NIMCET 2015 — Mathematics PYQ

Q: How do I solve NIMCET 2015 Mathematics questions?

Practice NIMCET 2015 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of NIMCET Mathematics questions?

NIMCET Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are NIMCET previous year papers available for free?

Yes. Mathem Solvex provides free access to NIMCET previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

NIMCET 2015 — Mathematics PYQ

NIMCET | Mathematics | 2015

If $\vec{a} = \hat{i} - \hat{k}$ , $\vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k}$ and $\vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k}$ , then $[\vec{a} \ \vec{b} \ \vec{c}]$ depends on:

Choose the correct answer:

Explanation

1. Understanding Scalar Triple Product

The scalar triple product $[\vec{a} \ \vec{b} \ \vec{c}]$ is calculated using the determinant of the components of the three vectors.

Given vectors:

$\vec{a} = 1\hat{i} + 0\hat{j} - 1\hat{k}$
$\vec{b} = x\hat{i} + 1\hat{j} + (1 - x)\hat{k}$
$\vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k}$

2. Setting up the Determinant

[\vec{a} \ \vec{b} \ \vec{c}] = \begin{vmatrix} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y \end{vmatrix}

3. Expanding the Determinant

We expand along the first row:

[\vec{a} \ \vec{b} \ \vec{c}] = 1 \begin{vmatrix} 1 & 1-x \\ x & 1+x-y \end{vmatrix} - 0 \begin{vmatrix} x & 1-x \\ y & 1+x-y \end{vmatrix} + (-1) \begin{vmatrix} x & 1 \\ y & x \end{vmatrix}

Now, solve the $2 \times 2$ determinants:

= 1[ (1)(1+x-y) - (x)(1-x) ] - 0 + (-1)[ (x)(x) - (y)(1) ]

= [ 1 + x - y - (x - x^2) ] - [ x^2 - y ]

= 1 + x - y - x + x^2 - x^2 + y

4. Simplifying the Expression

Combine the like terms:

$x - x = 0$
$-y + y = 0$
$x^2 - x^2 = 0$

[\vec{a} \ \vec{b} \ \vec{c}] = 1

Conclusion

The value of the scalar triple product is a constant ( $1$ ). Since the final result does not contain $x$ or $y$ , the value depends on neither $x$ nor $y$ .

Correct Option: (a)