For any vector \(\vec{x}\), the value of \((\vec{x} \times \hat{i})^2+(\vec{x} \times \hat{j})^2+(\vec{x} \times \hat{k})^2\) equal the:

Let's denote the vectors as \(\vec{a}, \vec{b}, \) and \(\vec{c}\). Volume of the First Parallelepiped The volume of the parallelepiped formed by vectors \(\vec{a} \times \vec{b}, \vec{b} \times \vec{c},\) and \(\vec{c} \times \vec{a}\) can be expressed as the scalar triple product of these vectors:Read more

Let’s denote the vectors as \(\vec{a}, \vec{b}, \) and \(\vec{c}\).

### Volume of the First Parallelepiped

The volume of the parallelepiped formed by vectors \(\vec{a} \times \vec{b}, \vec{b} \times \vec{c},\) and \(\vec{c} \times \vec{a}\) can be expressed as the scalar triple product of these vectors:

\[

\text{Volume} = [(\vec{a} \times \vec{b}) \cdot ((\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a}))]

\]

Using the vector triple product identity, \(\vec{A} \times (\vec{B} \times \vec{C}) = \vec{B}(\vec{A} \cdot \vec{C}) – \vec{C}(\vec{A} \cdot \vec{B})\), we can simplify the expression:

\[

\text{Volume} = [(\vec{a} \times \vec{b}) \cdot (\vec{b}(\vec{b} \cdot \vec{a}) – \vec{a}(\vec{b} \cdot \vec{b}))]

\]

Expanding further:

\[

\text{Volume} = [(\vec{a} \times \vec{b}) \cdot \vec{b}(\vec{b} \cdot \vec{a})] – [(\vec{a} \times \vec{b}) \cdot \vec{a}(\vec{b} \cdot \vec{b})]

\]

Since \((\vec{a} \times \vec{b}) \cdot \vec{b} = 0\) and \((\vec{a} \times \vec{b}) \cdot \vec{a} = 0\) (as the cross product is perpendicular to both vectors), the volume of the first parallelepiped is 0.

### Volume of the Second Parallelepiped

For the second parallelepiped, we have the vectors \((\vec{a} \times \vec{b}) \times(\vec{b} \times \vec{c})\), \((\vec{b} \times \vec{c}) \times(\vec{c} \times \vec{a})\), and \((\vec{c} \times \vec{a}) \times(\vec{a} \times \vec{b})\). The volume can be similarly expressed as the scalar triple product:

\[

\text{Volume} = [((\vec{a} \times \vec{b}) \times(\vec{b} \times \vec{c})) \cdot (((\vec{b} \times \vec{c}) \times(\vec{c} \times \vec{a})) \times ((\vec{c} \times \vec{a}) \times(\vec{a} \times \vec{b})))]

\]

Using the same vector triple product identity and the fact that the cross product of any two vectors is perpendicular to both, we can deduce that the volume of the second parallelepiped is also 0.

In summary, the volume of the second parallelepiped with \((\vec{a} \times \vec{b}) \times(\vec{b} \times \vec{c})\), \((\vec{b} \times \vec{c}) \times(\vec{c} \times \vec{a})\), and \((\vec{c} \times \vec{a}) \times(\vec{a} \times \vec{b})\) on coterminal edges is \(0 \, \mathrm{cu. \, units}\).

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Let's denote the vector \(\vec{x}\) in its component form as \(\vec{x} = x\hat{i} + y\hat{j} + z\hat{k}\). Then, the cross products of \(\vec{x}\) with the unit vectors \(\hat{i}, \hat{j}, \) and \(\hat{k}\) are: 1. \(\vec{x} \times \hat{i} = (y\hat{j} + z\hat{k}) \times \hat{i} = y\hat{k} - z\hat{jRead more

Let’s denote the vector \(\vec{x}\) in its component form as \(\vec{x} = x\hat{i} + y\hat{j} + z\hat{k}\).

Then, the cross products of \(\vec{x}\) with the unit vectors \(\hat{i}, \hat{j}, \) and \(\hat{k}\) are:

1. \(\vec{x} \times \hat{i} = (y\hat{j} + z\hat{k}) \times \hat{i} = y\hat{k} – z\hat{j}\)

2. \(\vec{x} \times \hat{j} = (x\hat{i} + z\hat{k}) \times \hat{j} = z\hat{i} – x\hat{k}\)

3. \(\vec{x} \times \hat{k} = (x\hat{i} + y\hat{j}) \times \hat{k} = x\hat{j} – y\hat{i}\)

Now, we can find the squares of the magnitudes of these cross products:

1. \((\vec{x} \times \hat{i})^2 = (y\hat{k} – z\hat{j}) \cdot (y\hat{k} – z\hat{j}) = y^2 + z^2\)

2. \((\vec{x} \times \hat{j})^2 = (z\hat{i} – x\hat{k}) \cdot (z\hat{i} – x\hat{k}) = z^2 + x^2\)

3. \((\vec{x} \times \hat{k})^2 = (x\hat{j} – y\hat{i}) \cdot (x\hat{j} – y\hat{i}) = x^2 + y^2\)

Adding these up, we get:

\[

(\vec{x} \times \hat{i})^2 + (\vec{x} \times \hat{j})^2 + (\vec{x} \times \hat{k})^2 = (y^2 + z^2) + (z^2 + x^2) + (x^2 + y^2) = 2(x^2 + y^2 + z^2)

\]

Since \(x^2 + y^2 + z^2\) is the square of the magnitude of the vector \(\vec{x}\), denoted as \(|\vec{x}|^2\), we can write the final expression as:

\[

(\vec{x} \times \hat{i})^2 + (\vec{x} \times \hat{j})^2 + (\vec{x} \times \hat{k})^2 = 2|\vec{x}|^2

\]

Therefore, for any vector \(\vec{x}\), the value of \((\vec{x} \times \hat{i})^2 + (\vec{x} \times \hat{j})^2 + (\vec{x} \times \hat{k})^2\) is equal to \(2|\vec{x}|^2\).

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