## Question

The points A and B have position vectors \(\overrightarrow {{\text{OA}}} = \left\{ {\begin{array}{*{20}{c}} 1 \\ 2 \\ { – 2} \end{array}} \right\}\) and \(\overrightarrow {{\text{OB}}} = \left\{ {\begin{array}{*{20}{c}} 1 \\ 0 \\ 2 \end{array}} \right\}\).

a.Find \(\overrightarrow {{\text{OA}}} \times \overrightarrow {{\text{OB}}} \).[2]

b.Hence find the area of the triangle OAB.[2]

**▶️Answer/Explanation**

## Markscheme

\(\overrightarrow {{\text{OA}}} \times \overrightarrow {{\text{OB}}} = \left( {\begin{array}{*{20}{c}} 4 \\ { – 4} \\ { – 2} \end{array}} \right)\)** ***(M1)A1*

**Note: M1A0 **can be awarded for attempt at a correct method

**shown**, or correct method implied by the digits 4, 4, 2 found in the correct order.

*[2 marks]*

\({\text{area}} = \frac{1}{2}\sqrt {{4^2} + {4^2} + {2^2}} = 3\) *M1A1*

*[2 marks]*

## Examiners report

Generally well done. Most students were able to obtain full marks on this question. Most of the errors made were due to careless mistakes.

Generally well done. Most students were able to obtain full marks on this question. Most of the errors made were due to careless mistakes. A few students did not take notice of the “hence” in part (b) and were consequently not able to obtain the marks.

## Question

The points A and B have position vectors \(\overrightarrow {{\text{OA}}} = \left\{ {\begin{array}{*{20}{c}} 1 \\ 2 \\ { – 2} \end{array}} \right\}\) and \(\overrightarrow {{\text{OB}}} = \left\{ {\begin{array}{*{20}{c}} 1 \\ 0 \\ 2 \end{array}} \right\}\).

a.Find \(\overrightarrow {{\text{OA}}} \times \overrightarrow {{\text{OB}}} \).[2]

b.Hence find the area of the triangle OAB.[2]

**▶️Answer/Explanation**

## Markscheme

\(\overrightarrow {{\text{OA}}} \times \overrightarrow {{\text{OB}}} = \left( {\begin{array}{*{20}{c}} 4 \\ { – 4} \\ { – 2} \end{array}} \right)\)** ***(M1)A1*

**Note: M1A0 **can be awarded for attempt at a correct method

**shown**, or correct method implied by the digits 4, 4, 2 found in the correct order.

*[2 marks]*

\({\text{area}} = \frac{1}{2}\sqrt {{4^2} + {4^2} + {2^2}} = 3\) *M1A1*

*[2 marks]*

## Examiners report

Generally well done. Most students were able to obtain full marks on this question. Most of the errors made were due to careless mistakes.

Generally well done. Most students were able to obtain full marks on this question. Most of the errors made were due to careless mistakes. A few students did not take notice of the “hence” in part (b) and were consequently not able to obtain the marks.