WJEC Further Unit 3 Specimen — Question 3 9 marks

Exam BoardWJEC
ModuleFurther Unit 3 (Further Unit 3)
SessionSpecimen
Marks9
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TopicVectors 3D & Lines
TypeKinematics with position vectors
DifficultyChallenging +1.2 This is a standard Further Maths mechanics problem requiring position vectors as functions of time, finding when the distance is minimized by differentiating |AB|² and setting to zero, then calculating the minimum distance. While it involves multiple steps and 3D vectors, the method is algorithmic and well-practiced in Further Maths courses—harder than typical A-level due to the vector calculus involved, but not requiring novel insight.
Spec1.10f Distance between points: using position vectors
At time \(t = 0\) s, the position vector of an object \(A\) is \(\mathbf{i}\) m and the position vector of another object \(B\) is \(3\mathbf{i}\) m. The constant velocity vector of \(A\) is \(2\mathbf{i} + 5\mathbf{j} - 4k\) ms\(^{-1}\) and the constant velocity vector of \(B\) is \(\mathbf{i} + 3\mathbf{j} - 5k\) ms\(^{-1}\). Determine the value of \(t\) when \(A\) and \(B\) are closest together and find the least distance between \(A\) and \(B\). [9]
AnswerMarks Guidance
Position vectors: \(\mathbf{r} = \mathbf{p} + t\mathbf{v}\)M1 AO3
\(\mathbf{r}_A = (1 + 2t)\mathbf{i} + 5\mathbf{j} - 4t\mathbf{k}\)A1 AO2
\(\mathbf{r}_B = (3 + t)\mathbf{i} + 3t\mathbf{j} - 5t\mathbf{k}\)M1 AO3
\(\mathbf{r}_B - \mathbf{r}_A = (2 - t)\mathbf{i} - 2t\mathbf{j} - t\mathbf{k}\)A1 AO1
Distance squared: \(AB^2 = x^2 + y^2 + z^2 = (2-t)^2 + 4t^2 + t^2\)M1 A1 AO1 AO1
\(AB^2 = 6t^2 - 4t + 4\)M1 AO2
Differentiate: \(\frac{d(AB^2)}{dt} = 2(2-t)(-1) + 10t = 12t - 4\)m1 AO2
\(-4 + 2t + 10t = 0\) giving \(t = \frac{1}{3}\)A1 AO1
(Least distance)\(^2\) = \((2 - \frac{1}{3})^2 + 5(\frac{1}{3})^2 = \frac{10}{3}\)A1 AO1
Least distance = \(\sqrt{\frac{10}{3}} = 1.83\) (m)A1 AO1
Total: [9]
**Position vectors:** $\mathbf{r} = \mathbf{p} + t\mathbf{v}$ | M1 | AO3 | Used

$\mathbf{r}_A = (1 + 2t)\mathbf{i} + 5\mathbf{j} - 4t\mathbf{k}$ | A1 | AO2 | either correct, any form

$\mathbf{r}_B = (3 + t)\mathbf{i} + 3t\mathbf{j} - 5t\mathbf{k}$ | M1 | AO3

$\mathbf{r}_B - \mathbf{r}_A = (2 - t)\mathbf{i} - 2t\mathbf{j} - t\mathbf{k}$ | A1 | AO1

**Distance squared:** $AB^2 = x^2 + y^2 + z^2 = (2-t)^2 + 4t^2 + t^2$ | M1 A1 | AO1 AO1 | cao

$AB^2 = 6t^2 - 4t + 4$ | M1 | AO2 | at least 1 power reduced

**Differentiate:** $\frac{d(AB^2)}{dt} = 2(2-t)(-1) + 10t = 12t - 4$ | m1 | AO2

$-4 + 2t + 10t = 0$ giving $t = \frac{1}{3}$ | A1 | AO1 | equating to 0 cao

**(Least distance)$^2$ = $(2 - \frac{1}{3})^2 + 5(\frac{1}{3})^2 = \frac{10}{3}$** | A1 | AO1

**Least distance = $\sqrt{\frac{10}{3}} = 1.83$ (m)** | A1 | AO1 | cao

**Total: [9]**

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At time $t = 0$ s, the position vector of an object $A$ is $\mathbf{i}$ m and the position vector of another object $B$ is $3\mathbf{i}$ m. The constant velocity vector of $A$ is $2\mathbf{i} + 5\mathbf{j} - 4k$ ms$^{-1}$ and the constant velocity vector of $B$ is $\mathbf{i} + 3\mathbf{j} - 5k$ ms$^{-1}$. Determine the value of $t$ when $A$ and $B$ are closest together and find the least distance between $A$ and $B$. [9]

\hfill \mbox{\textit{WJEC Further Unit 3  Q3 [9]}}
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