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Mathematics Question on Vector Algebra

Let S be the set of all a∈ R for which the angle between the vectorsu→=a(logeb)i^−6j^+3k^ \begin{array}{l} \overrightarrow{u}=a\left(\text{log}_e b\right)\hat{i}-6\hat{j}+3\hat{k}\end{array} and v→=(logeb)i^+2j^+2a(logeb)k^,(b>1)\begin{array}{l} \overrightarrow{v}=\left(\text{log}_e b\right)\hat{i}+2\hat{j}+2a\left(\text{log}_e b\right)\hat{k},\left(b>1\right)\end{array} is acute. Then S is equal to

A

(∞,−43)(∞, −\frac{4}{3})

B

ϕ

C

(−43,0)(−\frac{4}{3}, 0)

D

(127,∞)(\frac{12}{7}, ∞)

Answer

ϕ

Explanation

Solution

u→=a(logeb)i^−6j^+3k^\begin{array}{l} \overrightarrow{u}=a\left(\text{log}_e b\right)\hat{i}-6\hat{j}+3\hat{k}\end{array}and v→=(logeb)i^+2j^+2a(logeb)k^\begin{array}{l} \overrightarrow{v}=\left(\text{log}_e b\right)\hat{i}+2\hat{j}+2a\left(\text{log}_e b\right)\hat{k} \end{array}

For acute angle, u→⋅v→>0\begin{array}{l} \overrightarrow{u}\cdot\overrightarrow{v}>0 \end{array}

⇒ a(logeb)2−12+6a(logeb)>0\begin{array}{l} \Rightarrow\ a\left(\text{log}_e b\right)^2-12+6a\left(\text{log}_e b\right)>0\end{array}

∵ b > 1 so, loge b=t⇒t>0 as b>1\begin{array}{l} \text{log}_e\ b=t\Rightarrow t>0 \text{ as }b>1 \end{array}

at2+6at−12>0  ∀t>0\begin{array}{l} at^2+6at-12>0~~\forall t>0 \end{array}

⇒a∈ϕ\begin{array}{l}\Rightarrow a \in \phi\end{array}