Mixing
(High School Vectors) Distance between Parallel Lines
Given $2$ lines
$$L_1 : r = (1, 2, 3) + lambda (1, 2, 1.5)\
L_2 : r = (4, 0, -1) + mu (2, 4, 3)$$
Are parallel to each other with points $A(1, 2, 3)$ and $B(4, 0, -1)$.
Determine the distance between the two lines.
Method 1: distance from a point B to line L1
$AB(to) = OB(to) - OA(to) = (3, -2, -4)$
the distance is
$$frac{|(3, -2, -4) times (1, 2, 1.5) |}{sqrt{1 + 4 + 2.25}}= 4.716$$
Method 2: foot of perpendicular
Let foot of perpendicular of $A$ on $L2$ be $F$
$$OF(→) = (4+2mu, 4mu, -1+3mu) text{ for some }muinmathbb R\
AF(→) = (3+2mu, 4mu-2, -4+3mu) $$
Since $AF$ perpendicular to $L2$ :
$$(3+2mu, 4mu-2, -4+3mu) cdot (2, 4, 3) = 0\
μ = frac{14}{29}\
AF(to) = left(frac{115}{29}, frac{56}{29}, -frac{74}{29}right)\
|AF(to)| = 5.095$$
Method 3: Length of projection and Pythagoras Theorem
Length of projection of vector $vec{AB}$ onto $L_1$:
$$frac{|(3, -2, -4)cdot (1, 2, 1.5)|}{sqrt{1 + 4 + 2.25}} = 2.90065391$$
Length of $vec{AB} = | (3, -2, -4) |$.
Distance between two lines by Pythagoras theorem:
$$sqrt{| (3, -2, -4) |^2 - 2.90065391^2 } = 4.537237 . $$
Why do I get different answers for methods? I believe the three methods are correct.
vectors
asked Sep 13 '15 at 8:39
NetUser5y62
391114
add a comment |
Given $2$ lines
$$L_1 : r = (1, 2, 3) + lambda (1, 2, 1.5)\
L_2 : r = (4, 0, -1) + mu (2, 4, 3)$$
Are parallel to each other with points $A(1, 2, 3)$ and $B(4, 0, -1)$.
Determine the distance between the two lines.
Method 1: distance from a point B to line L1
$AB(to) = OB(to) - OA(to) = (3, -2, -4)$
the distance is
$$frac{|(3, -2, -4) times (1, 2, 1.5) |}{sqrt{1 + 4 + 2.25}}= 4.716$$
Method 2: foot of perpendicular
Let foot of perpendicular of $A$ on $L2$ be $F$
$$OF(→) = (4+2mu, 4mu, -1+3mu) text{ for some }muinmathbb R\
AF(→) = (3+2mu, 4mu-2, -4+3mu) $$
Since $AF$ perpendicular to $L2$ :
$$(3+2mu, 4mu-2, -4+3mu) cdot (2, 4, 3) = 0\
μ = frac{14}{29}\
AF(to) = left(frac{115}{29}, frac{56}{29}, -frac{74}{29}right)\
|AF(to)| = 5.095$$
Method 3: Length of projection and Pythagoras Theorem
Length of projection of vector $vec{AB}$ onto $L_1$:
$$frac{|(3, -2, -4)cdot (1, 2, 1.5)|}{sqrt{1 + 4 + 2.25}} = 2.90065391$$
Length of $vec{AB} = | (3, -2, -4) |$.
Distance between two lines by Pythagoras theorem:
$$sqrt{| (3, -2, -4) |^2 - 2.90065391^2 } = 4.537237 . $$
Why do I get different answers for methods? I believe the three methods are correct.
vectors
asked Sep 13 '15 at 8:39
NetUser5y62
391114
Why would anyone downvote this question and want to close it? It's a perfectly legitimate question, OP is confused about a mathematical topic and wants help. It could use some latex, but that's hardly a reason to close it...
– 5xum
Sep 13 '15 at 8:54
add a comment |
Given $2$ lines
$$L_1 : r = (1, 2, 3) + lambda (1, 2, 1.5)\
L_2 : r = (4, 0, -1) + mu (2, 4, 3)$$
Are parallel to each other with points $A(1, 2, 3)$ and $B(4, 0, -1)$.
Determine the distance between the two lines.
Method 1: distance from a point B to line L1
$AB(to) = OB(to) - OA(to) = (3, -2, -4)$
the distance is
$$frac{|(3, -2, -4) times (1, 2, 1.5) |}{sqrt{1 + 4 + 2.25}}= 4.716$$
Method 2: foot of perpendicular
Let foot of perpendicular of $A$ on $L2$ be $F$
$$OF(→) = (4+2mu, 4mu, -1+3mu) text{ for some }muinmathbb R\
AF(→) = (3+2mu, 4mu-2, -4+3mu) $$
Since $AF$ perpendicular to $L2$ :
$$(3+2mu, 4mu-2, -4+3mu) cdot (2, 4, 3) = 0\
μ = frac{14}{29}\
AF(to) = left(frac{115}{29}, frac{56}{29}, -frac{74}{29}right)\
|AF(to)| = 5.095$$
Method 3: Length of projection and Pythagoras Theorem
Length of projection of vector $vec{AB}$ onto $L_1$:
$$frac{|(3, -2, -4)cdot (1, 2, 1.5)|}{sqrt{1 + 4 + 2.25}} = 2.90065391$$
Length of $vec{AB} = | (3, -2, -4) |$.
