Question 9 - A-Level Maths

WJEC Further Unit 1 2018 June — Question 9 14 marks

Exam Board	WJEC
Module	Further Unit 1 (Further Unit 1)
Year	2018
Session	June
Marks	14
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors: Lines & Planes
Type	Perpendicular distance point to line
Difficulty	Standard +0.3 This is a standard Further Maths vectors question covering routine techniques: deriving vector equations from two points, converting to Cartesian form, checking if lines intersect by solving simultaneous equations, and finding a common perpendicular using the cross product. All parts follow textbook methods with no novel insight required, though it requires careful algebraic manipulation across multiple steps.
Spec	4.04a Line equations: 2D and 3D, cartesian and vector forms 4.04e Line intersections: parallel, skew, or intersecting 4.04g Vector product: a x b perpendicular vector

The line $L_1$ passes through the points $A(1, 2, -3)$ and $B(-2, 1, 0)$.

1. Show that the vector equation of $L_1$ can be written as $$\mathbf{r} = (1 - 3\lambda)\mathbf{i} + (2 - \lambda)\mathbf{j} + (-3 + 3\lambda)\mathbf{k}.$$
2. Write down the equation of $L_1$ in Cartesian form. [4]

The vector equation of the line $L_2$ is given by $\mathbf{r} = 2\mathbf{i} - 4\mathbf{j} + \mu(4\mathbf{j} + 7\mathbf{k})$.

Show that $L_1$ and $L_2$ do not intersect. [5]
Find a vector in the direction of the common perpendicular to $L_1$ and $L_2$. [5]

Show mark scheme Show mark scheme source

Part a)

(i)

Answer	Marks	Guidance
Answer: $AB = (-2i + j) - (i + 2j - 3k) = -3i - j + 3k$; Therefore, $r = i + 2j - 3k + \lambda(-3i - j + 3k)$ or $r = (1-3\lambda)i + (2-\lambda)j + (-3+3\lambda)k$	B1 M1 A1	Accept equivalent convincing

(ii)

Answer	Marks
Answer: $\frac{x-1}{-3} = \frac{y-2}{-1} = \frac{z+3}{3}$	B1

Part b)

Answer	Marks	Guidance
Answer: If intersecting, $1 - 3\lambda = 2, 2 - \lambda = -4 + 4\mu, -3 + 3\lambda = 7\mu$	M1
Answer: Solving (first pair): $\lambda = -\frac{1}{3}$ and $\mu = \frac{19}{12}$	m1 A1	Accept solution which use different pair; 2nd and 3rd equation give $\lambda = 54/19, \mu = 15/19$ then $-143/19 \neq 2$
Answer: Verifying in third equation: $-4 \neq 133/12$ $\rightarrow$ they do not intersect	A1 E1	1st and 3rd equation give $\lambda = -1/3, \mu = -4/7$ then $7/3 \neq -44/7$

Part c)

Answer	Marks	Guidance
Answer: Let $n = pi + qj + rk$
Answer: Then, from $L_1$: $(pi + qj + rk) \cdot (-3i - j + 3k) = 0$ $\rightarrow -3p - q + 3r = 0$ and, from $L_2$: $(pi + qj + rk) \cdot (4j + 7k) = 0$ $\rightarrow 4q + 7r = 0$	M1 A1	Accept M1 here if not awarded previously
Answer: Let $r = t$, then $q = -7t/4$ and $p = 19i/12$; $n = (19/12)i - (7t/4)j + tk$ (with $t \neq 0$)	M1 A1	Accept eg. If $t = 12$, $n = 19i - 21j + 12k$

**Part a)**

**(i)**
Answer: $AB = (-2i + j) - (i + 2j - 3k) = -3i - j + 3k$; Therefore, $r = i + 2j - 3k + \lambda(-3i - j + 3k)$ or $r = (1-3\lambda)i + (2-\lambda)j + (-3+3\lambda)k$ | B1 M1 A1 | Accept equivalent convincing

