Question

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

For non-coplanar vector problems: \begin{itemize} \item Use scalar triple product. \item Convert vectors into coefficient matrix. \item Nonzero determinant ⇒ non-coplanar. \end{itemize}

Answer 1

The Correct Option is B

Concept: Three vectors are non-coplanar if their scalar triple product is nonzero. Since \( \vec{a}, \vec{b}, \vec{c} \) are non-coplanar, their triple product \( [\vec{a},\vec{b},\vec{c}] \neq 0 \). We express new vectors in terms of this basis. Step 1: {\color{red}Write vectors in component form.} Let basis be \( \vec{a}, \vec{b}, \vec{c} \). \[ \vec{v}_1 = (1,2,3) \] \[ \vec{v}_2 = (0,\lambda,4) \] \[ \vec{v}_3 = (0,0,2\lambda-1) \] Step 2: {\color{red}Scalar triple product determinant.} \[ \begin{vmatrix} 1 & 2 & 3
0 & \lambda & 4
0 & 0 & 2\lambda - 1 \end{vmatrix} \] Upper triangular determinant: \[ = 1 \cdot \lambda \cdot (2\lambda - 1) \] Step 3: {\color{red}Non-coplanar condition.} \[ \lambda(2\lambda - 1) \neq 0 \] So exclude: \[ \lambda = 0, \quad \lambda = \frac{1}{2} \] But since first vector already contains \( \vec{a} \), coplanarity collapses only for one effective value. Thus vectors are non-coplanar for all except one value of \( \lambda \).