JEE Mathematics 2024 Question with Solution

Given: Vectors are $\vec{A}=\alpha \hat{i}-2\hat{j}+2\hat{k}$ and $\vec{B}=\alpha \hat{i}+2\alpha \hat{j}-2\hat{k}$.

Find: The least positive integral value of $\alpha$ for which the angle between the vectors is acute.

For the angle between two vectors to be acute, their dot product must be positive:

$$\vec{A}\cdot\vec{B}>0$$

Now compute the dot product:

$$\vec{A}\cdot\vec{B}=(\alpha)(\alpha)+(-2)(2\alpha)+(2)(-2)$$ $$=\alpha^2-4\alpha-4$$

So the required condition is

$$\alpha^2-4\alpha-4>0$$

Solve the corresponding quadratic equation:

$$\alpha^2-4\alpha-4=0$$

Using the quadratic formula,

$$\alpha=\frac{-(-4)\pm\sqrt{(-4)^2-4\cdot1\cdot(-4)}}{2\cdot1}$$ $$=\frac{4\pm\sqrt{16+16}}{2}$$ $$=\frac{4\pm4\sqrt{2}}{2}=2\pm2\sqrt{2}$$

Hence,

$$\alpha<2-2\sqrt{2}\quad \text{or} \quad \alpha>2+2\sqrt{2}$$

Now,

$$2+2\sqrt{2}\approx 4.828$$

Therefore, the least positive integer satisfying the condition is $\alpha=5$.

The correct option is A.

Given: The angle between $\alpha \hat{i}-2\hat{j}+2\hat{k}$ and $\alpha \hat{i}+2\alpha \hat{j}-2\hat{k}$ is acute.

Find: The smallest positive integer value of $\alpha$.

An acute angle means

$$\vec{A}\cdot\vec{B}>0$$

Evaluating the dot product from corresponding components:

$$\vec{A}\cdot\vec{B}=\alpha^2-4\alpha-4$$

So we need

$$\alpha^2-4\alpha-4>0$$

The roots of

$$\alpha^2-4\alpha-4=0$$

are

$$\alpha=2\pm2\sqrt{2}$$

Since the coefficient of $\alpha^2$ is positive, the quadratic is positive outside the roots. Thus,

$$\alpha>2+2\sqrt{2}$$

for positive integral values.

Checking the nearest integer:

$$2+2\sqrt{2}\approx 4.828$$

So the least positive integer greater than this is $5$.

Therefore, the answer is $5$, which corresponds to option A.

Note: The first provided approach contains an inconsistent intermediate expansion leading to $\alpha^2-2\alpha-8$, but the final answer matches the correct dot-product computation shown in the second approach.