QJEEM24Ph6c91: JEE Main 2024 Physics - Class 11 JEEM24Ph6c91 Solution: If and makes an angle with each other, then for The integer…

To solve this problem, we will use the concepts of vector addition and magnitudes involving the dot product. Given the angle between $\vec{a}$ and $\vec{b}$ is $\cos^{-1}\left(\frac{5}{9}\right)$ , we can use the properties of dot products and magnitudes to find the required integer value of $n$ .

First, let's use the condition $|\vec{a} + \vec{b}| = \sqrt{2} |\vec{a} - \vec{b}|$ . Square both sides to remove the square roots:

$|\vec{a} + \vec{b}|^2 = 2 |\vec{a} - \vec{b}|^2$

Now we will expand both sides using the formula for the magnitude of the sum and difference of vectors:

$|\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = \vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b}$

And

$|\vec{a} - \vec{b}|^2 = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) = \vec{a} \cdot \vec{a} - 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b}$

Substitute these back into the original equation:

$\vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = 2 (\vec{a} \cdot \vec{a} - 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b})$

Expand the right-hand side:

$\vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = 2 \vec{a} \cdot \vec{a} - 4 \vec{a} \cdot \vec{b} + 2 \vec{b} \cdot \vec{b}$

Rearrange all terms to one side to combine like terms:

$\vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} - 2 \vec{a} \cdot \vec{a} + 4 \vec{a} \cdot \vec{b} - 2 \vec{b} \cdot \vec{b} = 0$

Combine like terms:

$- \vec{a} \cdot \vec{a} + 6 \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{b} = 0$

We know that:

$\vec{a} \cdot \vec{a} = |\vec{a}|^2 \quad \text{and} \quad \vec{b} \cdot \vec{b} = |\vec{b}|^2 \quad \text{and} \quad \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta)$

Since the angle between $\vec{a}$ and $\vec{b}$ is $\cos^{-1}\left(\frac{5}{9}\right)$ , we have:

$\cos(\theta) = \frac{5}{9}$

Substitute these back into the equation:

$-|\vec{a}|^2 + 6 |\vec{a}| |\vec{b}| \left(\frac{5}{9}\right) - |\vec{b}|^2 = 0$

Simplify it further:

$-|\vec{a}|^2 + \frac{30}{9} |\vec{a}| |\vec{b}| - |\vec{b}|^2 = 0$

$-|\vec{a}|^2 + \frac{10}{3} |\vec{a}| |\vec{b}| - |\vec{b}|^2 = 0$

We know that $|\vec{a}| = n |\vec{b}|$ . Substitute this into the equation:

$-(n |\vec{b}|)^2 + \frac{10}{3} (n |\vec{b}|) |\vec{b}| - |\vec{b}|^2 = 0$

Simplify it:

$-n^2 |\vec{b}|^2 + \frac{10}{3} n |\vec{b}|^2 - |\vec{b}|^2 = 0$

Factor out $|\vec{b}|^2$ :

$|\vec{b}|^2 \left(-n^2 + \frac{10}{3} n - 1\right) = 0$

Since $|\vec{b}|^2 \neq 0$ , we can solve:

$-n^2 + \frac{10}{3} n - 1 = 0$

Multiply through by 3 to clear the fraction:

$-3n^2 + 10n - 3 = 0$

Rearrange it to match standard quadratic form:

$3n^2 - 10n + 3 = 0$

Solve this quadratic equation using the quadratic formula:

$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 3$ , $b = -10$ , and $c = 3$ :

$n = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm \sqrt{64}}{6} = \frac{10 \pm 8}{6}$

This results in two possible solutions for $n$ :

$n = \frac{18}{6} = 3 \quad \text{and} \quad n = \frac{2}{6} = \frac{1}{3}$

However, since $n$ is given to be an integer, we take:

$\boxed{3}$