How can the area of a paralellogram formed by two 3D vectors be computed without using the cross product?
$begingroup$
How can the area of the parallelogram formed by two 3D vectors
$$mathbf{u} = [a, b, c], space mathbf{v} = [e, f, g]$$
be computed without using the cross product?
linear-algebra
$endgroup$
add a comment |
$begingroup$
How can the area of the parallelogram formed by two 3D vectors
$$mathbf{u} = [a, b, c], space mathbf{v} = [e, f, g]$$
be computed without using the cross product?
linear-algebra
$endgroup$
2
$begingroup$
You may perhaps calculate the angle $alpha$ between $u$ and $v$ via dot product and then use $A=|u||v|sinalpha$, but it is basically the ugly way to calculate the cross product.
$endgroup$
– A.Γ.
Jan 2 at 20:22
add a comment |
$begingroup$
How can the area of the parallelogram formed by two 3D vectors
$$mathbf{u} = [a, b, c], space mathbf{v} = [e, f, g]$$
be computed without using the cross product?
linear-algebra
$endgroup$
How can the area of the parallelogram formed by two 3D vectors
$$mathbf{u} = [a, b, c], space mathbf{v} = [e, f, g]$$
be computed without using the cross product?
linear-algebra
linear-algebra
edited Jan 2 at 20:23
Bernard
122k740116
122k740116
asked Jan 2 at 20:16
K. ClaessonK. Claesson
18110
18110
2
$begingroup$
You may perhaps calculate the angle $alpha$ between $u$ and $v$ via dot product and then use $A=|u||v|sinalpha$, but it is basically the ugly way to calculate the cross product.
$endgroup$
– A.Γ.
Jan 2 at 20:22
add a comment |
2
$begingroup$
You may perhaps calculate the angle $alpha$ between $u$ and $v$ via dot product and then use $A=|u||v|sinalpha$, but it is basically the ugly way to calculate the cross product.
$endgroup$
– A.Γ.
Jan 2 at 20:22
2
2
$begingroup$
You may perhaps calculate the angle $alpha$ between $u$ and $v$ via dot product and then use $A=|u||v|sinalpha$, but it is basically the ugly way to calculate the cross product.
$endgroup$
– A.Γ.
Jan 2 at 20:22
$begingroup$
You may perhaps calculate the angle $alpha$ between $u$ and $v$ via dot product and then use $A=|u||v|sinalpha$, but it is basically the ugly way to calculate the cross product.
$endgroup$
– A.Γ.
Jan 2 at 20:22
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The square of the area is the determinant
$$begin{vmatrix}
bf{u}cdotbf u&bf{u}cdotbf v\
bf{v}cdotbf u&
bf{v}cdotbf v
end{vmatrix}.$$
$endgroup$
$begingroup$
Could you explain why this is the square of the area?
$endgroup$
– K. Claesson
Jan 2 at 20:33
2
$begingroup$
@K.Claesson Recall that $A^2=|mathbf u|^2|mathbf v|^2sin^2alpha=|mathbf u|^2|mathbf v|^2(1-cos^2alpha)=|mathbf u|^2|mathbf v|^2-(mathbf{ucdot v})^2$
$endgroup$
– Shubham Johri
Jan 2 at 20:49
add a comment |
$begingroup$
Rewrite $mathbf v$ as the sum of two components, one parallel to $mathbf u$ and one orthogonal to $mathbf u.$
Let $mathbf v_perp$ be the orthogonal component.
Then the area of the parallelogram is
$$ lVert u rVert, lVert mathbf v_perp rVert. $$
Projection of a vector and its answers explain how to get $mathbf v_parallel,$ the parallel component of $mathbf v.$
And then $mathbf v_perp = mathbf v - mathbf v_parallel.$
The effect is the same as the formula in the helpful comment under this answer.
$endgroup$
$begingroup$
How can I find the orthogonal component $mathbf{v}_perp$ or the length of the component directly?
$endgroup$
– K. Claesson
Jan 2 at 20:31
2
$begingroup$
@K.Claesson Subtract from $bf v$ the component of $bf v$ along $bf u$. $$mathbf{v_perp}=mathbf v-frac{mathbf{vcdot u}}{|mathbf u|^2}mathbf u$$
$endgroup$
– Shubham Johri
Jan 2 at 20:41
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The square of the area is the determinant
$$begin{vmatrix}
bf{u}cdotbf u&bf{u}cdotbf v\
bf{v}cdotbf u&
bf{v}cdotbf v
end{vmatrix}.$$
$endgroup$
$begingroup$
Could you explain why this is the square of the area?
