NIMCET 2012 — Mathematics PYQ

Three vectors $\mathbf{u},\mathbf{v},\mathbf{w}$ are non-coplanar if their scalar triple product is non-zero:

Given that $\mathbf{a},\mathbf{b},\mathbf{c}$ are non-coplanar, we know that $[\mathbf{a}\mathbf{b}\mathbf{c}]\ne 0$.

Step 2: Expressing the vectors in terms of $\mathbf{a},\mathbf{b},\mathbf{c}$

Let the given vectors be:

• $\mathbf{u}=1\mathbf{a}+2\mathbf{b}+3\mathbf{c}$

• $\mathbf{v}=0\mathbf{a}+\lambda \mathbf{b}+4\mathbf{c}$

• $\mathbf{w}=0\mathbf{a}+0\mathbf{b}+(2\lambda -1)\mathbf{c}$

Step 3: Calculating the Scalar Triple Product

The scalar triple product of vectors expressed as linear combinations of $\mathbf{a},\mathbf{b},\mathbf{c}$ can be calculated using the determinant of their coefficients:

Since the matrix is upper triangular, its determinant is the product of the diagonal elements:

Step 4: Setting the condition for non-coplanarity

For the vectors to be non-coplanar, the determinant must not be zero:

This implies:

1. $\lambda \ne 0$

2. $2\lambda -1\ne 0⟹\lambda \ne \frac{1}{2}$

Conclusion:

The vectors are non-coplanar for all real values of $\lambda$ except for two specific values: $0$ and $\frac{1}{2}$.

Correct Option: (c) All except two values of $\lambda$