Question

$\large\underline\bold{ANSWER \:FAST\red{\huge{\checkmark}}}$

the sum of two unit vectors is also a vector of unit magnitude then the magnitude of the difference of the two unit vectors is?

a)1 unit
b)2 unit
c)√3 units
d)0​

Answer 1

Given :-

The sum of two unit vectors is also a vector of unit magnitude.

To Find :-

The magnitude of the difference of the two unit vectors.

Solution :-

A/q,

$\sf{|\overrightarrow{a}+\overrightarrow{b}|=1}$

$\sf{\implies \sqrt{ |\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2|\overrightarrow{a}||\overrightarrow{b}|cos\theta}=1 }$

$\sf{\implies 1+1+2cos\theta=1 }$

$\sf{\implies 2(1+cos\theta)=1}$

$\sf{\implies cos\theta=\dfrac{1}{2}-1 }$

$\sf{\implies cos\theta=\dfrac{-1}{2} }$

$\sf{\implies \theta=120^\circ}$

_______________________

$\sf{|\overrightarrow{a}-\overrightarrow{b}|=\sqrt{ |\overrightarrow{a}|^2+|\overrightarrow{b}|^2-2|\overrightarrow{a}||\overrightarrow{b}|cos\theta} }$

$\sf{=\sqrt{1+1-2cos120^\circ} }$

$\sf{=\sqrt{2+2\times \dfrac{1}{2}} }$

$\sf{=\sqrt{2+1} }$

$\sf{=\sqrt{3} }$

Therefore, the answer is √3.

Answer 2

$\huge\underline\bold{AnSwEr,}$

Given :-

The sum of two unit vectors is also a vector of unit magnitude.

To Find :-

The magnitude of the difference of the two unit vectors.

Solution :-

A/q,

$\sf{|\overrightarrow{a}+\overrightarrow{b}|=1}$

$\sf{\implies \sqrt{ |\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2|\overrightarrow{a}||\overrightarrow{b}|cos\theta}=1 }$

$\sf{\implies 1+1+2cos\theta=1 }$

$\sf{\implies 2(1+cos\theta)=1}$

$\sf{\implies cos\theta=\dfrac{1}{2}-1 }$

$\sf{\implies cos\theta=\dfrac{-1}{2} }$

$\sf{\implies \theta=120^\circ}$

_______________________

$\sf{|\overrightarrow{a}-\overrightarrow{b}|=\sqrt{ |\overrightarrow{a}|^2+|\overrightarrow{b}|^2-2|\overrightarrow{a}||\overrightarrow{b}|cos\theta} }$

$\sf{=\sqrt{1+1-2cos120^\circ} }$

$\sf{=\sqrt{2+2\times \dfrac{1}{2}} }$

$\sf{=\sqrt{2+1} }$

$\sf{=\sqrt{3} }$

Therefore, the answer is √3.

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