How can the area of a paralellogram formed by two 3D vectors be computed without using the cross product?












0












$begingroup$


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







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  • 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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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














  • 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










2 Answers
2






active

oldest

votes


















2












$begingroup$

The square of the area is the determinant
$$begin{vmatrix}
bf{u}cdotbf u&bf{u}cdotbf v\
bf{v}cdotbf u&
bf{v}cdotbf v
end{vmatrix}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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





















2












$begingroup$

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.






share|cite|improve 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













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

The square of the area is the determinant
$$begin{vmatrix}
bf{u}cdotbf u&bf{u}cdotbf v\
bf{v}cdotbf u&
bf{v}cdotbf v
end{vmatrix}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















2












$begingroup$

The square of the area is the determinant
$$begin{vmatrix}
bf{u}cdotbf u&bf{u}cdotbf v\
bf{v}cdotbf u&
bf{v}cdotbf v
end{vmatrix}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












2








2





$begingroup$

The square of the area is the determinant
$$begin{vmatrix}
bf{u}cdotbf u&bf{u}cdotbf v\
bf{v}cdotbf u&
bf{v}cdotbf v
end{vmatrix}.$$






share|cite|improve this answer









$endgroup$



The square of the area is the determinant
$$begin{vmatrix}
bf{u}cdotbf u&bf{u}cdotbf v\
bf{v}cdotbf u&
bf{v}cdotbf v
end{vmatrix}.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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













2












$begingroup$

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.






share|cite|improve 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


















2












$begingroup$

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.






share|cite|improve 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
















2












2








2





$begingroup$

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.






share|cite|improve 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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve 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




















  • $begingroup$
    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




















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