Question

If \[ \mathbf{a} = \hat{i} + \hat{j} + \hat{k}, \quad \mathbf{b} = \hat{i} - \hat{j} + 2\hat{k}, \quad \mathbf{c} = x\hat{i} + \hat{j} + (x - 1)\hat{k} \] If the vector \( \mathbf{c} \) lies in the plane of \( \mathbf{a} \) and \( \mathbf{b} \), then \( x = \)

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

Hint:

For a vector to lie in the plane of two other vectors, express it as a linear combination of those vectors and solve for the unknowns.

Answer 1

The Correct Option is C

Step 1: Condition for \( \mathbf{c} \) to lie in the plane of \( \mathbf{a} \) and \( \mathbf{b} \).
For \( \mathbf{c} \) to lie in the plane of \( \mathbf{a} \) and \( \mathbf{b} \), it must be a linear combination of \( \mathbf{a} \) and \( \mathbf{b} \). Thus, we express \( \mathbf{c} \) as: \[ \mathbf{c} = \lambda \mathbf{a} + \mu \mathbf{b} \]
Step 2: Solve for \( x \).
Substituting the components of \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \) into the equation and solving for \( x \), we find: \[ x = -\frac{2}{3} \]
Step 3: Conclusion.
Thus, \( x = -\frac{2}{3} \), corresponding to option (C).