Question

If \(\mathbf{a} \times \mathbf{b} = \mathbf{c}\), \(\mathbf{b} \times \mathbf{c} = \mathbf{a}\) and \(a, b, c\) be the moduli of the vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) respectively, then

• A. \(a = 1, b = 1, c = 1\)
• B. \(a = 1, b = 1, c = a\)
• C. \(a = c, b = 1\)
• D. \(a = 1, b = c\)

Hint:

If \(\mathbf{u} \times \mathbf{v} = \mathbf{w}\), then all three vectors are mutually perpendicular.

Answer 1

The Correct Option is D

Step 1: Use magnitude of cross product}
\[ |\mathbf{a} \times \mathbf{b}| = ab \sin \theta \] Given \(\mathbf{a} \times \mathbf{b} = \mathbf{c}\), so \[ c = ab \sin \theta_1 \] Similarly, from \(\mathbf{b} \times \mathbf{c} = \mathbf{a}\), \[ a = bc \sin \theta_2 \] Step 2: Use perpendicularity}
Since cross product gives a vector perpendicular to both vectors, \[ \mathbf{c} \perp \mathbf{a}, \mathbf{b} \quad \text{and} \quad \mathbf{a} \perp \mathbf{b}, \mathbf{c} \] Thus, angles are \(90^\circ\), so \(\sin 90^\circ = 1\) \[ c = ab,\quad a = bc \] Step 3: Solve equations}
Substitute \(c = ab\) into second equation: \[ a = b(ab) = ab^2 \] \[ a \neq 0 \Rightarrow b^2 = 1 \Rightarrow b = 1 \] \[ c = ab = a(1) = a \] Step 4: Final Answer
\[ a = 1,\quad b = c \]