NIMCET Mathematics Previous Year Questions (PYQs) – Page 1 of 95

Given:

\( \vec{a} = \hat{i} - \hat{k}, \quad \) \(\vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k},\) \( \quad \vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k} \)

Form the matrix:

\( M = \begin{bmatrix} 1 & x & y \\ 0 & 1 & x \\ -1 & 1 - x & 1 + x - y \end{bmatrix} \)

Find the determinant:

\( \det(M) = \begin{vmatrix} 1 & x & y \\ 0 & 1 & x \\ -1 & 1 - x & 1 + x - y \end{vmatrix} = 1 \)

Since the determinant is constant and non-zero, the vectors are linearly independent.

\( \boxed{\text{The matrix does not depend on } x \text{ or } y} \)

Quick Solution

Given: \( \vec{a}, \vec{b} \) are unit vectors and

\( 2\vec{a} + \vec{b} = 3 \)

Take magnitude on both sides:

\( |2\vec{a} + \vec{b}| = 3 \Rightarrow |2\vec{a} + \vec{b}|^2 = 9 \)

Use identity:

\[ |2\vec{a} + \vec{b}|^2 = 4|\vec{a}|^2 + |\vec{b}|^2 + 4(\vec{a} \cdot \vec{b}) = 4 + 1 + 4(\vec{a} \cdot \vec{b}) = 5 + 4(\vec{a} \cdot \vec{b}) \]

Set equal to 9:

\[ 5 + 4(\vec{a} \cdot \vec{b}) = 9 \Rightarrow \vec{a} \cdot \vec{b} = 1 \Rightarrow \cos\theta = 1 \Rightarrow \theta = 0^\circ \]