Prove the area of parallelogram
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How to prove the length of the cross product axb is equal to the area of parallelogram determined by a and b?
vectors
asked Nov 25 at 14:45
Amber
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up vote
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favorite
How to prove the length of the cross product axb is equal to the area of parallelogram determined by a and b?
vectors
asked Nov 25 at 14:45
Amber
11
What does projection of the vector $atimes b$ on to the plane of $a$ and $b$ give?
– Yadati Kiran
Nov 25 at 14:48
$a times b = |a||b|sintheta$
– Larry
Nov 25 at 15:02
Ap=|axb|. How to prove the a and b by visually, vectors?
– Amber
Nov 25 at 15:18
Depends on how you've defined the cross product.
– Michael Hoppe
Nov 25 at 16:32
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
How to prove the length of the cross product axb is equal to the area of parallelogram determined by a and b?
vectors
asked Nov 25 at 14:45
Amber
11
How to prove the length of the cross product axb is equal to the area of parallelogram determined by a and b?
vectors
vectors
asked Nov 25 at 14:45
Amber
11
asked Nov 25 at 14:45
Amber
11
asked Nov 25 at 14:45
Amber
11
asked Nov 25 at 14:45
Amber
11
asked Nov 25 at 14:45
Amber
11
11
What does projection of the vector $atimes b$ on to the plane of $a$ and $b$ give?
– Yadati Kiran
Nov 25 at 14:48
$a times b = |a||b|sintheta$
– Larry
Nov 25 at 15:02
Ap=|axb|. How to prove the a and b by visually, vectors?
– Amber
Nov 25 at 15:18
Depends on how you've defined the cross product.
– Michael Hoppe
Nov 25 at 16:32
add a comment |
What does projection of the vector $atimes b$ on to the plane of $a$ and $b$ give?
– Yadati Kiran
Nov 25 at 14:48
$a times b = |a||b|sintheta$
– Larry
Nov 25 at 15:02
Ap=|axb|. How to prove the a and b by visually, vectors?
– Amber
Nov 25 at 15:18
Depends on how you've defined the cross product.
– Michael Hoppe
Nov 25 at 16:32
What does projection of the vector $atimes b$ on to the plane of $a$ and $b$ give?
– Yadati Kiran
Nov 25 at 14:48
What does projection of the vector $atimes b$ on to the plane of $a$ and $b$ give?
– Yadati Kiran
Nov 25 at 14:48
$a times b = |a||b|sintheta$
– Larry
Nov 25 at 15:02
$a times b = |a||b|sintheta$
– Larry
Nov 25 at 15:02
Ap=|axb|. How to prove the a and b by visually, vectors?
– Amber
Nov 25 at 15:18
Ap=|axb|. How to prove the a and b by visually, vectors?
– Amber
Nov 25 at 15:18
Depends on how you've defined the cross product.
– Michael Hoppe
Nov 25 at 16:32
Depends on how you've defined the cross product.
– Michael Hoppe
Nov 25 at 16:32
add a comment |
1 Answer
1
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oldest
votes
up vote
0
down vote
Presumably you've defined
$$atimes b=detbegin{pmatrix}i&j&k\ a_1&a_2&a_3\ b_1&b_2&b_3end{pmatrix}.$$
Now write that out, calculate
$$|atimes b|^2=(a_2b_3-a_3b_2)^2+(a_1b_3-a_1b_3)^2+(a_1b_2-a_2b_1)^2$$
and show that this squared length equals $|a|^2|b|^2-langle a,brangle^2$, which is known (?) to be the squared area of the parallelogram spanned by $a$ and $b$.
answered Nov 25 at 16:50
Michael Hoppe
10.6k31733
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Presumably you've defined
$$atimes b=detbegin{pmatrix}i&j&k\ a_1&a_2&a_3\ b_1&b_2&b_3end{pmatrix}.$$
Now write that out, calculate
$$|atimes b|^2=(a_2b_3-a_3b_2)^2+(a_1b_3-a_1b_3)^2+(a_1b_2-a_2b_1)^2$$
and show that this squared length equals $|a|^2|b|^2-langle a,brangle^2$, which is known (?) to be the squared area of the parallelogram spanned by $a$ and $b$.
answered Nov 25 at 16:50
Michael Hoppe
10.6k31733
add a comment |
up vote
0
down vote
Presumably you've defined
$$atimes b=detbegin{pmatrix}i&j&k\ a_1&a_2&a_3\ b_1&b_2&b_3end{pmatrix}.$$
Now write that out, calculate
$$|atimes b|^2=(a_2b_3-a_3b_2)^2+(a_1b_3-a_1b_3)^2+(a_1b_2-a_2b_1)^2$$
and show that this squared length equals $|a|^2|b|^2-langle a,brangle^2$, which is known (?) to be the squared area of the parallelogram spanned by $a$ and $b$.
answered Nov 25 at 16:50
Michael Hoppe
10.6k31733
add a comment |
up vote
0
down vote
up vote
0
down vote
Presumably you've defined
$$atimes b=detbegin{pmatrix}i&j&k\ a_1&a_2&a_3\ b_1&b_2&b_3end{pmatrix}.$$
Now write that out, calculate
$$|atimes b|^2=(a_2b_3-a_3b_2)^2+(a_1b_3-a_1b_3)^2+(a_1b_2-a_2b_1)^2$$
and show that this squared length equals $|a|^2|b|^2-langle a,brangle^2$, which is known (?) to be the squared area of the parallelogram spanned by $a$ and $b$.
answered Nov 25 at 16:50
Michael Hoppe
10.6k31733
Presumably you've defined
$$atimes b=detbegin{pmatrix}i&j&k\ a_1&a_2&a_3\ b_1&b_2&b_3end{pmatrix}.$$
Now write that out, calculate
$$|atimes b|^2=(a_2b_3-a_3b_2)^2+(a_1b_3-a_1b_3)^2+(a_1b_2-a_2b_1)^2$$
and show that this squared length equals $|a|^2|b|^2-langle a,brangle^2$, which is known (?) to be the squared area of the parallelogram spanned by $a$ and $b$.
answered Nov 25 at 16:50
Michael Hoppe
10.6k31733
edited Nov 25 at 16:57
answered Nov 25 at 16:50
Michael Hoppe
10.6k31733
answered Nov 25 at 16:50
Michael Hoppe
10.6k31733
answered Nov 25 at 16:50
Michael Hoppe
10.6k31733
10.6k31733
add a comment |
add a comment |
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What does projection of the vector $atimes b$ on to the plane of $a$ and $b$ give?
– Yadati Kiran
Nov 25 at 14:48
$a times b = |a||b|sintheta$
– Larry
Nov 25 at 15:02
Ap=|axb|. How to prove the a and b by visually, vectors?
– Amber
Nov 25 at 15:18
Depends on how you've defined the cross product.
– Michael Hoppe
Nov 25 at 16:32