Question

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

• A. 3
• B. 5
• C. 1
• D. 4

Answer 1

Answer: (B)

5

Answer 2

Given $|\vec{b}| = 1$ and $|\vec{b} \times \vec{a}| = 2$, we need to find $|(\vec{b} \times \vec{a}) - \vec{b}|^2$.

Expanding $|(\vec{b} \times \vec{a}) - \vec{b}|^2$ using $|\vec{u} - \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2\vec{u} \times \vec{v}$:

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

Since $|\vec{b} \times \vec{a}| = 2$, we get $|\vec{b} \times \vec{a}|^2 = 4$, and $|\vec{b}|^2 = 1$.

The cross product $(\vec{b} \times \vec{a}) \times \vec{b} = 0$ because $\vec{b} \times \vec{a}$ is perpendicular to $\vec{b}$.

Substituting these values:

$|(\vec{b} \times \vec{a}) - \vec{b}|^2 = 4 + 1 = 5.$