '15 at 14:28
$begingroup$
math.stackexchange.com/questions/210848/…
$endgroup$
– user137731
Dec 1 '15 at 14:05
$begingroup$
math.stackexchange.com/questions/210848/…
$endgroup$
– user137731
Dec 1 '15 at 14:05
$begingroup$
Hint: to find two points on Line 1, you can look for values for (x;y;z) which result in 0 or 1. This leads to points (8; -9; -1) and (11; -25; -3). From there, you directly get an equation of the second form, including both direction vectors.
$endgroup$
– Axel Kemper
Dec 1 '15 at 14:09
$begingroup$
Hint: to find two points on Line 1, you can look for values for (x;y;z) which result in 0 or 1. This leads to points (8; -9; -1) and (11; -25; -3). From there, you directly get an equation of the second form, including both direction vectors.
$endgroup$
– Axel Kemper
Dec 1 '15 at 14:09
$begingroup$
What do you mean by second form??
$endgroup$
– Ivy
Dec 1 '15 at 14:17
$begingroup$
What do you mean by second form??
$endgroup$
– Ivy
Dec 1 '15 at 14:17
$begingroup$
The shortest distance between two lines is the length of projection of any vector between the two lines onto the cross product of the direction of the two lines.
$endgroup$
– Yiyuan Lee
Dec 1 '15 at 14:20
$begingroup$
The shortest distance between two lines is the length of projection of any vector between the two lines onto the cross product of the direction of the two lines.
$endgroup$
– Yiyuan Lee
Dec 1 '15 at 14:20
$begingroup$
The "second form" is the form used for Line 2: point plus multiple of direction vector. Once you have two points, you can use their difference as direction vector and one of the points as starting point.
$endgroup$
– Axel Kemper
Dec 1 '15 at 14:28
$begingroup$
The "second form" is the form used for Line 2: point plus multiple of direction vector. Once you have two points, you can use their difference as direction vector and one of the points as starting point.
$endgroup$
– Axel Kemper
Dec 1 '15 at 14:28
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
My answer is based on Lee Yiyuan's suggestion.
Rewrite the equation of line 1 as
$$left(begin{matrix}x\y\zend{matrix}right)
=left(begin{matrix}8\-9\-1end{matrix}right)
+left(begin{matrix}3\-16\-2end{matrix}right) t.
$$
A vector perpendicular to both lines is
$$begin{vmatrix}i&j&k\3&-16&-2\3&8&-5end{vmatrix}
=96i+9j+72k.$$
One of the vectors joining two lines is
$$left(begin{matrix}8\-9\-1end{matrix}right)
-left(begin{matrix}15\29\5end{matrix}right)
=left(begin{matrix}-7\-38\-6end{matrix}right).
$$
Calculate the projection of this vector to the vector perpendicular to both lines, and then take its absolute value as the final answer.
begin{split}
& mbox{Distance between $L_1$ and $L_2$}\
=&leftlvertfrac{langle(-7,-38,-6),(96,9,72)rangle}{lVert(96,9,72)rVert}rightrvert\
=&leftlvertfrac{-1446}{sqrt{14481}}rightrvert\
=&frac{482}{sqrt{1609}}
end{split}
answered Dec 1 '15 at 14:46
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.9k72445
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
My answer is based on Lee Yiyuan's suggestion.
Rewrite the equation of line 1 as
$$left(begin{matrix}x\y\zend{matrix}right)
=left(begin{matrix}8\-9\-1end{matrix}right)
+left(begin{matrix}3\-16\-2end{matrix}right) t.
$$
A vector perpendicular to both lines is
$$begin{vmatrix}i&j&k\3&-16&-2\3&8&-5end{vmatrix}
=96i+9j+72k.$$
One of the vectors joining two lines is
$$left(begin{matrix}8\-9\-1end{matrix}right)
-left(begin{matrix}15\29\5end{matrix}right)
=left(begin{matrix}-7\-38\-6end{matrix}right).
$$
Calculate the projection of this vector to the vector perpendicular to both lines, and then take its absolute value as the final answer.
begin{split}
& mbox{Distance between $L_1$ and $L_2$}\
=&leftlvertfrac{langle(-7,-38,-6),(96,9,72)rangle}{lVert(96,9,72)rVert}rightrvert\
=&leftlvertfrac{-1446}{sqrt{14481}}rightrvert\
=&frac{482}{sqrt{1609}}
end{split}
answered Dec 1 '15 at 14:46
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.9k72445
$endgroup$
add a comment |
$begingroup$
My answer is based on Lee Yiyuan's suggestion.
