Question

If the vectors \( 2\hat{i} + 2\hat{j} + 6\hat{k} \), \( 2\hat{i} + \lambda\hat{j} + 6\hat{k} \), \( 2\hat{i} - 3\hat{j} + \hat{k} \) are coplanar, then the value of \( \lambda \) is:

• A. \( -10 \)
• B. \( 1 \)
• C. \( 0 \)
• D. \( 10 \)
• E. \( 2 \)

Hint:

Notice that the first and second vectors are identical if \( \lambda = 2 \). Since any set of vectors where two vectors are identical is always linearly dependent (and thus coplanar), \( \lambda = 2 \) can be spotted immediately by observation.

Answer 1

Answer: (E)

Concept: Three vectors are coplanar if their scalar triple product is zero. This is equivalent to saying the determinant formed by their components must equal zero.

Step 1: Set up the determinant.
Let the vectors be \( \vec{a} = (2, 2, 6) \), \( \vec{b} = (2, \lambda, 6) \), and \( \vec{c} = (2, -3, 1) \). The condition for coplanarity is: \[ \begin{vmatrix} 2 & 2 & 6 \\ 2 & \lambda & 6 \\ 2 & -3 & 1 \end{vmatrix} = 0 \]

Step 2: Expand the determinant.
Expanding along the first row: \[ 2(\lambda(1) - 6(-3)) - 2(2(1) - 6(2)) + 6(2(-3) - \lambda(2)) = 0 \] \[ 2(\lambda + 18) - 2(2 - 12) + 6(-6 - 2\lambda) = 0 \] \[ 2\lambda + 36 - 2(-10) - 36 - 12\lambda = 0 \]

Step 3: Solve for \( \lambda \).
\[ 2\lambda + 36 + 20 - 36 - 12\lambda = 0 \] \[ -10\lambda + 20 = 0 \] \[ 10\lambda = 20 \implies \lambda = 2 \]