Mixing
Cross product as area of triangle and parallelogram
As area of triangle is equal to area of parallelogram when two vectors are represented by two diagonals of a parallelogram. How can I prove it?
Please let me know how prove it ??
vectors
edited Nov 20 '18 at 12:48
RAM_3R
527214
asked Nov 20 '18 at 12:25
tapus aggarwal
82
add a comment |
As area of triangle is equal to area of parallelogram when two vectors are represented by two diagonals of a parallelogram. How can I prove it?
Please let me know how prove it ??
vectors
edited Nov 20 '18 at 12:48
RAM_3R
527214
asked Nov 20 '18 at 12:25
tapus aggarwal
82
For a parallelogram $bsin theta = h$
– Phil H
Nov 20 '18 at 13:50
What have you tried ? Take any two arbitrary vectors and calculate both the areas. See if they are equal.
– PSG
Nov 20 '18 at 13:58
add a comment |
As area of triangle is equal to area of parallelogram when two vectors are represented by two diagonals of a parallelogram. How can I prove it?
Please let me know how prove it ??
vectors
edited Nov 20 '18 at 12:48
RAM_3R
527214
asked Nov 20 '18 at 12:25
tapus aggarwal
82
As area of triangle is equal to area of parallelogram when two vectors are represented by two diagonals of a parallelogram. How can I prove it?
Please let me know how prove it ??
vectors
vectors
edited Nov 20 '18 at 12:48
RAM_3R
527214
asked Nov 20 '18 at 12:25
tapus aggarwal
82
edited Nov 20 '18 at 12:48
RAM_3R
527214
asked Nov 20 '18 at 12:25
tapus aggarwal
82
edited Nov 20 '18 at 12:48
RAM_3R
527214
edited Nov 20 '18 at 12:48
RAM_3R
527214
edited Nov 20 '18 at 12:48
RAM_3R
527214
527214
asked Nov 20 '18 at 12:25
tapus aggarwal
82
asked Nov 20 '18 at 12:25
tapus aggarwal
82
asked Nov 20 '18 at 12:25
tapus aggarwal
82
82
For a parallelogram $bsin theta = h$
– Phil H
Nov 20 '18 at 13:50
What have you tried ? Take any two arbitrary vectors and calculate both the areas. See if they are equal.
– PSG
Nov 20 '18 at 13:58
add a comment |
For a parallelogram $bsin theta = h$
– Phil H
Nov 20 '18 at 13:50
What have you tried ? Take any two arbitrary vectors and calculate both the areas. See if they are equal.
– PSG
Nov 20 '18 at 13:58
For a parallelogram $bsin theta = h$
– Phil H
Nov 20 '18 at 13:50
For a parallelogram $bsin theta = h$
– Phil H
Nov 20 '18 at 13:50
What have you tried ? Take any two arbitrary vectors and calculate both the areas. See if they are equal.
– PSG
Nov 20 '18 at 13:58
What have you tried ? Take any two arbitrary vectors and calculate both the areas. See if they are equal.
– PSG
Nov 20 '18 at 13:58
add a comment |
1 Answer
1
active
oldest
votes
The area of a triangle can be written as half the cross product of any of its three sides i.e.
$$Areatriangle=frac{1}{2}vec{a}^{}timesvec{b}^{}=frac{1}{2}absintheta=frac{1}{2}base.height$$
Similarly, the area of a parallelogram can be written as $$Area=base.height=2.[frac{1}{2}diagonal_1times(frac{1}{2} diagonal_2)]=frac{1}{2}vec{diagonal_1}^{}times vec{diagonal_2}^{}$$ because diagonals of a parallelogram bisect each other so, in the 2 triangles created by a diagonal in a parallelogram the area can be written as above where height is just $frac{1}{2} diagonal_2sintheta$ where theta is the angle between the diagonals. This is just application of above area of triangle formula.
answered Nov 20 '18 at 14:57
Mustang
3067
Here I had used $times$ as the cross product between vectors sorry for any confusion.
