Calculate the lengths of the heights descending from triangle vertices

Question

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

Answer 1

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 $Area=\frac12bh$:

$$H_A = 2\frac{Area}{BC}\approx2\frac{15.780}{\sqrt{50}}\approx 4.463$$

$$H_B = 2\frac{Area}{AC}\approx2\frac{15.780}{\sqrt{20}}\approx 7.057$$

$$H_C = 2\frac{Area}{AB}\approx2\frac{15.780}{\sqrt{66}}\approx 3.885$$

Answer 2

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 ;-)

Answer 3

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.