Calculate the lengths of the heights descending from triangle vertices Given a triangle, calculate the lengths of the heights descending from triangle vertices  $A, B$ and $C$ respectively. 
$$A(-1,-1,5),~~ B(0,3,-2),~~ C(3,-1,3)$$
I don't get it with which forma i should use and how to solve this question
 A: 1.) Use the distance formula to determine the side lengths:
$$d = \sqrt {(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}$$
$$AB = \sqrt {(-1-0)^2+(-1-3)^2+(5-(-2))^2}=\sqrt{66}$$
$$AC = \sqrt {(-1-3)^2+(-1-(-1))^2+(5-3)^2}=\sqrt{20}$$
$$BC = \sqrt {(0-3)^2+(3-(-1))^2+(-2-3)^2}=\sqrt{50}$$
2.) Then use Heron's formula to calculate the area:
$$Area = \sqrt{s(s-a)(s-b)(s-c)}$$
Where $s = \frac {a+b+c} {2}$.
$$s=\frac {\sqrt{66}+\sqrt{20}+\sqrt{50}} {2}$$
$$Area=\sqrt{(\frac {\sqrt{66}+\sqrt{20}+\sqrt{50}} {2})((\frac {\sqrt{66}+\sqrt{20}+\sqrt{50}} {2})-\sqrt{66})((\frac {\sqrt{66}+\sqrt{20}+\sqrt{50}} {2})-\sqrt{20})((\frac {\sqrt{66}+\sqrt{20}+\sqrt{50}} {2})-\sqrt{50})} \approx 15.780$$
3.) Calculate each height using $Area=\frac12bh$:
$$H_A = 2\frac{Area}{BC}\approx2\frac{15.780}{\sqrt{50}}\approx 4.463$$
$$H_B = 2\frac{Area}{AC}\approx2\frac{15.780}{\sqrt{20}}\approx 7.057$$
$$H_C = 2\frac{Area}{AB}\approx2\frac{15.780}{\sqrt{66}}\approx 3.885$$
A: Hint:


*

*calculate the sides lengths,

*calculate the triangle area,

*use $|\triangle ABC| = \frac{1}{2}h_a\cdot a$ to calculate $h_a$.


I hope this helps ;-)
A: Hint: first, calculate the length of each side, then use Heron's formula to determine the area of the triangle. Finally, use the formula $\text{area}=\text{side}\times\text{height}/2$ to determine the heights.
