Science:Math Exam Resources/Courses/MATH152/April 2012/Question 01 (d)/Solution 1

The area of the parallelogram spanned by two vectors is the magnitude of their cross product. The area of a triangle is half the area of a parallelogram. The parallelogram form by A, B, and C is spanned by ${\vec {AB}}$ and ${\vec {AC}}$ which in 1(a), we already determined were

${\begin{aligned}{\vec {AB}}&=[-2,2,1]\\{\vec {AC}}&=[-1,1,1].\end{aligned}}$

We need the cross product of this, but as we saw in 1(a), this is just the normal vector to the plane $S_{2}$ and therefore ${\vec {AB}}\times {\vec {AC}}=\mathbf {m} =[1,1,0]$. Therefore the area of the triangle is

${\textrm {area}}={\frac {1}{2}}||{\vec {AB}}\times {\vec {AC}}||={\frac {1}{2}}||\mathbf {m} ||={\frac {1}{2}}{\sqrt {1^{2}+1^{2}+0^{2}}}={\frac {1}{2}}{\sqrt {2}}.$