Answer to Question #132570 in Analytic Geometry for Promise Omiponle

Question

1. Find the volume of the parallelpiped spanned by the vectors
A=i+j-k; B= i-j+k; C=-i+j+k.

Answer 1

The volume of a parallelepiped determined by the vectors a, b, and c, is the magnitude of their scalar triple product.

and the formula for finding the volume is given by;

Volume = (height) × (Area of base)

$Volume, V=\begin{vmatrix} 𝐚. (𝐛 ×𝐜) \\ \end{vmatrix}$

$⇒V=\begin{vmatrix} (1,1,-1).((1,-1,1)×(-1,1,1)) \\ \end{vmatrix}$

The scalar triple product can be calculated using the formula;

$𝐚.(𝐛 × 𝐜) = \begin{vmatrix} ai & aj & ak \\ bi & bj & bk \\ ci & cj & ck \end{vmatrix}$

$=\begin{vmatrix} 1 & 1 & -1 \\ 1 & -1 & 1 \\ -1 & 1 & 1 \end{vmatrix}$

$= -1(1×1-(-1)×(-1))-1(1×1-(-1)×1)+1(1×(-1)-1×1)$

$=-1(1-1)-1(1+1)+1(-1-1)$

$=-1(0)-1(2)+1(-2)$

$=0-2-2$

$= -4$

Since we want the modulus, i.e. the absolute value of the result, we therefore conclude that the parallelpiped has volume 4 cubic units.