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

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 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

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

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

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$2$

$1$

$\frac{1}{2}$

$4$

answer is B.

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Detailed Solution

$\vec{A} \cdot (α \vec{A} + \vec{B}) = \vec{B} \cdot (α \vec{A} + \vec{B}) \Rightarrow a s |\vec{A}| = |\vec{B}|$

$α + \vec{A} \cdot \vec{B} = α \vec{A} \cdot \vec{B} + 1$

$(α - 1) (\vec{A} \cdot \vec{B} - 1) = 0$

$\Rightarrow \vec{A} \cdot \vec{B} \neq 1 as \vec{A} \cdot \vec{B} = |\vec{A}| \cdot |\vec{B}| cos θ then θ=0 which means parallel but given non parallel \Rightarrow \propto = 1$