Question

Let \(\mathbf{a}, \mathbf{b}\) and \(\mathbf{c}\) be three vectors such that $\mathbf{a} \times \mathbf{b} = \mathbf{a} \times \mathbf{c} \text{ and } \mathbf{a} \times \mathbf{b} \neq 0 \text{ Show that } \mathbf{b} = \mathbf{c}$.

Hint:

Quick Tip: When vectors are equal to each other, their cross products with a common vector must also be equal. A non-zero cross product indicates that the vectors are not parallel to each other.

Answer 1

Given the conditions: \[ \mathbf{a} \times \mathbf{b} = \mathbf{a} \times \mathbf{c} \quad \text{and} \quad \mathbf{a} \times \mathbf{b} eq 0 \] Subtracting the two vector equations yields: \[ \mathbf{a} \times \mathbf{b} - \mathbf{a} \times \mathbf{c} = 0 \] Applying the distributive property of the cross product: \[ \mathbf{a} \times (\mathbf{b} - \mathbf{c}) = 0 \] For the cross product to be zero, vectors \( \mathbf{a} \) and \( \mathbf{b} - \mathbf{c} \) must be parallel. This means: \[ \mathbf{b} - \mathbf{c} = \lambda \mathbf{a} \quad \text{for some scalar} \ \lambda \] Rearranging, we get: \[ \mathbf{b} = \mathbf{c} + \lambda \mathbf{a} \] Since \( \mathbf{a} \times \mathbf{b} eq 0 \), \( \mathbf{a} \) is not perpendicular to \( \mathbf{b} \), which implies \( \mathbf{b} eq \mathbf{c} \). Consequently, the initial premise implies: \[ \mathbf{b} = \mathbf{c} \]