Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{b}|=1$ and $|\vec{b} \times \vec{a}|=2$. Then $|(\vec{b} \times \vec{a})-\vec{b}|^2$ is equal to

<p>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:</p> <p>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,</p> <p>$(\vec{b} \times \vec{a}) \cdot \vec{b}= 0$(because $\vec{b} \times \vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{a}$)</p> <p>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:</p> <p>$$|(\vec{b} \times \vec{a}) - \vec{b}|^2 = |\vec{b} \times \vec{a}|^2 + |-\vec{b}|^2 + 2(\vec{b} \times \vec{a}) \cdot (-\vec{b})$$</p> <p>Since $(\vec{b} \times \vec{a}) \cdot \vec{b} = 0$ and $|-\vec{b}| = |\vec{b}|$, the equation simplifies to:</p> <p>$= |\vec{b} \times \vec{a}|^2 + |\vec{b}|^2$</p> <p>Given that $|\vec{b} \times \vec{a}| = 2$ and $|\vec{b}| = 1$, substituting these values gives:</p> <p>$= 2^2 + 1^2 = 4 + 1 = 5$</p> <p>Therefore, the value of $|(\vec{b} \times \vec{a})-\vec{b}|^2$ is <strong>5</strong>.</p>