NIMCET 2013 — Mathematics PYQ

Q: How do I solve NIMCET 2013 Mathematics questions?

Practice NIMCET 2013 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 2013 — Mathematics PYQ

NIMCET | Mathematics | 2013

Let $\vec{a} = \hat{j} - \hat{k}$ and $\vec{c} = \hat{i} - \hat{j} - \hat{k}$ . Then the vector $\vec{b}$ satisfying $(\vec{a} \times \vec{b}) + \vec{c} = 0$ and $\vec{a} \cdot \vec{b} = 3$ , is:

Choose the correct answer:

Explanation

Solution

Step 1: Write down the given information

$\vec{a} = 0\hat{i} + 1\hat{j} - 1\hat{k}$
$\vec{c} = 1\hat{i} - 1\hat{j} - 1\hat{k}$
$(\vec{a} \times \vec{b}) = -\vec{c} = -\hat{i} + \hat{j} + \hat{k}$
$\vec{a} \cdot \vec{b} = 3$

Step 2: Use the Vector Triple Product formula

To isolate $\vec{b}$ , we take the cross product of $\vec{a}$ with the equation $(\vec{a} \times \vec{b}) = -\vec{c}$ :

\vec{a} \times (\vec{a} \times \vec{b}) = \vec{a} \times (-\vec{c})

Using the Vector Triple Product identity $\vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z}$ :

(\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b} = -(\vec{a} \times \vec{c})

Step 3: Calculate the components

$\vec{a} \cdot \vec{b} = 3$ (given)
$\vec{a} \cdot \vec{a} = (0)^2 + (1)^2 + (-1)^2 = 2$
$\vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & -1 \\ 1 & -1 & -1 \end{vmatrix} = \hat{i}(-1-1) - \hat{j}(0+1) + \hat{k}(0-1) = -2\hat{i} - \hat{j} - \hat{k}$

Step 4: Substitute and solve for $\vec{b}$

Substitute the values into the identity from Step 2:

3\vec{a} - 2\vec{b} = -(-2\hat{i} - \hat{j} - \hat{k})

3(0\hat{i} + \hat{j} - \hat{k}) - 2\vec{b} = 2\hat{i} + \hat{j} + \hat{k}

3\hat{j} - 3\hat{k} - 2\vec{b} = 2\hat{i} + \hat{j} + \hat{k}

Isolate $2\vec{b}$ :

-2\vec{b} = 2\hat{i} + \hat{j} + \hat{k} - 3\hat{j} + 3\hat{k}

-2\vec{b} = 2\hat{i} - 2\hat{j} + 4\hat{k}

Divide by $-2$ :

\vec{b} = -\hat{i} + \hat{j} - 2\hat{k}

Correct Option: 1. $-\hat{i} + \hat{j} - 2\hat{k}$

Choose the correct answer:

Explore More