Rewrite the equation of line 1 as
$$left(begin{matrix}x\y\zend{matrix}right)
=left(begin{matrix}8\-9\-1end{matrix}right)
+left(begin{matrix}3\-16\-2end{matrix}right) t.
$$
A vector perpendicular to both lines is
$$begin{vmatrix}i&j&k\3&-16&-2\3&8&-5end{vmatrix}
=96i+9j+72k.$$
One of the vectors joining two lines is
$$left(begin{matrix}8\-9\-1end{matrix}right)
-left(begin{matrix}15\29\5end{matrix}right)
=left(begin{matrix}-7\-38\-6end{matrix}right).
$$
Calculate the projection of this vector to the vector perpendicular to both lines, and then take its absolute value as the final answer.
begin{split}
& mbox{Distance between $L_1$ and $L_2$}\
=&leftlvertfrac{langle(-7,-38,-6),(96,9,72)rangle}{lVert(96,9,72)rVert}rightrvert\
=&leftlvertfrac{-1446}{sqrt{14481}}rightrvert\
=&frac{482}{sqrt{1609}}
end{split}
answered Dec 1 '15 at 14:46
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.9k72445
$endgroup$
add a comment |
$begingroup$
My answer is based on Lee Yiyuan's suggestion.
Rewrite the equation of line 1 as
$$left(begin{matrix}x\y\zend{matrix}right)
=left(begin{matrix}8\-9\-1end{matrix}right)
+left(begin{matrix}3\-16\-2end{matrix}right) t.
$$
A vector perpendicular to both lines is
$$begin{vmatrix}i&j&k\3&-16&-2\3&8&-5end{vmatrix}
=96i+9j+72k.$$
One of the vectors joining two lines is
$$left(begin{matrix}8\-9\-1end{matrix}right)
-left(begin{matrix}15\29\5end{matrix}right)
=left(begin{matrix}-7\-38\-6end{matrix}right).
$$
Calculate the projection of this vector to the vector perpendicular to both lines, and then take its absolute value as the final answer.
begin{split}
& mbox{Distance between $L_1$ and $L_2$}\
=&leftlvertfrac{langle(-7,-38,-6),(96,9,72)rangle}{lVert(96,9,72)rVert}rightrvert\
=&leftlvertfrac{-1446}{sqrt{14481}}rightrvert\
=&frac{482}{sqrt{1609}}
end{split}
answered Dec 1 '15 at 14:46
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.9k72445
$endgroup$
My answer is based on Lee Yiyuan's suggestion.
Rewrite the equation of line 1 as
$$left(begin{matrix}x\y\zend{matrix}right)
=left(begin{matrix}8\-9\-1end{matrix}right)
+left(begin{matrix}3\-16\-2end{matrix}right) t.
$$
A vector perpendicular to both lines is
$$begin{vmatrix}i&j&k\3&-16&-2\3&8&-5end{vmatrix}
=96i+9j+72k.$$
One of the vectors joining two lines is
$$left(begin{matrix}8\-9\-1end{matrix}right)
-left(begin{matrix}15\29\5end{matrix}right)
=left(begin{matrix}-7\-38\-6end{matrix}right).
$$
Calculate the projection of this vector to the vector perpendicular to both lines, and then take its absolute value as the final answer.
begin{split}
& mbox{Distance between $L_1$ and $L_2$}\
=&leftlvertfrac{langle(-7,-38,-6),(96,9,72)rangle}{lVert(96,9,72)rVert}rightrvert\
=&leftlvertfrac{-1446}{sqrt{14481}}rightrvert\
=&frac{482}{sqrt{1609}}
end{split}
answered Dec 1 '15 at 14:46
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.9k72445
answered Dec 1 '15 at 14:46
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.9k72445
answered Dec 1 '15 at 14:46
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.9k72445
answered Dec 1 '15 at 14:46
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
12.9k72445
12.9k72445
add a comment |
add a comment |
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$begingroup$
math.stackexchange.com/questions/210848/…
$endgroup$
– user137731
Dec 1 '15 at 14:05
$begingroup$
Hint: to find two points on Line 1, you can look for values for (x;y;z) which result in 0 or 1. This leads to points (8; -9; -1) and (11; -25; -3). From there, you directly get an equation of the second form, including both direction vectors.
$endgroup$
– Axel Kemper
Dec 1 '15 at 14:09
$begingroup$
What do you mean by second form??
$endgroup$
– Ivy
Dec 1 '15 at 14:17
$begingroup$
The shortest distance between two lines is the length of projection of any vector between the two lines onto the cross product of the direction of the two lines.
$endgroup$
– Yiyuan Lee
Dec 1 '15 at 14:20
$begingroup$
The "second form" is the form used for Line 2: point plus multiple of direction vector. Once you have two points, you can use their difference as direction vector and one of the points as starting point.
$endgroup$
– Axel Kemper
Dec 1 '15 at 14:28