$endgroup$
– K. Claesson
Jan 2 at 20:33
2
$begingroup$
@K.Claesson Recall that $A^2=|mathbf u|^2|mathbf v|^2sin^2alpha=|mathbf u|^2|mathbf v|^2(1-cos^2alpha)=|mathbf u|^2|mathbf v|^2-(mathbf{ucdot v})^2$
$endgroup$
– Shubham Johri
Jan 2 at 20:49
add a comment |
$begingroup$
The square of the area is the determinant
$$begin{vmatrix}
bf{u}cdotbf u&bf{u}cdotbf v\
bf{v}cdotbf u&
bf{v}cdotbf v
end{vmatrix}.$$
$endgroup$
$begingroup$
Could you explain why this is the square of the area?
$endgroup$
– K. Claesson
Jan 2 at 20:33
2
$begingroup$
@K.Claesson Recall that $A^2=|mathbf u|^2|mathbf v|^2sin^2alpha=|mathbf u|^2|mathbf v|^2(1-cos^2alpha)=|mathbf u|^2|mathbf v|^2-(mathbf{ucdot v})^2$
$endgroup$
– Shubham Johri
Jan 2 at 20:49
add a comment |
$begingroup$
The square of the area is the determinant
$$begin{vmatrix}
bf{u}cdotbf u&bf{u}cdotbf v\
bf{v}cdotbf u&
bf{v}cdotbf v
end{vmatrix}.$$
$endgroup$
The square of the area is the determinant
$$begin{vmatrix}
bf{u}cdotbf u&bf{u}cdotbf v\
bf{v}cdotbf u&
bf{v}cdotbf v
end{vmatrix}.$$
answered Jan 2 at 20:30
Lord Shark the UnknownLord Shark the Unknown
105k1160133
105k1160133
$begingroup$
Could you explain why this is the square of the area?
$endgroup$
– K. Claesson
Jan 2 at 20:33
2
$begingroup$
@K.Claesson Recall that $A^2=|mathbf u|^2|mathbf v|^2sin^2alpha=|mathbf u|^2|mathbf v|^2(1-cos^2alpha)=|mathbf u|^2|mathbf v|^2-(mathbf{ucdot v})^2$
$endgroup$
– Shubham Johri
Jan 2 at 20:49
add a comment |
$begingroup$
Could you explain why this is the square of the area?
$endgroup$
– K. Claesson
Jan 2 at 20:33
2
$begingroup$
@K.Claesson Recall that $A^2=|mathbf u|^2|mathbf v|^2sin^2alpha=|mathbf u|^2|mathbf v|^2(1-cos^2alpha)=|mathbf u|^2|mathbf v|^2-(mathbf{ucdot v})^2$
$endgroup$
– Shubham Johri
Jan 2 at 20:49
$begingroup$
Could you explain why this is the square of the area?
$endgroup$
– K. Claesson
Jan 2 at 20:33
$begingroup$
Could you explain why this is the square of the area?
$endgroup$
– K. Claesson
Jan 2 at 20:33
2
2
$begingroup$
@K.Claesson Recall that $A^2=|mathbf u|^2|mathbf v|^2sin^2alpha=|mathbf u|^2|mathbf v|^2(1-cos^2alpha)=|mathbf u|^2|mathbf v|^2-(mathbf{ucdot v})^2$
$endgroup$
– Shubham Johri
Jan 2 at 20:49
$begingroup$
@K.Claesson Recall that $A^2=|mathbf u|^2|mathbf v|^2sin^2alpha=|mathbf u|^2|mathbf v|^2(1-cos^2alpha)=|mathbf u|^2|mathbf v|^2-(mathbf{ucdot v})^2$
$endgroup$
– Shubham Johri
Jan 2 at 20:49
add a comment |
$begingroup$
Rewrite $mathbf v$ as the sum of two components, one parallel to $mathbf u$ and one orthogonal to $mathbf u.$
Let $mathbf v_perp$ be the orthogonal component.
Then the area of the parallelogram is
$$ lVert u rVert, lVert mathbf v_perp rVert. $$
Projection of a vector and its answers explain how to get $mathbf v_parallel,$ the parallel component of $mathbf v.$
And then $mathbf v_perp = mathbf v - mathbf v_parallel.$
The effect is the same as the formula in the helpful comment under this answer.
$endgroup$
$begingroup$
How can I find the orthogonal component $mathbf{v}_perp$ or the length of the component directly?
