JEE 2023 — Mathematics PYQ

Q: How do I solve JEE 2023 Mathematics questions?

Practice JEE 2023 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 JEE Mathematics questions?

JEE 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 JEE previous year papers available for free?

Yes. Mathem Solvex provides free access to JEE 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

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $\vec{a} = -\hat{i} - \hat{j} + \hat{k}$ , $\vec{a} \cdot \vec{b} = 1$ and $\vec{a} \times \vec{b} = \hat{i} - \hat{j}$ . Then $\vec{a} - 6\vec{b}$ is equal to

Choose the correct answer:

Explanation

Solving

1. Use the vector triple product identity

We know that:

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

2. Calculate the components

Given $\vec{a} = (-1, -1, 1)$
Given $\vec{a} \cdot \vec{b} = 1$
Given $\vec{a} \times \vec{b} = (1, -1, 0)$
Calculate $|\vec{a}|^2$ (which is $\vec{a} \cdot \vec{a}$ ):

$\vec{a} \cdot \vec{a} = (-1)^2 + (-1)^2 + (1)^2 = 1 + 1 + 1 = 3$

3. Calculate $\vec{a} \times (\vec{a} \times \vec{b})$

\vec{a} \times (\vec{a} \times \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & 1 \\ 1 & -1 & 0 \end{vmatrix}

= \hat{i}(0 - (-1)) - \hat{j}(0 - 1) + \hat{k}(1 - (-1))

= \hat{i} + \hat{j} + 2\hat{k}

4. Substitute values into the identity

\vec{i} + \vec{j} + 2\vec{k} = (1)\vec{a} - (3)\vec{b}

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

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

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

Multiply by 2 to find $6\vec{b}$ :

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

5. Calculate $\vec{a} - 6\vec{b}$

\vec{a} - 6\vec{b} = (-\hat{i} - \hat{j} + \hat{k}) - (-4\hat{i} - 4\hat{j} - 2\hat{k})

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

= 3\hat{i} + 3\hat{j} + 3\hat{k}

= 3(\hat{i} + \hat{j} + \hat{k})

Correct Option: (3)

Choose the correct answer:

Explore More