Calculate the height of the tetrahedron OABC with respect to O, given the points O(0,0,0) , A(2,4,0), B(0,2,4), C(6,0,2).

Question

Calculate the height of the tetrahedron OABC with respect to O, given the points O(0,0,0) , A(2,4,0), B(0,2,4), C(6,0,2) .

Volume is $$V = \frac{1}{3}S_{base}h = \frac{|(A \cdot (B \times C) )|}{6} = \frac{88}{6}$$

And the area of triangle ABC is

$$S_{base} = \frac{|\vec{AB} \times \vec{BC}|}{2} = 10\sqrt2$$

And according to my calculations this should result in $h = \frac{11\sqrt2}{5}$.

$$\frac{10\sqrt2}{3}h = \frac{88}{6} \rightarrow h = \frac{11\sqrt2}{5} $$

But my textbook says this should result in $h = \frac{13\sqrt2}{5}$. So who's correct?

Answer 1

Your book is correct.

The volume of tetrahedron is $\frac{104}{6}$, not $\frac{88}{6}$. This is where you're wrong.

_{(Always be careful while calculating the value of determinant, sometimes we forget to multiply by negative sign, while opening through first row, in the case of center element, I.e. $a_{12}$.)}