AQA Further Paper 2 2019 June — Question 7 6 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2019
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Cross Product & Distances
TypeArea of triangle using cross product
DifficultyStandard +0.3 This is a straightforward Further Maths vector geometry question testing standard techniques: computing a vector product to find triangle area, using the result to find a plane equation, then calculating perpendicular distance. All steps are routine applications of formulas with no conceptual challenges or novel problem-solving required, making it slightly easier than average even for Further Maths.
Spec4.04b Plane equations: cartesian and vector forms4.04g Vector product: a x b perpendicular vector
The points \(A\), \(B\) and \(C\) have coordinates \(A(4, 5, 2)\), \(B(-3, 2, -4)\) and \(C(2, 6, 1)\)
  1. Use a vector product to show that the area of triangle \(ABC\) is \(\frac{5\sqrt{11}}{2}\) [4 marks]
  2. The points \(A\), \(B\) and \(C\) lie in a plane. Find a vector equation of the plane in the form \(\mathbf{r} \cdot \mathbf{n} = k\) [1 mark]
  3. Hence find the exact distance of the plane from the origin. [1 mark]
Question 7:
AnswerMarks
7(a)Forms two vectors from
A, B and C, at least one
AnswerMarks Guidance
correctAO1.1a M1
     
−3 × 1 = 5
     
     
−6 −1 −13
𝐴𝐴����𝐴𝐴�⃗×𝐴𝐴����𝐴𝐴�⃗ =   
Area =
1
2�𝐴𝐴����𝐴𝐴�⃗×�𝐴𝐴���𝐴𝐴�⃗� 5 11
�9 2 +5 2 +(−13) 2 2
= 2 =
Obtains the correct vector
AnswerMarks Guidance
product.AO1.1b A1
Uses their vector product
correctly in a formula for
AnswerMarks Guidance
the area of a triangleAO1.2 M1
Uses a rigorous argument
to show the required
result, including stating
AnswerMarks Guidance
“Area =” oeAO2.1 R1
(b)States the correct
equationAO1.1b B1
(c)Divides their non-zerok
by the magnitude of their
normal vector to obtain
AnswerMarks Guidance
their exact valueAO3.1a B1F
7 11
−13
11
AnswerMarks Guidance
Total6
QMarking Instructions AO 
Question 7:
--- 7(a) ---
7(a) | Forms two vectors from
A, B and C, at least one
correct | AO1.1a | M1 | −7 −2  9 
     
−3 × 1 = 5
     
     
−6 −1 −13
𝐴𝐴����𝐴𝐴�⃗×𝐴𝐴����𝐴𝐴�⃗ =   
Area =
1
2�𝐴𝐴����𝐴𝐴�⃗×�𝐴𝐴���𝐴𝐴�⃗� 5 11
�9 2 +5 2 +(−13) 2 2
= 2 =
Obtains the correct vector
product. | AO1.1b | A1
Uses their vector product
correctly in a formula for
the area of a triangle | AO1.2 | M1
Uses a rigorous argument
to show the required
result, including stating
“Area =” oe | AO2.1 | R1
(b) | States the correct
equation | AO1.1b | B1 | 9
(c) | Divides their non-zerok
by the magnitude of their
normal vector to obtain
their exact value | AO3.1a | B1F | 𝐫𝐫.� 5 �= 35
7 11
−13
11
Total | 6
Q | Marking Instructions | AO | Marks | Typical Solution
The points $A$, $B$ and $C$ have coordinates $A(4, 5, 2)$, $B(-3, 2, -4)$ and $C(2, 6, 1)$

\begin{enumerate}[label=(\alph*)]
\item Use a vector product to show that the area of triangle $ABC$ is $\frac{5\sqrt{11}}{2}$
[4 marks]

\item The points $A$, $B$ and $C$ lie in a plane.

Find a vector equation of the plane in the form $\mathbf{r} \cdot \mathbf{n} = k$
[1 mark]

\item Hence find the exact distance of the plane from the origin.
[1 mark]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 2 2019 Q7 [6]}}
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