Find the area of the parallelogram ABCD.

By using a suitable scalar product of two vectors, determine whether \({\rm{A\hat BC}}\) is acute or obtuse.

\(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AD}}} = - \)i \( + 10\)j – 7k     M1A1

\({\text{area}} = \left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AD}}} } \right|{\text{ = }}\sqrt {{1^2} + {{10}^2} + {7^2}} \)

 \( = 5\sqrt 6 \left( {\sqrt {150} } \right)\)     A1

[3 marks]

METHOD 1

\(\overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AD}}} = - 4 - 2 - 6\)     M1A1

\( = - 12\)

considering the sign of the answer

\(\overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AD}}} < 0\), therefore angle \({\rm{D\hat AB}}\) is obtuse     M1

(as it is a parallelogram), \({\rm{A\hat BC}}\) is acute     A1

[4 marks]

METHOD 2

\(\overrightarrow {{\text{BA}}} \bullet \overrightarrow {{\text{BC}}} = + 4 + 2 + 6\)     M1A1

\( = 12\) considering the sign of the answer     M1

\(\overrightarrow {{\text{BA}}} \bullet \overrightarrow {{\text{BC}}} > 0 \Rightarrow {\rm{A\hat BC}}\) is acute     A1

[4 marks]