How do we find the set of values of $c$ for which the angle between the vectors $cx\hat{i}-6\hat{j}+3\hat{k}$ & $x\hat{i}-2\hat{j}+cx\hat{k}$ is acute for every $x$ belongs to $\mathbb{R}$ ????
I tried to apply $\cos\theta$ formula but not getting the correct answer. Please give proper and full explanation.
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The angle between two vectors $\vec{u}$ and $\vec{v}$ is acute if and only if $\vec{u} \cdot \vec{v} > 0$.
So, the angle between $\vec{u} = cx\hat{i}-6\hat{j}+3\hat{k}$ and $\vec{v} = x\hat{i}-2\hat{j}+cx\hat{k}$ is acute iff $\vec{u} \cdot \vec{v} = (cx)(x)+(-6)(-2)+(3)(cx) = cx^2+3cx+12 > 0$.
For the quadratic $cx^2+3cx+12$ to be positive for all real $x$, we need either the leading coefficient $c$ to be positive and the discriminant of the quadratic to be negative (i.e. the graph of the quadratic is concave up and there are no zeros) or for $c = 0$ (in this case, $cx^2+3cx+12 = 12$ which is always positive). Can you calculate the discriminant and figure out what values of $c > 0$ make the discriminant negative?
EDIT: Thanks to Siong Thye Goh for catching the fact that $c \ge 0$.
For two vectors to be acute, the inner product has to be positive.
$$\langle (cx,-6,3),(x,-2,cx) \rangle>0$$ which is equivalent to $$cx^2+12+3cx>0$$
Rearranging the terms:
$$cx^2+3cx+12>0$$
For this condition to hold, we require $c\geq0$ otherwise the quadratic function is strictly concave and will attain and some negative value. Also, we require the discriminant $(3c)^2-48c=9c^2-48c$ to be negative.
Credit: Thanks for JimmyK4542 for catching my mistake in the sign of the discriminant.