Question

If \( \vec{a}, \vec{b}, \vec{c} \) are non-coplanar vectors and the points \[ \lambda \vec{a} + 3\vec{b} - \vec{c}, \quad \vec{a} - \lambda \vec{b} + 3\vec{c}, \quad 3\vec{a} + 4\vec{b} - \lambda \vec{c}, \quad \vec{a} - 6\vec{b} + 6\vec{c} \] are coplanar, then one of the values of \( \lambda \) is:

• A. \(7\)
• B. \(5\)
• C. \(2\)
• D. \(1\)

Hint:

To check coplanarity of four points, set up a determinant equation using three relative vectors and solve.

Answer 1

The Correct Option is C

The given four points are:

\[ \lambda \vec{a} + 3\vec{b} - \vec{c}, \quad \vec{a} - \lambda \vec{b} + 3\vec{c}, \quad 3\vec{a} + 4\vec{b} - \lambda \vec{c}, \quad \vec{a} - 6\vec{b} + 6\vec{c} \]

Since these points are coplanar, their position vectors must be linearly dependent. That is, the determinant of the coefficient matrix formed by subtracting the first vector from the others must be zero.

Compute the vectors relative to the first point:

\[ \vec{v_1} = (\vec{a} - \lambda \vec{b} + 3\vec{c}) - (\lambda \vec{a} + 3\vec{b} - \vec{c}) = (1 - \lambda) \vec{a} + (-\lambda - 3) \vec{b} + (3 + 1) \vec{c} \]

\[ \vec{v_2} = (3\vec{a} + 4\vec{b} - \lambda \vec{c}) - (\lambda \vec{a} + 3\vec{b} - \vec{c}) = (3 - \lambda) \vec{a} + (4 - 3) \vec{b} + (-\lambda + 1) \vec{c} \]

\[ \vec{v_3} = (\vec{a} - 6\vec{b} + 6\vec{c}) - (\lambda \vec{a} + 3\vec{b} - \vec{c}) = (1 - \lambda) \vec{a} + (-6 - 3) \vec{b} + (6 + 1) \vec{c} \]

Since these three vectors are linearly dependent, their determinant must be zero:

\[ \begin{vmatrix} 1 - \lambda & -\lambda - 3 & 4 \\ 3 - \lambda & 1 & -\lambda + 1 \\ 1 - \lambda & -9 & 7 \end{vmatrix} = 0 \]

Expanding along the first row:

\[ (1 - \lambda) \begin{vmatrix} 1 & -\lambda + 1 \\ -9 & 7 \end{vmatrix} - (-\lambda - 3) \begin{vmatrix} 3 - \lambda & -\lambda + 1 \\ 1 - \lambda & 7 \end{vmatrix} + 4 \begin{vmatrix} 3 - \lambda & 1 \\ 1 - \lambda & -9 \end{vmatrix} = 0 \]

Solving the determinant and simplifying gives:

\[ \lambda = 2 \quad \text{(one of the possible values)} \]

Answer: \( \lambda = 2 \)