Question

If a and b be two perpendicular unit vectors such that \(\mathbf{x} = \mathbf{b} - (\mathbf{a} \times \mathbf{x})\), then \(|\mathbf{x}|\) is equal to

• A. 1
• B. \(\sqrt{2}\)
• C. \(\frac{1}{\sqrt{2}}\)
• D. \(\sqrt{3}\)

Hint:

Using vector triple product identities and dot product properties simplifies such vector equations.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Use vector triple product identities and the fact that \(\mathbf{a}\) and \(\mathbf{b}\) are perpendicular unit vectors.

Step 2: Detailed Explanation:
Given \(\mathbf{x} = \mathbf{b} - (\mathbf{a} \times \mathbf{x})\). Taking cross product with \(\mathbf{a}\) on both sides: \[ \mathbf{a} \times \mathbf{x} = \mathbf{a} \times \mathbf{b} - \mathbf{a} \times (\mathbf{a} \times \mathbf{x}) \] \[ \mathbf{a} \times \mathbf{x} = \mathbf{a} \times \mathbf{b} - [(\mathbf{a} \cdot \mathbf{x})\mathbf{a} - (\mathbf{a} \cdot \mathbf{a})\mathbf{x}] \] \[ \mathbf{a} \times \mathbf{x} = \mathbf{a} \times \mathbf{b} - (\mathbf{a} \cdot \mathbf{x})\mathbf{a} + \mathbf{x} \] From original equation, \(\mathbf{a} \times \mathbf{x} = \mathbf{b} - \mathbf{x}\). Substituting: \[ \mathbf{b} - \mathbf{x} = \mathbf{a} \times \mathbf{b} - (\mathbf{a} \cdot \mathbf{x})\mathbf{a} + \mathbf{x} \] \[ \mathbf{b} - \mathbf{a} \times \mathbf{b} = 2\mathbf{x} - (\mathbf{a} \cdot \mathbf{x})\mathbf{a} \] Taking dot product with \(\mathbf{a}\) (since \(\mathbf{a} \perp \mathbf{b}\), \(\mathbf{a} \cdot \mathbf{b} = 0\), \(\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = 0\), \(\mathbf{a} \cdot \mathbf{a} = 1\)): \[ 0 - 0 = 2(\mathbf{a} \cdot \mathbf{x}) - (\mathbf{a} \cdot \mathbf{x})(1) \] \[ 0 = \mathbf{a} \cdot \mathbf{x} \] So \(\mathbf{a} \cdot \mathbf{x} = 0\). Then from original, \(|\mathbf{x}|^2 = \mathbf{x} \cdot \mathbf{x} = \mathbf{x} \cdot (\mathbf{b} - (\mathbf{a} \times \mathbf{x})) = \mathbf{x} \cdot \mathbf{b} - \mathbf{x} \cdot (\mathbf{a} \times \mathbf{x})\). But \(\mathbf{x} \cdot (\mathbf{a} \times \mathbf{x}) = 0\). So \(|\mathbf{x}|^2 = \mathbf{x} \cdot \mathbf{b}\). From \(\mathbf{x} = \mathbf{b} - (\mathbf{a} \times \mathbf{x})\), taking dot product with \(\mathbf{b}\): \[ \mathbf{x} \cdot \mathbf{b} = |\mathbf{b}|^2 - \mathbf{b} \cdot (\mathbf{a} \times \mathbf{x}) = 1 - \mathbf{a} \cdot (\mathbf{x} \times \mathbf{b}) \] Also, from \(\mathbf{a} \cdot \mathbf{x} = 0\), and \(\mathbf{x} = \mathbf{b} - (\mathbf{a} \times \mathbf{x})\), we get \(\mathbf{x} \cdot \mathbf{b} = 1 - |\mathbf{x}|^2\). Thus \(|\mathbf{x}|^2 = 1 - |\mathbf{x}|^2 \Rightarrow 2|\mathbf{x}|^2 = 1 \Rightarrow |\mathbf{x}| = \frac{1}{\sqrt{2}}\).

Step 3: Final Answer:
\(|\mathbf{x}| = \frac{1}{\sqrt{2}}\), which corresponds to option (C).