QJEEM24Ma6421: JEE Main 2024 Physics - Class 11 JEEM24Ma6421 Solution: Let and be two vectors such that and . Then is equal to

To find the value of $|(\vec{b} \times \vec{a})-\vec{b}|^2$ , we can use properties of vector operations and magnitudes. Given $|\vec{b}| = 1$ and $|\vec{b} \times \vec{a}| = 2$ , let's break down the calculation step by step:

Firstly, we observe that the cross product of two vectors $\vec{b} \times \vec{a}$ is orthogonal (perpendicular) to both $\vec{b}$ and $\vec{a}$ . This means that when we take the dot product of $\vec{b} \times \vec{a}$ with either $\vec{b}$ or $\vec{a}$ , the result will be zero due to the orthogonal property. Specifically,

$(\vec{b} \times \vec{a}) \cdot \vec{b}= 0$ (because $\vec{b} \times \vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{a}$ )

Next, to find the magnitude squared of the vector $(\vec{b} \times \vec{a}) - \vec{b}$ , we apply the formula for the magnitude squared of a vector subtraction, which can be expressed as:

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

Since $(\vec{b} \times \vec{a}) \cdot \vec{b} = 0$ and $|-\vec{b}| = |\vec{b}|$ , the equation simplifies to:

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

Given that $|\vec{b} \times \vec{a}| = 2$ and $|\vec{b}| = 1$ , substituting these values gives:

$= 2^2 + 1^2 = 4 + 1 = 5$

Therefore, the value of $|(\vec{b} \times \vec{a})-\vec{b}|^2$ is 5.