If the sum of two unit vectors is a unit vector, prove that the magnitude of their difference is \sqrt{3}

 

 

 

 
 
 
 
 

Answers (1)

Let \vec{a} and \vec{b} be two unit vectors.

Given : |\vec{a}|=1, |\vec{b}|=1, |\vec{a}+\vec{b}|=1  [Sum of two unit vectors is a unit vector]

To prove:  |\vec{a}-\vec{b}|=\sqrt{3}

LHS : |\vec{a}-\vec{b}|

(\vec{a}-\vec{b})\cdot (\vec{a}-\vec{b})=|\vec{a}-\vec{b}|\cdot |\vec{a}-\vec{b}|\cos 0\\ \Rightarrow \vec{a}(\vec{a}-\vec{b})-\vec{b}(\vec{a}-\vec{b})=|\vec{a}-\vec{b}|^2\cdot 1\\\Rightarrow \vec{a}\cdot \vec{a}-\vec{a}\cdot \vec{b}-\vec{b}\cdot \vec{a}-\vec{b}\cdot (-\vec{b})=|\vec{a}-\vec{b}|^2\\ \Rightarrow |\vec{a}|^2-\vec{a}\cdot \vec{b}-\vec{b}\cdot \vec{a}+|\vec{b}|^2=|\vec{a}-\vec{b}|^2\\\Rightarrow |\vec{a}-\vec{b}|^2=1-2|\vec{a}\cdot \vec{b}|+1= 2(1-\vec{a}\cdot \vec{b})\: \: \: \: \: -(1)

To find the value \vec{a}\cdot \vec{b}

(\vec{a}+\vec{b})\cdot (\vec{a}+\vec{b})=|\vec{a}+\vec{b}|\cdot |\vec{a}+\vec{b}|\cos 0\\ \Rightarrow \vec{a}(\vec{a}+\vec{b})+\vec{b}(\vec{a}+\vec{b})=1\\\Rightarrow \vec{a}\cdot \vec{a}+\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{a}+\vec{b}\cdot \vec{b}=1\\ \Rightarrow |\vec{a}|^2+2\vec{a}\cdot \vec{b}+|\vec{b}|^2=1\\\Rightarrow 2(\vec{a}\cdot \vec{b})=-1\, \, \: \: \: \: \: \: -(2)

Substituting  \vec{a}\cdot \vec{b} value in equation 1

|\vec{a}-\vec{b}|^2=2\left ( 1-\left ( -\frac{1}{2} \right ) \right )=3

\therefore |\vec{a}-\vec{b}|=\sqrt{3}

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