Let $x \in R$ and $\log _2 x>0$. Then, the vectors $\mathbf{A}=\left(2, \log _2 x, s\right)$ and…
Let $x \in R$ and $\log _2 x>0$. Then, the vectors $\mathbf{A}=\left(2, \log _2 x, s\right)$ and $\mathbf{B}=\left(\log _2 x, s, \log _2 x\right)$ include an acute angle if
$s>1$
$s>-1$
$s=-1$
$s < -1$
Solution
$
\begin{aligned}
& \text { (b) Given, } A=2 \hat{\mathbf{i}}+\log _2 x \hat{\mathbf{j}}+s \hat{\mathbf{k}} \\
& \mathbf{B}=\log _2 x \hat{\mathbf{i}}+s \hat{\mathbf{j}}+\log _2 x \hat{\mathbf{k}}
\end{aligned}
$
Let the angle between $A$ and $B$ be $\theta$, then
$
\cos \theta=\frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}||\mathbf{B}|}=\frac{2 \log _2 x+s \log _2 x+s \log _2 x}{|A||B|}
$
' $\theta$ ' will be acute angle if $\cos \theta>0$,
$
\begin{array}{lc}
\text { i.e. } & \frac{2 \log _2 x+2 s \log _2 x}{|A||B|}>0 \\
\Rightarrow & 2 \log _2 x+2 s \log _2 x>0 \Rightarrow 2 \log _2 x^{s+1}>0 \\
\Rightarrow & x^{s+1}>2^{\circ} \\
\Rightarrow & x^{s+1}>1
\end{array}
$
\begin{array}{ll}
\Rightarrow & x^{s+1}>x^{\circ} \\
\Rightarrow & 1+s>0 \quad\left[\because 2 \log _2 x>0\right. always ] \\
\Rightarrow & \multicolumn{2}{c}{s>-1}
\end{array}
$