| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line intersection with line |
| Difficulty | Standard +0.3 This is a standard three-part vectors question requiring routine techniques: (a) dot product of direction vectors equals zero, (b) equating components and solving simultaneous equations, (c) using the perpendicular distance formula. All methods are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| Use correct method to evaluate the scalar product of relevant vectors | M1 | \((-4-2+6)\) |
| Obtain answer zero and deduce the given statement | A1 | Need a conclusion or a statement in advance that the scalar product will be zero. |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Express general point of \(l\) or \(m\) in component form, e.g. \((3+4s,\,2-s,\,5+3s)\) or \((1-t,\,-1+2t,\,-2+2t)\) | B1 | |
| Equate at least two pairs of components and solve for \(s\) or for \(t\) | M1 | |
| Obtain correct answer \(s=-1\) and \(t=2\) | A1 | |
| Verify that all three equations are satisfied | A1 | |
| State position vector of the intersection \(-\mathbf{i}+3\mathbf{j}+2\mathbf{k}\), or equivalent | A1 | Can come from 1 correct value and no contradictory statement. |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Taking a general point \(P\) on \(m\), form an equation in \(t\) by either equating a relevant scalar product to zero, or equating the derivative of \( | \overrightarrow{OP} | \) to zero, or taking a specific point \(Q\) on \(m\), e.g. \((1,-1,-2)\), using Pythagoras in triangle \(OPQ\) |
| Obtain \(t = \frac{7}{9}\) | A1 | |
| Carry out correct method to find \(OP\) | DM1 | |
| Obtain \(\dfrac{\sqrt{5}}{3}\) | A1 | Obtain the given answer from full and correct working. |
| Alternative method for 9(c): | ||
| Take a specific point \(Q\) on \(m\), e.g. \((-1,3,2)\) and use a scalar product to find \(QN\), the projection of \(\overrightarrow{OQ}\) on \(m\) | \*M1 | |
| Obtain \(QN = \frac{11}{3}\), or equivalent | A1 | |
| Use Pythagoras to obtain \(ON\) | DM1 | |
| Obtain the given answer correctly | A1 | |
| Total: 4 |
## Question 9(a):
| Use correct method to evaluate the scalar product of relevant vectors | M1 | $(-4-2+6)$ |
| Obtain answer zero and deduce the given statement | A1 | Need a conclusion or a statement in advance that the scalar product will be zero. |
| **Total: 2** | | |
## Question 9(b):
| Express general point of $l$ or $m$ in component form, e.g. $(3+4s,\,2-s,\,5+3s)$ or $(1-t,\,-1+2t,\,-2+2t)$ | B1 | |
| Equate at least two pairs of components and solve for $s$ or for $t$ | M1 | |
| Obtain correct answer $s=-1$ and $t=2$ | A1 | |
| Verify that all three equations are satisfied | A1 | |
| State position vector of the intersection $-\mathbf{i}+3\mathbf{j}+2\mathbf{k}$, or equivalent | A1 | Can come from 1 correct value and no contradictory statement. |
| **Total: 5** | | |
## Question 9(c):
| Taking a general point $P$ on $m$, form an equation in $t$ by **either** equating a relevant scalar product to zero, **or** equating the derivative of $|\overrightarrow{OP}|$ to zero, **or** taking a specific point $Q$ on $m$, e.g. $(1,-1,-2)$, using Pythagoras in triangle $OPQ$ | \*M1 | e.g. $\begin{pmatrix}1-t\\-1+2t\\-2+2t\end{pmatrix}\cdot\begin{pmatrix}-1\\2\\2\end{pmatrix}=0$ |
| Obtain $t = \frac{7}{9}$ | A1 | |
| Carry out correct method to find $OP$ | DM1 | |
| Obtain $\dfrac{\sqrt{5}}{3}$ | A1 | Obtain the **given answer** from full and correct working. |
| **Alternative method for 9(c):** | | |
| Take a specific point $Q$ on $m$, e.g. $(-1,3,2)$ and use a scalar product to find $QN$, the projection of $\overrightarrow{OQ}$ on $m$ | \*M1 | |
| Obtain $QN = \frac{11}{3}$, or equivalent | A1 | |
| Use Pythagoras to obtain $ON$ | DM1 | |
| Obtain the given answer correctly | A1 | |
| **Total: 4** | | |
---9 Two lines $l$ and $m$ have equations $\mathbf { r } = 3 \mathbf { i } + 2 \mathbf { j } + 5 \mathbf { k } + s ( 4 \mathbf { i } - \mathbf { j } + 3 \mathbf { k } )$ and $\mathbf { r } = \mathbf { i } - \mathbf { j } - 2 \mathbf { k } + t ( - \mathbf { i } + 2 \mathbf { j } + 2 \mathbf { k } )$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Show that $l$ and $m$ are perpendicular.
\item Show that $l$ and $m$ intersect and state the position vector of the point of intersection.
\item Show that the length of the perpendicular from the origin to the line $m$ is $\frac { 1 } { 3 } \sqrt { 5 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2021 Q9 [11]}}