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: Set up the condition for non-coplanarity.
Three 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. Since \( \vec{a}, \vec{b}, \vec{c} \) are non-coplanar (hence linearly independent), we can express the vectors in the basis \( \vec{a}, \vec{b}, \vec{c} \). Form the matrix with these vectors as rows: \[ \begin{pmatrix} 1 & 2 & 3
0 & \lambda & 4
0 & 0 & 2\lambda - 1 \end{pmatrix}. \] The vectors are non-coplanar if the determinant of this matrix is non-zero.
Step 2: Compute the determinant.
The matrix is upper triangular, so the determinant is the product of the diagonal elements: \[ \text{Det} = 1 \cdot \lambda \cdot (2\lambda - 1) = \lambda (2\lambda - 1). \] The vectors are linearly dependent (hence coplanar) when the determinant is zero: \[ \lambda (2\lambda - 1) = 0 \implies \lambda = 0 \text{ or } \lambda = \frac{1}{2}. \]
Step 3: Analyze the result.
If \( \lambda = 0 \), the vectors become \( \vec{a} + 2\vec{b} + 3\vec{c} \), \( 4\vec{c} \), \( -1\vec{c} \). The second and third vectors are scalar multiples of each other (\( 4\vec{c} \) and \( -1\vec{c} \)), so they are coplanar.
If \( \lambda = \frac{1}{2} \), the vectors become \( \vec{a} + 2\vec{b} + 3\vec{c} \), \( \frac{1}{2} \vec{b} + 4\vec{c} \), \( 0\vec{c} = \vec{0} \). The third vector is the zero vector, making 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: Determine the number of exceptional values.
There are two values of \( \lambda \) (\( \lambda = 0 \) and \( \lambda = \frac{1}{2} \)) where the vectors are coplanar. The vectors are non-coplanar for all \( \lambda \) except these two values.
Step 5: Select the correct answer.
The vectors are non-coplanar for all \( \lambda \) except \( \lambda = 0 \) and \( \lambda = \frac{1}{2} \), which matches option (C) "all except two values of \( \lambda \)".