Standard +0.3 This is a straightforward application of the standard formula for triangle area using cross product: Area = ½|AB × AC|. Students need to find two vectors, compute their cross product (a routine procedure), find its magnitude, and divide by 2. While it involves several computational steps, it requires no problem-solving insight—just direct application of a well-practiced technique from the Further Maths syllabus.
6.
The points \(A , B\) and \(C\) have coordinates \(A ( 4,5,2 ) , B ( - 3,2 , - 4 )\) and \(C ( 2,6,1 )\)
Use a vector product to show that the area of triangle \(A B C\) is \(\frac { 5 \sqrt { 11 } } { 2 }\)
[0pt]
[4 marks]
6.
The points $A , B$ and $C$ have coordinates $A ( 4,5,2 ) , B ( - 3,2 , - 4 )$ and $C ( 2,6,1 )$\\
Use a vector product to show that the area of triangle $A B C$ is $\frac { 5 \sqrt { 11 } } { 2 }$\\[0pt]
[4 marks]\\
\hfill \mbox{\textit{SPS SPS FM 2020 Q6 [4]}}