Distance between two lines by Pythagoras theorem:
$$sqrt{| (3, -2, -4) |^2 - 2.90065391^2 } = 4.537237 . $$
Why do I get different answers for methods? I believe the three methods are correct.
vectors
asked Sep 13 '15 at 8:39
NetUser5y62
391114
Given $2$ lines
$$L_1 : r = (1, 2, 3) + lambda (1, 2, 1.5)\
L_2 : r = (4, 0, -1) + mu (2, 4, 3)$$
Are parallel to each other with points $A(1, 2, 3)$ and $B(4, 0, -1)$.
Determine the distance between the two lines.
Method 1: distance from a point B to line L1
$AB(to) = OB(to) - OA(to) = (3, -2, -4)$
the distance is
$$frac{|(3, -2, -4) times (1, 2, 1.5) |}{sqrt{1 + 4 + 2.25}}= 4.716$$
Method 2: foot of perpendicular
Let foot of perpendicular of $A$ on $L2$ be $F$
$$OF(→) = (4+2mu, 4mu, -1+3mu) text{ for some }muinmathbb R\
AF(→) = (3+2mu, 4mu-2, -4+3mu) $$
Since $AF$ perpendicular to $L2$ :
$$(3+2mu, 4mu-2, -4+3mu) cdot (2, 4, 3) = 0\
μ = frac{14}{29}\
AF(to) = left(frac{115}{29}, frac{56}{29}, -frac{74}{29}right)\
|AF(to)| = 5.095$$
Method 3: Length of projection and Pythagoras Theorem
Length of projection of vector $vec{AB}$ onto $L_1$:
$$frac{|(3, -2, -4)cdot (1, 2, 1.5)|}{sqrt{1 + 4 + 2.25}} = 2.90065391$$
Length of $vec{AB} = | (3, -2, -4) |$.
Distance between two lines by Pythagoras theorem:
$$sqrt{| (3, -2, -4) |^2 - 2.90065391^2 } = 4.537237 . $$
Why do I get different answers for methods? I believe the three methods are correct.
vectors
vectors
asked Sep 13 '15 at 8:39
NetUser5y62
391114
asked Sep 13 '15 at 8:39
NetUser5y62
391114
edited Nov 20 '18 at 9:23
asked Sep 13 '15 at 8:39
NetUser5y62
391114
asked Sep 13 '15 at 8:39
NetUser5y62
391114
asked Sep 13 '15 at 8:39
NetUser5y62
391114
391114
Why would anyone downvote this question and want to close it? It's a perfectly legitimate question, OP is confused about a mathematical topic and wants help. It could use some latex, but that's hardly a reason to close it...
– 5xum
Sep 13 '15 at 8:54
add a comment |
Why would anyone downvote this question and want to close it? It's a perfectly legitimate question, OP is confused about a mathematical topic and wants help. It could use some latex, but that's hardly a reason to close it...
– 5xum
Sep 13 '15 at 8:54
Why would anyone downvote this question and want to close it? It's a perfectly legitimate question, OP is confused about a mathematical topic and wants help. It could use some latex, but that's hardly a reason to close it...
– 5xum
Sep 13 '15 at 8:54
Why would anyone downvote this question and want to close it? It's a perfectly legitimate question, OP is confused about a mathematical topic and wants help. It could use some latex, but that's hardly a reason to close it...
– 5xum
Sep 13 '15 at 8:54
add a comment |
1 Answer
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votes
The vector $AF$ should be $$left(frac{115}{29}, -frac{2}{29}, -frac{74}{29}right)$$
This vector has a norm of approximately $4.716$ which is what you hot with method 1.
Your steps were correct up to the point where you inserted $mu$ into the formula
$$AF = (3+2mu, 4mu - 2, -4+3mu)$$
since
$$3+2cdot frac{14}{29} = frac{87 + 28}{29} = frac{115}{29}\
4cdot frac{14}{29} - 2 = frac{56 - 58}{29} = -frac{2}{29}\
-4+3cdotfrac{14}{29} = frac{-116 + 42}{29} = -frac{74}{29}$$
answered Sep 13 '15 at 9:32
5xum
89.5k393161
@GuoJinLong Please format it correctly, like the rest of your question. Also, do this in future questions.
– 5xum
Sep 13 '15 at 9:46
@GuoJinLong You made a mistake in calculating the length of projection. Redo the calculation to get the right result.
– 5xum
Sep 13 '15 at 11:53
add a comment |
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1 Answer
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The vector $AF$ should be $$left(frac{115}{29}, -frac{2}{29}, -frac{74}{29}right)$$
This vector has a norm of approximately $4.716$ which is what you hot with method 1.
