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Visualisation of Subtraction of Polynomials

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Question

If $\vec{a} and \vec{b}$ are two unit vectors such that $|\vec{a} + \vec{b}|$ is also a unit vector, then $|\overset{⇀}{a} - \overset{⇀}{b}|______________________.$

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Solution

Given:
$\vec{a} and \vec{b}$ are unit vectors
$|\vec{a} + \vec{b}|$ is also a unit vector

$if (\vec{a} + \vec{b}) is a unit vector \Rightarrow |\vec{a} + \vec{b}| = 1 \Rightarrow {|\vec{a} + \vec{b}|}^{2} = 1 \Rightarrow (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = 1 \Rightarrow \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + 2 \vec{a} \cdot \vec{b} = 1 \Rightarrow {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 2 \vec{a} \cdot \vec{b} = 1 \Rightarrow 1 + 1 + 2 \vec{a} \cdot \vec{b} = 1 (∵ |\vec{a}| = 1, |\vec{b}| = 1) \Rightarrow 2 \vec{a} \cdot \vec{b} = - 1 . . . (1) Now, {|\vec{a} - \vec{b}|}^{2} = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} - 2 \vec{a} \cdot \vec{b} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} - (- 1) (From (1)) = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} + 1 = 1 + 1 + 1 (∵ |\vec{a}| = 1, |\vec{b}| = 1) = 3 \Rightarrow |\vec{a} - \vec{b}| = \sqrt{3}$

Hence, $|\vec{a} - \vec{b}| = \sqrt{3}$ .

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