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]