For all real x,the vectors Cx hat{i} - 6 hat{ | vedclass

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(C) Let $\vec{a} = Cx \hat{i} - 6 \hat{j} - 3 \hat{k}$ and $\vec{b} = x \hat{i} + 2 \hat{j} + 2Cx \hat{k}$.Since the angle between $\vec{a}$ and $\vec

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

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(C) Let $\vec{a} = Cx \hat{i} - 6 \hat{j} - 3 \hat{k}$ and $\vec{b} = x \hat{i} + 2 \hat{j} + 2Cx \hat{k}$.Since the angle between $\vec{a}$ and $\vec{b}$ is obtuse,their dot product must be negative,i.e.,$\vec{a} \cdot \vec{b} Calculating the dot product: $(Cx)(x) + (-6)(2) + (-3)(2Cx) $Cx^2 - 12 - 6Cx $Cx^2 - 6Cx - 12 For this quadratic expression in $x$ to be negative for all real $x$,the coefficient of $x^2$ must be negative $(C $D = (-6C)^2 - 4(C)(-12) = 36C^2 + 48C Dividing by $12$: $3C^2 + 4C $C(3C + 4) This inequality holds when $-\frac{4}{3} Since we also require $C < 0$,the intersection is $(-\frac{4}{3}, 0)$.

For all real $x$,the vectors $Cx \hat{i} - 6 \hat{j} - 3 \hat{k}$ and $x \hat{i} + 2 \hat{j} + 2Cx \hat{k}$ make an obtuse angle with each other. Then the value of $C$ can be in:

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