Scalar Triple Product Area/Volume
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I am not currently in school, but need to relearn some vector algebra for an electrodynamics class. I have a textbook that I am using to teach myself the basics.
Here's the problem, in one section on the scalar triple product (A · (B x C)), it states that geometrically the magnitude of this (|A · (B x C)|) represents the volume of the parallelpiped generated by the 3 vectors A, B and C since |B x C| is the area of the base and |A cos ϴ|
is the height.
I get what they are saying, but mathematically it does not work out that way. Assume A = (1x + 0y + 1z) and B = (0x + 1y + 1z). The length (magnitude) of both A and B is √2. So the area of the base formed by these vectors would be 2; however, A x B = √3.
Can someone explain where I am going wrong? Or is the book wrong?
vector-analysis
asked Jan 10 at 12:47
Mity EltuMity Eltu
1
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add a comment |
$begingroup$
I am not currently in school, but need to relearn some vector algebra for an electrodynamics class. I have a textbook that I am using to teach myself the basics.
Here's the problem, in one section on the scalar triple product (A · (B x C)), it states that geometrically the magnitude of this (|A · (B x C)|) represents the volume of the parallelpiped generated by the 3 vectors A, B and C since |B x C| is the area of the base and |A cos ϴ|
is the height.
I get what they are saying, but mathematically it does not work out that way. Assume A = (1x + 0y + 1z) and B = (0x + 1y + 1z). The length (magnitude) of both A and B is √2. So the area of the base formed by these vectors would be 2; however, A x B = √3.
Can someone explain where I am going wrong? Or is the book wrong?
vector-analysis
asked Jan 10 at 12:47
Mity EltuMity Eltu
1
$endgroup$
add a comment |
$begingroup$
I am not currently in school, but need to relearn some vector algebra for an electrodynamics class. I have a textbook that I am using to teach myself the basics.
Here's the problem, in one section on the scalar triple product (A · (B x C)), it states that geometrically the magnitude of this (|A · (B x C)|) represents the volume of the parallelpiped generated by the 3 vectors A, B and C since |B x C| is the area of the base and |A cos ϴ|
is the height.
I get what they are saying, but mathematically it does not work out that way. Assume A = (1x + 0y + 1z) and B = (0x + 1y + 1z). The length (magnitude) of both A and B is √2. So the area of the base formed by these vectors would be 2; however, A x B = √3.
Can someone explain where I am going wrong? Or is the book wrong?
vector-analysis
asked Jan 10 at 12:47
Mity EltuMity Eltu
1
$endgroup$
I am not currently in school, but need to relearn some vector algebra for an electrodynamics class. I have a textbook that I am using to teach myself the basics.
Here's the problem, in one section on the scalar triple product (A · (B x C)), it states that geometrically the magnitude of this (|A · (B x C)|) represents the volume of the parallelpiped generated by the 3 vectors A, B and C since |B x C| is the area of the base and |A cos ϴ|
is the height.
I get what they are saying, but mathematically it does not work out that way. Assume A = (1x + 0y + 1z) and B = (0x + 1y + 1z). The length (magnitude) of both A and B is √2. So the area of the base formed by these vectors would be 2; however, A x B = √3.
Can someone explain where I am going wrong? Or is the book wrong?
vector-analysis
vector-analysis
asked Jan 10 at 12:47
Mity EltuMity Eltu
1
asked Jan 10 at 12:47
Mity EltuMity Eltu
1
asked Jan 10 at 12:47
Mity EltuMity Eltu
1
asked Jan 10 at 12:47
Mity EltuMity Eltu
1
asked Jan 10 at 12:47
Mity EltuMity Eltu
1
1
add a comment |
add a comment |
1 Answer
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oldest
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The area of the base is indeed a cross product's modulus. Your mistake is in multiplying vectors' lengths as if the base is a rectangle. However, the angle between sides of a parallelogram clearly matters. As a sanity check, note that with an extremely small angle there should be next to no area. As for why the cross product works, note the true height is a side multiplied by a sine (draw a diagram if it helps you see why). When $vec{A},,vec{B}$ are angles separated by an angle $theta$, $|vec{A}timesvec{B}|=|vec{A}||vec{B}||sintheta|$.
answered Jan 10 at 12:52
J.G.J.G.
