# How do we find circumcenter and orthocenter in 3D and vectors, if the vector of vertices of the triangle are given?

The position vectors of the vertices of triangle are $$3 \hat i + 4 \hat j + 5 \hat k$$, $$\hat i + 7 \hat k$$ and $$5 \hat i + 5 \hat j$$. The distance between the circumcentre and the orthocenter is?

I found the orthocenter using triangle properties and formula. But how do I find orthocenter and circumcenter using vectors. Can I find it using midpoint theorem of two vectors, and then scalar triple product?

• You only need the side lengths of the triangle. Once you have the side lengths, you can compute position of the two centers using formula on wiki. Commented Oct 18, 2020 at 14:23
• I'm not sure why you felt compelled to use an image to convey the problem statement. Using your own words is better, and the notation here is not all that esoteric. See this introduction to posting "typeset" mathematics notation. Even posting your own wording and getting help with the $\LaTeX$ syntax would be an improvement. Commented Oct 18, 2020 at 15:23

$$\vec{a}=(3,4,5);\vec{b}=(1,0,7);\vec{c}=(5,5,0)$$ sides of the triangle are $$x=||\vec{a}-\vec{b}||=2\sqrt{6}\\y=||\vec{a}-\vec{c}||=\sqrt{30}\\ z=||\vec{c}-\vec{b}||=3\sqrt{10}$$ Half perimeter is $$p=\frac{x+y+z}{2}$$ and area $$S=\sqrt{p(p-a)(p-b)(p-c)}$$ Circumscribed circle radius is $$R=\frac{xyz}{4S}$$ distance between circumcenter and orthocenter is $$d=\sqrt{9 R^2-\left(x^2+y^2+z^2\right)}=3 \sqrt{\frac{274}{11}}$$