$endgroup$
– K. Claesson
Jan 2 at 20:31
2
$begingroup$
@K.Claesson Subtract from $bf v$ the component of $bf v$ along $bf u$. $$mathbf{v_perp}=mathbf v-frac{mathbf{vcdot u}}{|mathbf u|^2}mathbf u$$
$endgroup$
– Shubham Johri
Jan 2 at 20:41
add a comment |
$begingroup$
Rewrite $mathbf v$ as the sum of two components, one parallel to $mathbf u$ and one orthogonal to $mathbf u.$
Let $mathbf v_perp$ be the orthogonal component.
Then the area of the parallelogram is
$$ lVert u rVert, lVert mathbf v_perp rVert. $$
Projection of a vector and its answers explain how to get $mathbf v_parallel,$ the parallel component of $mathbf v.$
And then $mathbf v_perp = mathbf v - mathbf v_parallel.$
The effect is the same as the formula in the helpful comment under this answer.
$endgroup$
$begingroup$
How can I find the orthogonal component $mathbf{v}_perp$ or the length of the component directly?
$endgroup$
– K. Claesson
Jan 2 at 20:31
2
$begingroup$
@K.Claesson Subtract from $bf v$ the component of $bf v$ along $bf u$. $$mathbf{v_perp}=mathbf v-frac{mathbf{vcdot u}}{|mathbf u|^2}mathbf u$$
$endgroup$
– Shubham Johri
Jan 2 at 20:41
add a comment |
$begingroup$
Rewrite $mathbf v$ as the sum of two components, one parallel to $mathbf u$ and one orthogonal to $mathbf u.$
Let $mathbf v_perp$ be the orthogonal component.
Then the area of the parallelogram is
$$ lVert u rVert, lVert mathbf v_perp rVert. $$
Projection of a vector and its answers explain how to get $mathbf v_parallel,$ the parallel component of $mathbf v.$
And then $mathbf v_perp = mathbf v - mathbf v_parallel.$
The effect is the same as the formula in the helpful comment under this answer.
$endgroup$
Rewrite $mathbf v$ as the sum of two components, one parallel to $mathbf u$ and one orthogonal to $mathbf u.$
Let $mathbf v_perp$ be the orthogonal component.
Then the area of the parallelogram is
$$ lVert u rVert, lVert mathbf v_perp rVert. $$
Projection of a vector and its answers explain how to get $mathbf v_parallel,$ the parallel component of $mathbf v.$
And then $mathbf v_perp = mathbf v - mathbf v_parallel.$
The effect is the same as the formula in the helpful comment under this answer.
edited Jan 2 at 20:46
answered Jan 2 at 20:20
David KDavid K
54.9k344120
54.9k344120
$begingroup$
How can I find the orthogonal component $mathbf{v}_perp$ or the length of the component directly?
$endgroup$
– K. Claesson
Jan 2 at 20:31
2
$begingroup$
@K.Claesson Subtract from $bf v$ the component of $bf v$ along $bf u$. $$mathbf{v_perp}=mathbf v-frac{mathbf{vcdot u}}{|mathbf u|^2}mathbf u$$
$endgroup$
– Shubham Johri
Jan 2 at 20:41
add a comment |
$begingroup$
How can I find the orthogonal component $mathbf{v}_perp$ or the length of the component directly?
$endgroup$
– K. Claesson
Jan 2 at 20:31
2
$begingroup$
@K.Claesson Subtract from $bf v$ the component of $bf v$ along $bf u$. $$mathbf{v_perp}=mathbf v-frac{mathbf{vcdot u}}{|mathbf u|^2}mathbf u$$
$endgroup$
– Shubham Johri
Jan 2 at 20:41
$begingroup$
How can I find the orthogonal component $mathbf{v}_perp$ or the length of the component directly?
$endgroup$
– K. Claesson
Jan 2 at 20:31
$begingroup$
How can I find the orthogonal component $mathbf{v}_perp$ or the length of the component directly?
$endgroup$
– K. Claesson
Jan 2 at 20:31
2
2
$begingroup$
@K.Claesson Subtract from $bf v$ the component of $bf v$ along $bf u$. $$mathbf{v_perp}=mathbf v-frac{mathbf{vcdot u}}{|mathbf u|^2}mathbf u$$
$endgroup$
– Shubham Johri
Jan 2 at 20:41
$begingroup$
@K.Claesson Subtract from $bf v$ the component of $bf v$ along $bf u$. $$mathbf{v_perp}=mathbf v-frac{mathbf{vcdot u}}{|mathbf u|^2}mathbf u$$
$endgroup$
– Shubham Johri
Jan 2 at 20:41
add a comment |
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$begingroup$
You may perhaps calculate the angle $alpha$ between $u$ and $v$ via dot product and then use $A=|u||v|sinalpha$, but it is basically the ugly way to calculate the cross product.
$endgroup$
– A.Γ.
Jan 2 at 20:22