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:
For any vector `x`, the value of `(x x i)^2 + (x x j)^2 + (x x k)^2` equal the:
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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. \(\vec{x} \times \hat{i} = (y\hat{j} + z\hat{k}) \times \hat{i} = y\hat{k} – z\hat{j}\)
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 \(2|\vec{x}|^2\).