**(ii)**
Answer: $\frac{x-1}{-3} = \frac{y-2}{-1} = \frac{z+3}{3}$ | B1 |

**Part b)**
Answer: If intersecting, $1 - 3\lambda = 2, 2 - \lambda = -4 + 4\mu, -3 + 3\lambda = 7\mu$ | M1 |

Answer: Solving (first pair): $\lambda = -\frac{1}{3}$ and $\mu = \frac{19}{12}$ | m1 A1 | Accept solution which use different pair; 2nd and 3rd equation give $\lambda = 54/19, \mu = 15/19$ then $-143/19 \neq 2$

Answer: Verifying in third equation: $-4 \neq 133/12$ $\rightarrow$ they do not intersect | A1 E1 | 1st and 3rd equation give $\lambda = -1/3, \mu = -4/7$ then $7/3 \neq -44/7$

**Part c)**
Answer: Let $n = pi + qj + rk$ | 

Answer: Then, from $L_1$: $(pi + qj + rk) \cdot (-3i - j + 3k) = 0$ $\rightarrow -3p - q + 3r = 0$ and, from $L_2$: $(pi + qj + rk) \cdot (4j + 7k) = 0$ $\rightarrow 4q + 7r = 0$ | M1 A1 | Accept M1 here if not awarded previously

Answer: Let $r = t$, then $q = -7t/4$ and $p = 19i/12$; $n = (19/12)i - (7t/4)j + tk$ (with $t \neq 0$) | M1 A1 | Accept eg. If $t = 12$, $n = 19i - 21j + 12k$

Show LaTeX source

The line $L_1$ passes through the points $A(1, 2, -3)$ and $B(-2, 1, 0)$.

\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the vector equation of $L_1$ can be written as
$$\mathbf{r} = (1 - 3\lambda)\mathbf{i} + (2 - \lambda)\mathbf{j} + (-3 + 3\lambda)\mathbf{k}.$$
\item Write down the equation of $L_1$ in Cartesian form. [4]
\end{enumerate}
\end{enumerate}

The vector equation of the line $L_2$ is given by $\mathbf{r} = 2\mathbf{i} - 4\mathbf{j} + \mu(4\mathbf{j} + 7\mathbf{k})$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that $L_1$ and $L_2$ do not intersect. [5]

\item Find a vector in the direction of the common perpendicular to $L_1$ and $L_2$. [5]
\end{enumerate}

\hfill \mbox{\textit{WJEC Further Unit 1 2018 Q9 [14]}}

This paper (9 questions)

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Q1 6 Q2 6 Q3 8 Q4 7 Q5 8 Q6 7 Q7 5 Q8 9 Q9 14

Answer	Marks	Guidance
Answer: If intersecting, \(1 - 3\lambda = 2, 2 - \lambda = -4 + 4\mu, -3 + 3\lambda = 7\mu\)	M1
Answer: Solving (first pair): \(\lambda = -\frac{1}{3}\) and \(\mu = \frac{19}{12}\)	m1 A1	Accept solution which use different pair; 2nd and 3rd equation give \(\lambda = 54/19, \mu = 15/19\) then \(-143/19 \neq 2\)
Answer: Verifying in third equation: \(-4 \neq 133/12\) \(\rightarrow\) they do not intersect	A1 E1	1st and 3rd equation give \(\lambda = -1/3, \mu = -4/7\) then \(7/3 \neq -44/7\)

Answer	Marks	Guidance
Answer: Let \(n = pi + qj + rk\)
Answer: Then, from \(L_1\): \((pi + qj + rk) \cdot (-3i - j + 3k) = 0\) \(\rightarrow -3p - q + 3r = 0\) and, from \(L_2\): \((pi + qj + rk) \cdot (4j + 7k) = 0\) \(\rightarrow 4q + 7r = 0\)	M1 A1	Accept M1 here if not awarded previously
Answer: Let \(r = t\), then \(q = -7t/4\) and \(p = 19i/12\); \(n = (19/12)i - (7t/4)j + tk\) (with \(t \neq 0\))	M1 A1	Accept eg. If \(t = 12\), \(n = 19i - 21j + 12k\)