Question

The number of distinct values of \(a\) for which the vectors \( \lambda^2 \hat{i} + \hat{j} + \hat{k} \), \( \hat{i} + \lambda^2 \hat{j} + \hat{k} \), and \( \hat{i} + \hat{j} + \lambda^2 \hat{k} \) are coplanar is:

• A. 1
• B. 2
• C. 3
• D. 6

Hint:

To check for coplanarity of vectors, calculate their scalar triple product. If it equals zero, the vectors are coplanar.

Answer 1

The Correct Option is B

Step 1: Vectors are coplanar if their scalar triple product is zero. The scalar triple product of three vectors \( \vec{A}, \vec{B}, \vec{C} \) is given by: \[ \vec{A} \cdot (\vec{B} \times \vec{C}) = 0 \] Let the three vectors be: \[ \vec{A} = \lambda^2 \hat{i} + \hat{j} + \hat{k}, \quad \vec{B} = \hat{i} + \lambda^2 \hat{j} + \hat{k}, \quad \vec{C} = \hat{i} + \hat{j} + \lambda^2 \hat{k} \] We need to compute the scalar triple product and solve for \(a\).