Question

Find the volume of tetrahedron whose vertices are A(1,1,0) B(-4,3,6) C(-1,0,3) and D(2,4,-5).

Answer 1

Hint: Here, we will use the concept that volume of tetrahedron is given as one – sixth of the modulus of the products of the vectors from which it is formed. So, first we will find the vectors and then calculate the scalar triple product which is equal to the determinant of the coefficients of the vectors.

Complete step-by-step answer:
We know that a tetrahedron is one kind of pyramid, which is a polyhedron with a flat polygon base and triangular faces connecting the base to a common point. In the case of a tetrahedron the base is a triangle and any of the four faces can be considered as the base , so a tetrahedron is also known as a triangular pyramid.
A tetrahedron has four faces, six edges and four vertices.
Its three edges meet at each vertex.

The four vertices that we have been given in the question are A(1,1,0) B(-4,3,6) C(-1,0,3) and D(2,4,-5).
So, the vector AB will be = (-4-1) i + (3-1) j + (6-0) k = -5i + 2j +6k
Similarly vector AC is given as = -2i –j + 3k
And, vector AD = i + 3j – 5k

Now, the formula for the volume of the tetrahedron will be = ${\displaystyle \frac{1}{6}}\times |scalartripleproductofthesethreevectors|$
$$={\displaystyle \frac{1}{6}}\times |\{(\overrightarrow{AB}\times \overrightarrow{AC}).\overrightarrow{AD}\}|$$
Scalar triple product = $\left(\begin{array}{ccc}-5& 2& 6\\ -2& -1& 3\\ 1& 3& -5\end{array}\right)$
$\begin{array}{rl}& =-5(5-9)-2(10-3)+6\{-6-(-1)\}\\ & =-5(-4)-2\left(7\right)+6(-6+1)\\ & =20-14-30\\ & =-6-30=-36\end{array}$

Therefore, volume of the tetrahedron = ${\displaystyle \frac{1}{6}}\times |-36|={\displaystyle \frac{36}{6}}=6$
Hence, the volume of the given tetrahedron is 6 cubic units.

Note: Here, it should be noted that we always apply modulus while finding the volume of the tetrahedron. If the scalar triple product is negative, we have to make it positive because volume is always a positive quantity.