Let $\overline{AD},$ $\overline{BE},$ $\overline{CF}$ be the altitudes of acute triangle $ABC.$ If $9 \overrightarrow{AD} + 4 \overrightarrow{BE} + 7 \overrightarrow{CF} = \mathbf{0},$ compute $\angle ACB,$ in degrees.
I know that, in general, $\overrightarrow{XY} = \overrightarrow{Y}-\overrightarrow{X}$, so then I changed the first equation to $$9\overrightarrow{D}+4\overrightarrow{E}+7\overrightarrow{F}-9\overrightarrow{A}-4\overrightarrow{B}-7\overrightarrow{C} = \mathbf{0}.$$
Additionally, $\overline{AD}$ and $\overline{BC}$ are orthogonal, and same with the other two altitudes. So that means $$\overrightarrow{AD}\cdot \overrightarrow{BC} = \mathbf{0}$$$$(\overrightarrow{D}-\overrightarrow{A})(\overrightarrow{C}-\overrightarrow{B})=\mathbf{0}$$$$\overrightarrow{DC}-\overrightarrow{DB}-\overrightarrow{AC}+\overrightarrow{AB}=\mathbf{0}.$$
However, I can't seem to relate this equation to the original one. I also know that $\cos{\angle{ACB}} = \frac{\overrightarrow{AC}\cdot \overrightarrow{BC}}{||AC||\text{ }||BC||}$, but I'm not sure how to find the magnitudes of $\overrightarrow{AC}$ and $\overrightarrow{BC}$, since there are no concrete side lengths. I could also find the $\sin$ of the angle using cross product, but I don't think that will come up either.
Help is greatly appreciated, thanks!
Let $\triangle ABC$ have sides ${a,b,c}$. Let the corresponding unit altitude vectors ${\hat h_a, \hat h_b, \hat h_c }$. Also let area of $\triangle ABC = \Delta$.
We use the following result : $$a\hat h_a+ b\hat h_b+ c\hat h_c =0 \tag{1}$$
(To prove this bring these three vectors with tails at a common point, mark the angles and compare the components. One should obtain the familiar sine-rule and projection-rule.)
Given is $$9\cdot \frac{2\Delta}{a} \, \hat h_a+ 4\cdot \frac{2\Delta}{b} \, \hat h_b+ 7\cdot \frac{2\Delta}{c} \, \hat h_c =0 $$
Comparing this with (1), we get $$a:b:c = 3:2:\sqrt 7$$
Now use cosine-rule to obtain $$\angle C=60^{\circ}$$