QJEEM23Ma0e10: JEE Main 2023 Physics - Class 11 JEEM23Ma0e10 Solution: Let and . If is a vector such that and , then is equal to…

Given that $\vec{a} = \vec{b} \times \vec{c}$ , we can find the magnitudes of $\vec{a}$ and $\vec{c}$ :

$|\vec{a}| = \sqrt{3^2 + 1^2 + (-1)^2} = \sqrt{11}$

$|\vec{c}| = \sqrt{2^2 + (-3)^2 + 3^2} = \sqrt{22}$

We know that the magnitude of the cross product of two vectors is equal to the product of the magnitudes of the vectors and the sine of the angle between them :

$|\vec{a}| = |\vec{b} \times \vec{c}| = |\vec{b}||\vec{c}|\sin\theta$

Plugging in the known values :

$\sqrt{11} = \sqrt{50}\sqrt{22}\sin\theta$

Solving for the sine of the angle between the vectors :

$\sin\theta = \frac{1}{10}$

Now we can find $|\vec{b} + \vec{c}|^2$ using the formula :

$|\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b} \cdot \vec{c}$

We have the dot product $\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}|\cos\theta$ , and we can use the relationship between sine and cosine: $\cos\theta = \sqrt{1 - \sin^2\theta} = \frac{\sqrt{99}}{10}$ .

Substitute the values into the formula :

$|\vec{b} + \vec{c}|^2 = 50 + 22 + 2\sqrt{50}\sqrt{22}\frac{\sqrt{99}}{10} = 72 + 66$

Finally, we need to find the absolute value of the difference :

$|72 - |\vec{b} + \vec{c}|^2| = |72 - (72 + 66)| = 66$