– Mustang
Nov 20 '18 at 15:57
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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active
oldest
votes
The area of a triangle can be written as half the cross product of any of its three sides i.e.
$$Areatriangle=frac{1}{2}vec{a}^{}timesvec{b}^{}=frac{1}{2}absintheta=frac{1}{2}base.height$$
Similarly, the area of a parallelogram can be written as $$Area=base.height=2.[frac{1}{2}diagonal_1times(frac{1}{2} diagonal_2)]=frac{1}{2}vec{diagonal_1}^{}times vec{diagonal_2}^{}$$ because diagonals of a parallelogram bisect each other so, in the 2 triangles created by a diagonal in a parallelogram the area can be written as above where height is just $frac{1}{2} diagonal_2sintheta$ where theta is the angle between the diagonals. This is just application of above area of triangle formula.
answered Nov 20 '18 at 14:57
Mustang
3067
Here I had used $times$ as the cross product between vectors sorry for any confusion.
– Mustang
Nov 20 '18 at 15:57
add a comment |
The area of a triangle can be written as half the cross product of any of its three sides i.e.
$$Areatriangle=frac{1}{2}vec{a}^{}timesvec{b}^{}=frac{1}{2}absintheta=frac{1}{2}base.height$$
Similarly, the area of a parallelogram can be written as $$Area=base.height=2.[frac{1}{2}diagonal_1times(frac{1}{2} diagonal_2)]=frac{1}{2}vec{diagonal_1}^{}times vec{diagonal_2}^{}$$ because diagonals of a parallelogram bisect each other so, in the 2 triangles created by a diagonal in a parallelogram the area can be written as above where height is just $frac{1}{2} diagonal_2sintheta$ where theta is the angle between the diagonals. This is just application of above area of triangle formula.
answered Nov 20 '18 at 14:57
Mustang
3067
Here I had used $times$ as the cross product between vectors sorry for any confusion.
– Mustang
Nov 20 '18 at 15:57
add a comment |
The area of a triangle can be written as half the cross product of any of its three sides i.e.
$$Areatriangle=frac{1}{2}vec{a}^{}timesvec{b}^{}=frac{1}{2}absintheta=frac{1}{2}base.height$$
Similarly, the area of a parallelogram can be written as $$Area=base.height=2.[frac{1}{2}diagonal_1times(frac{1}{2} diagonal_2)]=frac{1}{2}vec{diagonal_1}^{}times vec{diagonal_2}^{}$$ because diagonals of a parallelogram bisect each other so, in the 2 triangles created by a diagonal in a parallelogram the area can be written as above where height is just $frac{1}{2} diagonal_2sintheta$ where theta is the angle between the diagonals. This is just application of above area of triangle formula.
answered Nov 20 '18 at 14:57
Mustang
3067
The area of a triangle can be written as half the cross product of any of its three sides i.e.
$$Areatriangle=frac{1}{2}vec{a}^{}timesvec{b}^{}=frac{1}{2}absintheta=frac{1}{2}base.height$$
Similarly, the area of a parallelogram can be written as $$Area=base.height=2.[frac{1}{2}diagonal_1times(frac{1}{2} diagonal_2)]=frac{1}{2}vec{diagonal_1}^{}times vec{diagonal_2}^{}$$ because diagonals of a parallelogram bisect each other so, in the 2 triangles created by a diagonal in a parallelogram the area can be written as above where height is just $frac{1}{2} diagonal_2sintheta$ where theta is the angle between the diagonals. This is just application of above area of triangle formula.
answered Nov 20 '18 at 14:57
Mustang
3067
edited Nov 21 '18 at 12:16
answered Nov 20 '18 at 14:57
Mustang
3067
answered Nov 20 '18 at 14:57
Mustang
3067
answered Nov 20 '18 at 14:57
Mustang
3067
3067
Here I had used $times$ as the cross product between vectors sorry for any confusion.
– Mustang
Nov 20 '18 at 15:57
add a comment |
Here I had used $times$ as the cross product between vectors sorry for any confusion.
– Mustang
Nov 20 '18 at 15:57
Here I had used $times$ as the cross product between vectors sorry for any confusion.
– Mustang
Nov 20 '18 at 15:57
Here I had used $times$ as the cross product between vectors sorry for any confusion.
– Mustang
Nov 20 '18 at 15:57
add a comment |
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For a parallelogram $bsin theta = h$
– Phil H
Nov 20 '18 at 13:50
What have you tried ? Take any two arbitrary vectors and calculate both the areas. See if they are equal.
– PSG
Nov 20 '18 at 13:58