QCE Vault / Specialist Maths Vectors and matrices — Question 14 QCAA 2025, Paper 2 · 8 marks
Browse all questions Vectors revision Original practice exam Report an issue Q14 · 2025 · Technology-active Simple familiar 8 marks
QUESTION 14 (8 marks) The origin, O O O , is joined to points A ( 1 , 2 , 5 ) A(1,2,5) A ( 1 , 2 , 5 ) and B ( − 3 , 4 , 0 ) B(-3,4,0) B ( − 3 , 4 , 0 ) to form triangle O A B OAB O A B .
Point C C C is the point on O B OB O B such that A C AC A C is perpendicular to O B OB O B , as shown. a) Given the length of side O B OB O B is 5 units, show that the vector projection of O A → \overrightarrow{OA} O A on O B → \overrightarrow{OB} O B is
1 5 ( − 3 4 0 ) . \frac15\begin{pmatrix}-3\\4\\0\end{pmatrix}. 5 1 − 3 4 0 . [2 marks] b) Use your result from Question 14a) to determine the length of O C OC O C . [1 mark] c) Determine the length of side O A OA O A . [1 mark] d) Use Pythagoras’ theorem to determine the length of A C AC A C . [1 mark] e) Use your result from Question 14d) to determine the area of triangle O A B OAB O A B . [1 mark] f) Use a vector product method to verify your result from Question 14e).
[2 marks] WORKED SOLUTION
QCAA guide · typeset solution 8 marks ANSWER a) proj O B → O A → = 1 5 ( − 3 , 4 , 0 ) \operatorname{proj}_{\overrightarrow{OB}}\overrightarrow{OA}=\frac15(-3,4,0) proj O B O A = 5 1 ( − 3 , 4 , 0 ) . b) O C = 1 OC=1 O C = 1 . c) O A = 30 OA=\sqrt{30} O A = 30 . d) A C = 29 AC=\sqrt{29} A C = 29 . e) Area = 5 29 2 =\frac{5\sqrt{29}}2 = 2 5 29 units 2 ^2 2 . f) O A → × O B → = ( − 20 , − 15 , 10 ) \overrightarrow{OA}\times\overrightarrow{OB}=(-20,-15,10) O A × O B = ( − 20 , − 15 , 10 ) verifies the same area. Worked solution
a) Using b ^ = O B → / ∣ O B ∣ = 1 5 ( − 3 , 4 , 0 ) \hat{\mathbf b}=\overrightarrow{OB}/|OB|=\frac15(-3,4,0) b ^ = O B /∣ O B ∣ = 5 1 ( − 3 , 4 , 0 ) ,
proj O B → O A → = ( O A → ⋅ b ^ ) b ^ = 1 5 ( − 3 4 0 ) . \operatorname{proj}_{\overrightarrow{OB}}\overrightarrow{OA}=(\overrightarrow{OA}\cdot\hat{\mathbf b})\hat{\mathbf b}=\frac15\begin{pmatrix}-3\\4\\0\end{pmatrix}. proj O B O A = ( O A ⋅ b ^ ) b ^ = 5 1 − 3 4 0 . b) The projection is O C → \overrightarrow{OC} O C , so O C = 1 OC=1 O C = 1 unit. c)
O A = 1 2 + 2 2 + 5 2 = 30 . OA=\sqrt{1^2+2^2+5^2}=\sqrt{30}. O A = 1 2 + 2 2 + 5 2 = 30 . d) By Pythagoras,
A C = O A 2 − O C 2 = 30 − 1 = 29 . AC=\sqrt{OA^2-OC^2}=\sqrt{30-1}=\sqrt{29}. A C = O A 2 − O C 2 = 30 − 1 = 29 . e)
Area ( △ O A B ) = 1 2 ( O B ) ( A C ) = 1 2 ( 5 ) ( 29 ) = 5 29 2 units 2 . \text{Area}(\triangle OAB)=\frac12(OB)(AC)=\frac12(5)(\sqrt{29})=\frac{5\sqrt{29}}2\ \text{units}^2. Area ( △ O A B ) = 2 1 ( O B ) ( A C ) = 2 1 ( 5 ) ( 29 ) = 2 5 29 units 2 . f)
O A → × O B → = ( 1 2 5 ) × ( − 3 4 0 ) = ( − 20 − 15 10 ) . \overrightarrow{OA}\times\overrightarrow{OB}=\begin{pmatrix}1\\2\\5\end{pmatrix}\times\begin{pmatrix}-3\\4\\0\end{pmatrix}=\begin{pmatrix}-20\\-15\\10\end{pmatrix}. O A × O B = 1 2 5 × − 3 4 0 = − 20 − 15 10 .
Thus
1 2 ∣ O A → × O B → ∣ = 1 2 400 + 225 + 100 = 5 29 2 , \frac12\left|\overrightarrow{OA}\times\overrightarrow{OB}\right|=\frac12\sqrt{400+225+100}=\frac{5\sqrt{29}}2, 2 1 O A × O B = 2 1 400 + 225 + 100 = 2 5 29 ,
which verifies the result. correctly substitutes into a suitable rule that represents the vector projection
[1 mark] determines the vector projection having demonstrated accurate calculation of the scalar product
[1 mark] determines the length of side O C OC O C [1 mark] correctly determines the length of side O A OA O A [1 mark] determines the length of A C AC A C [1 mark] determines the area of triangle O A B OAB O A B [1 mark] correctly calculates a suitable vector product
[1 mark] verifies the result from Question 14e) based on the vector product result
[1 mark] One QCAA sample method typeset for web; criterion wording is adapted from the official marking guide.
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