Let x in R and log_2 x 0. Then,the vectors | vedclass

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(B) Given,$A = 2 \hat{i} + \log_2 x \hat{j} + s \hat{k}$ and $B = \log_2 x \hat{i} + s \hat{j} + \log_2 x \hat{k}$.Let the angle between $A$ and $B$ b

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$s > -1$

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(B) Given,$A = 2 \hat{i} + \log_2 x \hat{j} + s \hat{k}$ and $B = \log_2 x \hat{i} + s \hat{j} + \log_2 x \hat{k}$.Let the angle between $A$ and $B$ be $	heta$. Then,$\cos 	heta = \frac{A \cdot B}{|A| |B|}$.$A \cdot B = (2)(\log_2 x) + (\log_2 x)(s) + (s)(\log_2 x) = 2 \log_2 x + 2s \log_2 x = 2 \log_2 x (1 + s)$.For $	heta$ to be an acute angle,we must have $\cos 	heta > 0$.Since $|A| > 0$ and $|B| > 0$,the condition $\cos 	heta > 0$ implies $A \cdot B > 0$.Therefore,$2 \log_2 x (1 + s) > 0$.Given $\log_2 x > 0$,we can divide by $2 \log_2 x$ without changing the inequality sign.Thus,$1 + s > 0$,which implies $s > -1$.

Let $x \in R$ and $\log_2 x > 0$. Then,the vectors $A = (2, \log_2 x, s)$ and $B = (\log_2 x, s, \log_2 x)$ include an acute angle if

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