NIMCET 2023 — Mathematics PYQ

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Question

NIMCET 2023 — Mathematics PYQ

NIMCET | Mathematics | 2023

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 the scalar triple product $[\vec{a} \ \vec{b} \ \vec{c}]$ depends on:

Choose the correct answer:

Explanation

Step 1: Write the vectors in component form

$\vec{a} = (1, 0, -1)$
$\vec{b} = (x, 1, 1-x)$
$\vec{c} = (y, x, 1+x-y)$

Step 2: Set 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}

Step 3: Expand the determinant along the first row

Expanding along $R_1$ :

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

Step 4: Simplify the calculations

First part:

1 \cdot [(1)(1+x-y) - (x)(1-x)]

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

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

= 1-y+x^2

Second part:

-1 \cdot [(x)(x) - (y)(1)]

= -1 \cdot (x^2 - y)

= -x^2 + y

Step 5: Combine the results

[\vec{a} \ \vec{b} \ \vec{c}] = (1 - y + x^2) + (-x^2 + y)

[\vec{a} \ \vec{b} \ \vec{c}] = 1 - y + x^2 - x^2 + y

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

Conclusion:

The value of the scalar triple product is a constant ( $1$ ). It does not contain the variables $x$ or $y$ . Therefore, it depends on neither $x$ nor $y$ .

Correct Option:

(A) neither $x$ nor $y$

Choose the correct answer:

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