Question

If \( \vec{a}, \vec{b}, \vec{c} \) are non-coplanar vectors and \( \lambda \) is a real number, then the vectors \( \vec{a} + 2\vec{b} + 3\vec{c} \), \( \lambda \vec{b} + 4\vec{c} \), and \( (2\lambda - 1)\vec{c} \) are non-coplanar for:

• A. no value of \( \lambda \).
• B. all except one value of \( \lambda \).
• C. all except two values of \( \lambda \).
• D. all values of \( \lambda \).

Hint:

To check if vectors are non-coplanar, form a matrix with their coefficients in a basis and compute the determinant. Non-zero determinant indicates linear independence.

Answer 1

The Correct Option is C

Step 1: Non-coplanarity condition.
Vectors \( \vec{v_1} = \vec{a} + 2\vec{b} + 3\vec{c} \), \( \vec{v_2} = \lambda \vec{b} + 4\vec{c} \), and \( \vec{v_3} = (2\lambda - 1)\vec{c} \) are non-coplanar if they are linearly independent. Express the vectors in the basis \( \vec{a}, \vec{b}, \vec{c} \) and form the matrix: \[\n\begin{pmatrix}\n1 & 2 & 3
\n0 & \lambda & 4
\n0 & 0 & 2\lambda - 1\n\end{pmatrix}.\n\] Non-coplanarity means the determinant is non-zero.
Step 2: Determinant calculation.
The upper triangular matrix's determinant is the product of diagonal elements: \[\n\text{Det} = 1 \cdot \lambda \cdot (2\lambda - 1) = \lambda (2\lambda - 1).\n\] The vectors are coplanar (linearly dependent) when the determinant is zero: \[\n\lambda (2\lambda - 1) = 0 \implies \lambda = 0 \text{ or } \lambda = \frac{1}{2}.\n\]
Step 3: Result analysis.
If \( \lambda = 0 \), vectors are \( \vec{a} + 2\vec{b} + 3\vec{c} \), \( 4\vec{c} \), \( -1\vec{c} \). The last two are scalar multiples, thus coplanar.
If \( \lambda = \frac{1}{2} \), vectors are \( \vec{a} + 2\vec{b} + 3\vec{c} \), \( \frac{1}{2} \vec{b} + 4\vec{c} \), \( 0\vec{c} = \vec{0} \). The zero vector makes them coplanar.
For all other \( \lambda \), the determinant \( \lambda (2\lambda - 1) \neq 0 \), so the vectors are linearly independent and non-coplanar.

Step 4: Exceptional values.
There are two values of \( \lambda \) (\( \lambda = 0 \) and \( \lambda = \frac{1}{2} \)) where vectors are coplanar. Vectors are non-coplanar for all other \( \lambda \).
Step 5: Answer selection.
Vectors are non-coplanar for all \( \lambda \) except \( \lambda = 0 \) and \( \lambda = \frac{1}{2} \), which corresponds to option (C) "all except two values of \( \lambda \)".