Question:
If \(\vec{a}, \vec{b}, \vec{c}\) are three non-coplanar vectors, then \([\vec{a}\times\vec{b},\ \vec{b}\times\vec{c},\ \vec{c}\times\vec{a}]\) is equal to:
If \(\vec{a}, \vec{b}, \vec{c}\) are three non-coplanar vectors, then \([\vec{a}\times\vec{b},\ \vec{b}\times\vec{c},\ \vec{c}\times\vec{a}]\) is equal to:
Show Hint
For non-coplanar vectors, the scalar triple product of cross products equals the square of the original scalar triple product.
Updated On: Apr 16, 2026
- \([\vec{a}\ \vec{b}\ \vec{c}]^3\)
- \([\vec{a}\ \vec{b}\ \vec{c}]^2\)
- 0
- None of these
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The Correct Option is B
Solution and Explanation
Concept:
The scalar triple product satisfies the identity:
\[
[\vec{a}\times\vec{b},\ \vec{b}\times\vec{c},\ \vec{c}\times\vec{a}] = [\vec{a}\ \vec{b}\ \vec{c}]^2
\]
Step 1: Proof outline. \[ \vec{a}\times\vec{b} = \vec{a} \times \vec{b} \] Using properties of scalar triple product: \[ (\vec{a}\times\vec{b}) \cdot [(\vec{b}\times\vec{c}) \times (\vec{c}\times\vec{a})] = [\vec{a}\ \vec{b}\ \vec{c}]^2 \] The result is the square of the original scalar triple product.
Step 1: Proof outline. \[ \vec{a}\times\vec{b} = \vec{a} \times \vec{b} \] Using properties of scalar triple product: \[ (\vec{a}\times\vec{b}) \cdot [(\vec{b}\times\vec{c}) \times (\vec{c}\times\vec{a})] = [\vec{a}\ \vec{b}\ \vec{c}]^2 \] The result is the square of the original scalar triple product.
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