Let a→×r→+(r→⋅b→)b→=a→ and |a→||b→|2⩽a→⋅b→⩽|a→||b→|2 . If f(θ)=[r→ a→ b→], where θ in angle between vectors a→ and b→ , also given that a→ and b→ are unit vectors, then which of the following option(s) is/are true?

Local minimum value of f ( θ ) is 1 2 , Local maximum value of f ( θ ) is 3 2

Let a→×r→+(r→⋅b→)b→=a→ and |a→||b→|2⩽a→⋅b→⩽|a→||b→|2 . If f(θ)=[r→ a→ b→], where θ in angle between vectors a→ and b→ , also given that a→ and b→ are unit vectors, then which of the following option(s) is/are true?

Take dot product with a → , we get ( r → ⋅ b → ) = | a → | 2 a → ⋅ b → Take cross product with b → we get | a → | 2 a → ⋅ b ⇀ a → − ( a → . b → ) r → = ( b → × a → ) | a → | 2 a → ⋅ b → a → − ( a → ⋅ b → ) r → = ( b → × a → ) ⇒ r → = | a → | 2 ( a → ⋅ b → ) 2 a → + a → × b → a → ⋅ b → ⇒ f ( θ ) = [ r → a → b → ] = | a → × b → | 2 a → . b → = sin 2 θ cos θ

Maximum value of f ( θ ) is 3 2

Local minimum value of f ( θ ) is 1 2

Local maximum value of f ( θ ) is 3 2

Minimum value of f ( θ ) is 1 2

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Let $\vec{a} \times \vec{r} + (\vec{r} \cdot \vec{b}) \vec{b} = \vec{a}$ and $\frac{| \vec{a} | | \vec{b} |}{2} ⩽ \vec{a} \cdot \vec{b} ⩽ \frac{| \vec{a} | | \vec{b} |}{\sqrt{2}}$ . If $f (θ) = [\vec{r} \vec{a} \vec{b}],$ where $θ$ in angle between vectors $\vec{a}$ and $\vec{b}$ , also given that $\vec{a}$ and $\vec{b}$ are unit vectors, then which of the following option(s) is/are true?

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Maximum value of $f (θ)$ is $\frac{3}{2}$

Local minimum value of $f (θ)$ is $\frac{1}{\sqrt{2}}$

Local maximum value of $f (θ)$ is $\frac{3}{2}$

Minimum value of $f (θ)$ is $\frac{1}{\sqrt{2}}$

answer is B, C.

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Detailed Solution

Take dot product with $\vec{a}$ , we get $(\vec{r} \cdot \vec{b}) = \frac{| \vec{a} |^{2}}{\vec{a} \cdot \vec{b}}$
Take cross product with $\vec{b}$ we get $\frac{{| \vec{a} |}^{2}}{\vec{a} \cdot \overset{⇀}{b}} \vec{a} - (\vec{a} . \vec{b}) \vec{r} = (\vec{b} \times \vec{a})$

$\frac{| \vec{a} |^{2}}{\vec{a} \cdot \vec{b}} \vec{a} - (\vec{a} \cdot \vec{b}) \vec{r} = (\vec{b} \times \vec{a}) \Rightarrow \vec{r} = \frac{{| \vec{a} |}^{2}}{{(\vec{a} \cdot \vec{b})}^{2}} \vec{a} + \frac{\vec{a} \times \vec{b}}{\vec{a} \cdot \vec{b}} \Rightarrow f (θ) = [\vec{r} \vec{a} \vec{b}] = \frac{{| \vec{a} \times \vec{b} |}^{2}}{\vec{a} . \vec{b}} = \frac{\sin^{2} θ}{\cos θ}$