Let A→ and B→ be two $$non-parallel$$ unit vectors in a plane. If $$($$αA→$$+B$$→$$)$$ bisects the internal angle between A→ and B→, then ∝ is equal to

Question

Let A→ and B→ be two $$non-parallel$$ unit vectors in a plane. If $$($$αA→$$+B$$→$$)$$ bisects the  internal angle between A→ and B→, then ∝ is equal to

Infinity Learn · Accepted Answer

A→  ⋅ $$($$ α  A→   $$+$$ B→  $$)$$  $$=$$  B→  ⋅ $$($$ α  A→   $$+$$ B→  $$)$$  ⇒as A→$$=B$$→    α $$+$$ A→ ⋅ B→  $$=$$   α  A→   ⋅ B→ $$+$$ $$1$$ α$$-1A$$→·B→$$-1=0$$ ⇒   A→   ⋅ B→   ≠   $$1$$     as A→   ⋅ B→ $$=A$$→   ⋅ B→  $$cos$$ θ then θ$$=0$$ which means parallel but given non parallel  ⇒  ∝    $$=$$  $$1$$

$Let \vec{A} and \vec{B} be two non-parallel unit vectors in a plane. If (α \vec{A} + \vec{B}) bisects the$ $internal angle between \vec{A} and \vec{B}, then \propto is equal to$

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Let A→ and B→ be two non-parallel unit vectors in a plane. If (αA→+B→) bisects the internal angle between A→ and B→, then ∝ is equal to

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$Let \vec{A} and \vec{B} be two non-parallel unit vectors in a plane. If (α \vec{A} + \vec{B}) bisects the$ $internal angle between \vec{A} and \vec{B}, then \propto is equal to$