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Patrick I-Z
Patrick I-Z
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Hi Dick, I don't know if there is one or another way to give a formal meaning to this formula, but I understand it this way:

The volume of three vectors $X$, $Y$, $Z$ of ${\bf R}^3$ is given by ${\rm vol}(X,Y,Z) = \det[X\ Y\ Z]$, but weWe have the identity $\det[X\ Y\ Z] = \langle X, Y \times Z\rangle$, where the brackets denotes the scalar product. So, $Y \times Z = (Y \times Z)_1 e_1 + (Y \times Z)_2 e_2 + (Y \times Z)_3 e_3$$Y \times Z = \sum_{i=1}^3 \langle e_i, Y \times Z \rangle e_i = \sum_{i=1}^3 \det[e_i \ Y\ Z]e_i$, where $e_i$ are the vectors of the canonical basis, that is $Y \times Z = \sum_{i=1}^3 \langle e_i, Y \times Z \rangle e_i = \sum_{i=1}^3 \det[e_i \ Y\ Z]e_i$ which is your formula, I think.

-- Note: It seems to be the same remark as Deane Yang's

Hi Dick, I don't know if there is one or another way to give a formal meaning to this formula, but I understand it this way:

The volume of three vectors $X$, $Y$, $Z$ of ${\bf R}^3$ is given by ${\rm vol}(X,Y,Z) = \det[X\ Y\ Z]$, but we have the identity $\det[X\ Y\ Z] = \langle X, Y \times Z\rangle$, where the brackets denotes the scalar product. So, $Y \times Z = (Y \times Z)_1 e_1 + (Y \times Z)_2 e_2 + (Y \times Z)_3 e_3$, where $e_i$ are the vectors of the canonical basis, that is $Y \times Z = \sum_{i=1}^3 \langle e_i, Y \times Z \rangle e_i = \sum_{i=1}^3 \det[e_i \ Y\ Z]e_i$ which is your formula, I think.

-- Note: It seems to be the same remark as Deane Yang's

Hi Dick, I don't know if there is one or another way to give a formal meaning to this formula, but I understand it this way:

We have the identity $\det[X\ Y\ Z] = \langle X, Y \times Z\rangle$, where the brackets denotes the scalar product. So, $Y \times Z = \sum_{i=1}^3 \langle e_i, Y \times Z \rangle e_i = \sum_{i=1}^3 \det[e_i \ Y\ Z]e_i$, where $e_i$ are the vectors of the canonical basis, which is your formula, I think.

-- Note: It seems to be the same remark as Deane Yang's

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Patrick I-Z
Patrick I-Z
  • 2.3k
  • 28
  • 23

Hi Dick, I don't know if there is one or another way to give a formal meaning to this formula, but I understand it this way:

The volume of three vectors $X$, $Y$, $Z$ of ${\bf R}^3$ is given by ${\rm vol}(X,Y,Z) = \det[X\ Y\ Z]$, but we have the identity $\det[X\ Y\ Z] = \langle X, Y \times Z\rangle$, where the brackets denotes the scalar product. So, $Y \times Z = (Y \times Z)_1 e_1 + (Y \times Z)_2 e_2 + (Y \times Z)_3 e_3$, where $e_i$ are the vectors of the canonical basis, that is $Y \times Z = \sum_{i=1}^3 \langle e_i, Y \times Z \rangle e_i = \sum_{i=1}^3 \det[e_i \ Y\ Z]e_i$ which is your formula, I think.

-- Note: It seems to be the same remark as Deane Yang's

Hi Dick, I don't know if there is one or another way to give a formal meaning to this formula, but I understand it this way:

The volume of three vectors $X$, $Y$, $Z$ of ${\bf R}^3$ is given by ${\rm vol}(X,Y,Z) = \det[X\ Y\ Z]$, but we have the identity $\det[X\ Y\ Z] = \langle X, Y \times Z\rangle$, where the brackets denotes the scalar product. So, $Y \times Z = (Y \times Z)_1 e_1 + (Y \times Z)_2 e_2 + (Y \times Z)_3 e_3$, where $e_i$ are the vectors of the canonical basis, that is $Y \times Z = \sum_{i=1}^3 \langle e_i, Y \times Z \rangle e_i = \sum_{i=1}^3 \det[e_i \ Y\ Z]e_i$ which is your formula, I think.

Hi Dick, I don't know if there is one or another way to give a formal meaning to this formula, but I understand it this way:

The volume of three vectors $X$, $Y$, $Z$ of ${\bf R}^3$ is given by ${\rm vol}(X,Y,Z) = \det[X\ Y\ Z]$, but we have the identity $\det[X\ Y\ Z] = \langle X, Y \times Z\rangle$, where the brackets denotes the scalar product. So, $Y \times Z = (Y \times Z)_1 e_1 + (Y \times Z)_2 e_2 + (Y \times Z)_3 e_3$, where $e_i$ are the vectors of the canonical basis, that is $Y \times Z = \sum_{i=1}^3 \langle e_i, Y \times Z \rangle e_i = \sum_{i=1}^3 \det[e_i \ Y\ Z]e_i$ which is your formula, I think.

-- Note: It seems to be the same remark as Deane Yang's

Source Link
Patrick I-Z
Patrick I-Z
  • 2.3k
  • 28
  • 23

Hi Dick, I don't know if there is one or another way to give a formal meaning to this formula, but I understand it this way:

The volume of three vectors $X$, $Y$, $Z$ of ${\bf R}^3$ is given by ${\rm vol}(X,Y,Z) = \det[X\ Y\ Z]$, but we have the identity $\det[X\ Y\ Z] = \langle X, Y \times Z\rangle$, where the brackets denotes the scalar product. So, $Y \times Z = (Y \times Z)_1 e_1 + (Y \times Z)_2 e_2 + (Y \times Z)_3 e_3$, where $e_i$ are the vectors of the canonical basis, that is $Y \times Z = \sum_{i=1}^3 \langle e_i, Y \times Z \rangle e_i = \sum_{i=1}^3 \det[e_i \ Y\ Z]e_i$ which is your formula, I think.