JEE Mathematics 2024 Question with Solution

Given: The 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 them is acute.

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

Now compute the dot product:

So,

Hence the required condition is:

Solve the corresponding quadratic equation:

Using the quadratic formula,

Therefore,

Now,

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

Therefore, the correct option is C.

The solution contains one inconsistent intermediate approach, but the correct working gives the least positive integral value as $5$.

Given: We need the angle between $\vec{A}=\alpha \hat{i}-2\hat{j}+2\hat{k}$ and $\vec{B}=\alpha \hat{i}+2\alpha \hat{j}-2\hat{k}$ to be acute.

Find: The smallest positive integer $\alpha$.

An acute angle means:

Substituting components,

Thus we solve:

The roots are:

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

So the valid values satisfy:

because we need a positive integer. As $2+2\sqrt{2}\approx4.828$, the first positive integer greater than this is $5$.

Therefore, the least positive integral value is $5$, so the correct option is C.