Let a and b be two vectors such that |b|=1 and |b a|=2.

Question

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

Accepted Answer

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

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

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

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

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

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

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

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

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

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

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