Hello, Oasis1993!

The points A,B and C have position vectors given respectively:

. . $\displaystyle a = \langle7,4,-2\rangle,\;\;b = \langle5,3,-3\rangle, \;c = \langle6, 5,-4\rangle$

b) Find the area of $\displaystyle \Delta ABC.$ You are expected to know this fact:

The area of the parallelogram determined by two vectors

. . is the magnitude of the cross product of the two vectors. Code:

* . . . . . .
/ .
v / .
/ .
* - - - - - *
u
Area = |u x v|

Hence, the area of the *triangle* determined by two vectors is: .$\displaystyle \tfrac{1}{2}|\vec u \times \vec v|$

Let: .$\displaystyle \begin{array}{ccccc}\vec u &=& \overrightarrow{ab} &=& \langle \text{-}2,\text{-}1,-1\rangle \\ \vec v &=& \overrightarrow{bc} &=& \langle 1,2,\text{-}1\rangle \end{array}$

Then: .$\displaystyle \vec u \times \vec v \;-\;\begin{vmatrix}i&j&k \\ \text{-}2&\text{-}1&\text{-}1 \\ 1&2&\text{-}1 \end{vmatrix} \;=\;3i - 3j - 3k \;=\;\langle3,\text{-}3,\text{-}3\rangle$

$\displaystyle \text{Area of parallelogram}\;=\;\sqrt{3^2 + (\text{-}3)^2+(\text{-}3)^2} \;=\;\sqrt{27} \;=\;3\sqrt{3}$

. . $\displaystyle \text{Therefore: area of triangle} \:=\:\frac{3\sqrt{3}}{2}$