If the angle between the vectors vec{a} = (2, | vedclass

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(D) For the angle between vectors $\vec{a} = 2\hat{i} + (\log_3 x)\hat{j} + a\hat{k}$ and $\vec{b} = -3\hat{i} + (a \log_3 x)\hat{j} + (\log_3 x)\hat{

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(D) For the angle between vectors $\vec{a} = 2\hat{i} + (\log_3 x)\hat{j} + a\hat{k}$ and $\vec{b} = -3\hat{i} + (a \log_3 x)\hat{j} + (\log_3 x)\hat{k}$ to be acute,the dot product must be positive: $\vec{a} \cdot \vec{b} > 0$.Calculating the dot product: $(2)(-3) + (\log_3 x)(a \log_3 x) + (a)(\log_3 x) > 0$.$-6 + a(\log_3 x)^2 + a(\log_3 x) > 0$.Let $y = \log_3 x$. Since $x > 0$,$y$ can take any real value $y \in \mathbb{R}$.The inequality becomes $ay^2 + ay - 6 > 0$ for all $y \in \mathbb{R}$.For a quadratic $Ay^2 + By + C > 0$ to hold for all $y \in \mathbb{R}$,we must have $A > 0$ and the discriminant $D = B^2 - 4AC Here,$A = a$,$B = a$,and $C = -6$.Condition $1$: $a > 0$.Condition $2$: $D = a^2 - 4(a)(-6) = a^2 + 24a Solving $a(a + 24) Combining $a > 0$ and $a \in (-24, 0)$,there is no real value of $a$ that satisfies both conditions. Thus,there is no such $a$.

If the angle between the vectors $\vec{a} = (2, \log_3 x, a)$ and $\vec{b} = (-3, a \log_3 x, \log_3 x)$ is acute,then...

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