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(C) The angle between two vectors $\vec{a}$ and $\vec{b}$ is obtuse if and only if their dot product is negative,i.e.,$\vec{a} \cdot \vec{b} < 0$. Cal

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$\left(-\frac{4}{3}, 0\right)$

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(C) The angle between two vectors $\vec{a}$ and $\vec{b}$ is obtuse if and only if their dot product is negative,i.e.,$\vec{a} \cdot \vec{b} Calculating the dot product: $\vec{a} \cdot \vec{b} = (cx)(x) + (-6)(2) + (3)(2cx) = cx^2 - 12 + 6cx$. We require $cx^2 + 6cx - 12 For a quadratic expression $Ax^2 + Bx + C Here,$A = c$,$B = 6c$,and $C = -12$. Condition $1$: $c Condition $2$: $D = B^2 - 4AC = (6c)^2 - 4(c)(-12) = 36c^2 + 48c $12c(3c + 4) The roots are $c = 0$ and $c = -4/3$. For the inequality to hold,$c$ must lie between the roots: $-4/3 Since $c < 0$ is already satisfied by this interval,the set of all real values of $c$ is $\left(-\frac{4}{3}, 0\right)$.

The set of all real values of $c$ such that the angle between the vectors $\vec{a} = cx \hat{i} - 6 \hat{j} + 3 \hat{k}$ and $\vec{b} = x \hat{i} + 2 \hat{j} + 2cx \hat{k}$ is an obtuse angle for all real $x$ is:

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