Your steps were correct up to the point where you inserted $mu$ into the formula
$$AF = (3+2mu, 4mu - 2, -4+3mu)$$
since
$$3+2cdot frac{14}{29} = frac{87 + 28}{29} = frac{115}{29}\
4cdot frac{14}{29} - 2 = frac{56 - 58}{29} = -frac{2}{29}\
-4+3cdotfrac{14}{29} = frac{-116 + 42}{29} = -frac{74}{29}$$
answered Sep 13 '15 at 9:32
5xum
89.5k393161
@GuoJinLong Please format it correctly, like the rest of your question. Also, do this in future questions.
– 5xum
Sep 13 '15 at 9:46
@GuoJinLong You made a mistake in calculating the length of projection. Redo the calculation to get the right result.
– 5xum
Sep 13 '15 at 11:53
add a comment |
The vector $AF$ should be $$left(frac{115}{29}, -frac{2}{29}, -frac{74}{29}right)$$
This vector has a norm of approximately $4.716$ which is what you hot with method 1.
Your steps were correct up to the point where you inserted $mu$ into the formula
$$AF = (3+2mu, 4mu - 2, -4+3mu)$$
since
$$3+2cdot frac{14}{29} = frac{87 + 28}{29} = frac{115}{29}\
4cdot frac{14}{29} - 2 = frac{56 - 58}{29} = -frac{2}{29}\
-4+3cdotfrac{14}{29} = frac{-116 + 42}{29} = -frac{74}{29}$$
answered Sep 13 '15 at 9:32
5xum
89.5k393161
@GuoJinLong Please format it correctly, like the rest of your question. Also, do this in future questions.
– 5xum
Sep 13 '15 at 9:46
@GuoJinLong You made a mistake in calculating the length of projection. Redo the calculation to get the right result.
– 5xum
Sep 13 '15 at 11:53
add a comment |
The vector $AF$ should be $$left(frac{115}{29}, -frac{2}{29}, -frac{74}{29}right)$$
This vector has a norm of approximately $4.716$ which is what you hot with method 1.
Your steps were correct up to the point where you inserted $mu$ into the formula
$$AF = (3+2mu, 4mu - 2, -4+3mu)$$
since
$$3+2cdot frac{14}{29} = frac{87 + 28}{29} = frac{115}{29}\
4cdot frac{14}{29} - 2 = frac{56 - 58}{29} = -frac{2}{29}\
-4+3cdotfrac{14}{29} = frac{-116 + 42}{29} = -frac{74}{29}$$
answered Sep 13 '15 at 9:32
5xum
89.5k393161
The vector $AF$ should be $$left(frac{115}{29}, -frac{2}{29}, -frac{74}{29}right)$$
This vector has a norm of approximately $4.716$ which is what you hot with method 1.
Your steps were correct up to the point where you inserted $mu$ into the formula
$$AF = (3+2mu, 4mu - 2, -4+3mu)$$
since
$$3+2cdot frac{14}{29} = frac{87 + 28}{29} = frac{115}{29}\
4cdot frac{14}{29} - 2 = frac{56 - 58}{29} = -frac{2}{29}\
-4+3cdotfrac{14}{29} = frac{-116 + 42}{29} = -frac{74}{29}$$
answered Sep 13 '15 at 9:32
5xum
89.5k393161
edited Sep 13 '15 at 9:37
answered Sep 13 '15 at 9:32
5xum
89.5k393161
answered Sep 13 '15 at 9:32
5xum
89.5k393161
answered Sep 13 '15 at 9:32
5xum
89.5k393161
89.5k393161
@GuoJinLong Please format it correctly, like the rest of your question. Also, do this in future questions.
– 5xum
Sep 13 '15 at 9:46
@GuoJinLong You made a mistake in calculating the length of projection. Redo the calculation to get the right result.
– 5xum
Sep 13 '15 at 11:53
add a comment |
@GuoJinLong Please format it correctly, like the rest of your question. Also, do this in future questions.
– 5xum
Sep 13 '15 at 9:46
@GuoJinLong You made a mistake in calculating the length of projection. Redo the calculation to get the right result.
– 5xum
Sep 13 '15 at 11:53
@GuoJinLong Please format it correctly, like the rest of your question. Also, do this in future questions.
– 5xum
Sep 13 '15 at 9:46
@GuoJinLong Please format it correctly, like the rest of your question. Also, do this in future questions.
– 5xum
Sep 13 '15 at 9:46
@GuoJinLong You made a mistake in calculating the length of projection. Redo the calculation to get the right result.
– 5xum
Sep 13 '15 at 11:53
@GuoJinLong You made a mistake in calculating the length of projection. Redo the calculation to get the right result.
– 5xum
Sep 13 '15 at 11:53
add a comment |
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Why would anyone downvote this question and want to close it? It's a perfectly legitimate question, OP is confused about a mathematical topic and wants help. It could use some latex, but that's hardly a reason to close it...
– 5xum
Sep 13 '15 at 8:54