24.6k22539
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$begingroup$
I missed that. I forgot what the real area of a parallelogram is (B x h) and not side x side (like a square). Thank you. I should have done a bit more thinking on that one.
$endgroup$
– Mity Eltu
Jan 10 at 12:59
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The area of the base is indeed a cross product's modulus. Your mistake is in multiplying vectors' lengths as if the base is a rectangle. However, the angle between sides of a parallelogram clearly matters. As a sanity check, note that with an extremely small angle there should be next to no area. As for why the cross product works, note the true height is a side multiplied by a sine (draw a diagram if it helps you see why). When $vec{A},,vec{B}$ are angles separated by an angle $theta$, $|vec{A}timesvec{B}|=|vec{A}||vec{B}||sintheta|$.
answered Jan 10 at 12:52
J.G.J.G.
24.6k22539
$endgroup$
$begingroup$
I missed that. I forgot what the real area of a parallelogram is (B x h) and not side x side (like a square). Thank you. I should have done a bit more thinking on that one.
$endgroup$
– Mity Eltu
Jan 10 at 12:59
add a comment |
$begingroup$
The area of the base is indeed a cross product's modulus. Your mistake is in multiplying vectors' lengths as if the base is a rectangle. However, the angle between sides of a parallelogram clearly matters. As a sanity check, note that with an extremely small angle there should be next to no area. As for why the cross product works, note the true height is a side multiplied by a sine (draw a diagram if it helps you see why). When $vec{A},,vec{B}$ are angles separated by an angle $theta$, $|vec{A}timesvec{B}|=|vec{A}||vec{B}||sintheta|$.
answered Jan 10 at 12:52
J.G.J.G.
24.6k22539
$endgroup$
$begingroup$
I missed that. I forgot what the real area of a parallelogram is (B x h) and not side x side (like a square). Thank you. I should have done a bit more thinking on that one.
$endgroup$
– Mity Eltu
Jan 10 at 12:59
add a comment |
$begingroup$
The area of the base is indeed a cross product's modulus. Your mistake is in multiplying vectors' lengths as if the base is a rectangle. However, the angle between sides of a parallelogram clearly matters. As a sanity check, note that with an extremely small angle there should be next to no area. As for why the cross product works, note the true height is a side multiplied by a sine (draw a diagram if it helps you see why). When $vec{A},,vec{B}$ are angles separated by an angle $theta$, $|vec{A}timesvec{B}|=|vec{A}||vec{B}||sintheta|$.
answered Jan 10 at 12:52
J.G.J.G.
24.6k22539
$endgroup$
The area of the base is indeed a cross product's modulus. Your mistake is in multiplying vectors' lengths as if the base is a rectangle. However, the angle between sides of a parallelogram clearly matters. As a sanity check, note that with an extremely small angle there should be next to no area. As for why the cross product works, note the true height is a side multiplied by a sine (draw a diagram if it helps you see why). When $vec{A},,vec{B}$ are angles separated by an angle $theta$, $|vec{A}timesvec{B}|=|vec{A}||vec{B}||sintheta|$.
answered Jan 10 at 12:52
J.G.J.G.
24.6k22539
answered Jan 10 at 12:52
J.G.J.G.
24.6k22539
answered Jan 10 at 12:52
J.G.J.G.
24.6k22539
answered Jan 10 at 12:52
J.G.J.G.
24.6k22539
24.6k22539
$begingroup$
I missed that. I forgot what the real area of a parallelogram is (B x h) and not side x side (like a square). Thank you. I should have done a bit more thinking on that one.
$endgroup$
– Mity Eltu
Jan 10 at 12:59
add a comment |
$begingroup$
I missed that. I forgot what the real area of a parallelogram is (B x h) and not side x side (like a square). Thank you. I should have done a bit more thinking on that one.
$endgroup$
– Mity Eltu
Jan 10 at 12:59
$begingroup$
I missed that. I forgot what the real area of a parallelogram is (B x h) and not side x side (like a square). Thank you. I should have done a bit more thinking on that one.
$endgroup$
– Mity Eltu
Jan 10 at 12:59
$begingroup$
I missed that. I forgot what the real area of a parallelogram is (B x h) and not side x side (like a square). Thank you. I should have done a bit more thinking on that one.
$endgroup$
– Mity Eltu
Jan 10 at 12:59
add a comment |
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