Question

If the sum of two unit vector is a unit vector, find the angle between these two vectors.​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm \: Let \: \vec{a} \: and \: \vec{b} \: are \: two \: units \: vector \: and \: \theta \: be \: angle \: between \: them.$

$\rm\implies \: |\vec{a} | = |\vec{b} | = 1$

Further given that

$\rm \: |\vec{a} + \vec{b} | = 1$

$\rm \: |\vec{a} + \vec{b} |^{2} = {1}^{2}$

$\rm \: (\vec{a} + \vec{b} ).(\vec{a} + \vec{b} ) = 1$

$\rm \: \vec{a} .\vec{a} + \vec{a} .\vec{b} + \vec{b} .\vec{a} + \vec{b} .\vec{b} = 1$

$\rm \: { |\vec{a} | }^{2} + 2(\vec{a} .\vec{b} ) + { |\vec{b} | }^{2} = 1$

$\rm \: {1}^{2} + 2( |\vec{a} | |\vec{b} | \: cos\theta) + {1}^{2} = 1$

$\rm \: 1 + 2(1 \times 1 \times cos\theta) + 1 = 1$

$\rm \: 1 + 2 \: cos\theta = 0$

$\rm \: 2 \: cos\theta = - 1$

$\rm \: cos\theta = - \dfrac{1}{2}$

$\rm \: cos\theta = cos \dfrac{2\pi}{3}$

$\rm\implies \:\boxed{ \rm{ \:\rm \:\theta = \dfrac{2\pi}{3} \: \: }}$

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Formulae Used :-

$\boxed{ \rm{ \: |\vec{a} | \: is \: unit \: vector \: \rm\implies \: |\vec{a} | = 1 \: }}$

$\boxed{ \rm{ \: |\vec{a} |^{2} = \vec{a} \: . \: \vec{a} \: }}$

$\boxed{ \rm{ \:\vec{a} \: . \: \vec{b} \: = \: |\vec{a} | |\vec{b} | \: cos\theta \: }}$

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Additional information :-

$\boxed{ \rm{ \:\vec{a} \: . \: \vec{b} \: = \: \vec{b} \: . \: \vec{a} \: }}$

$\boxed{ \rm{ \:\vec{a} \: \times \: \vec{b} \: = \: - \: \vec{b} \: \times \: \vec{a} \: }}$

$\boxed{ \rm{ \:\vec{a} \: \times \: \vec{a} \: = \: 0 \: }}$

$\boxed{ \rm{ \:\vec{a} \: \times \: \vec{b} \: = \: 0 \: \rm\implies \:\vec{a} \: \parallel \: \vec{b} }}$

$\boxed{ \rm{ \:\vec{a} \: . \: \vec{b} \: = \: 0 \: \rm\implies \:\vec{a} \: \perp\: \vec{b} }}$

$\boxed{ \rm{ \: |\vec{a} \times \vec{b} | = |\vec{a} | \: |\vec{b} | \: sin\theta \: }}$