| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Line lies in or parallel to plane |
| Difficulty | Standard +0.3 This is a standard Further Maths vectors question with routine techniques: (i) uses the perpendicular distance formula from point to line (cross product method or projection), and (ii) requires checking that the line's direction vector is perpendicular to the plane's normal and that a point on the line satisfies the plane equation. Both parts are textbook exercises requiring methodical application of learned formulas rather than problem-solving insight. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Find \(\overrightarrow{PQ}\) for a general point \(Q\) on \(l\), e.g. \(-3\mathbf{i}+6\mathbf{k}+\mu(2\mathbf{i}-\mathbf{j}-2\mathbf{k})\) | B1 | |
| Calculate scalar product of \(\overrightarrow{PQ}\) and a direction vector for \(l\) and equate the result to zero | M1 | |
| Solve for \(\mu\) and obtain \(\mu = 2\) | A1 | |
| Carry out a complete method for finding the length of \(\overrightarrow{PQ}\) | M1 | |
| Obtain answer 3 | A1 | |
| Alternative: Calling the point \((1,2,3)\) \(A\), state \(\overrightarrow{AP}\) (or \(\overrightarrow{PA}\)) in component form, e.g. \(3\mathbf{i}-6\mathbf{k}\) | B1 | |
| Use scalar product with direction vector of \(l\) to find projection of \(\overrightarrow{AP}\) (or \(\overrightarrow{PA}\)) on \(l\) | M1 | |
| Obtain correct answer in any form, e.g. \(\frac{18}{\sqrt{9}}\) | A1 | |
| Use Pythagoras to find the perpendicular | M1 | |
| Obtain answer 3 | A1 | |
| Alternative: State \(\overrightarrow{AP}\) (or \(\overrightarrow{PA}\)) in component form | B1 | |
| Calculate a vector product with a direction vector for \(l\) | M1 | |
| Obtain correct answer, e.g. \(6\mathbf{i}-6\mathbf{j}-3\mathbf{k}\) | A1 | |
| Divide modulus of the product by that of the direction vector | M1 | |
| Obtain answer 3 | A1 | |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substitute coordinates of a general point of \(l\) in the plane equation and equate constant terms | M1 | |
| Obtain a correct equation, e.g. \(a + 2b + 6 = 13\) | A1 | |
| Equate the coefficient of \(\mu\) to zero | M1 | |
| Obtain a correct equation, e.g. \(2a - b - 4 = 0\) | A1 | |
| Substitute \((1, 2, 3)\) in the plane equation | M1 | |
| Obtain a correct equation, e.g. \(a + 2b + 6 = 13\) | A1 | |
| Alternative: Find a second point on \(l\) and obtain an equation in \(a\) and/or \(b\) | M1 | |
| Obtain a correct equation, e.g. \(5a - 2 = 13\) | A1 | |
| Equate scalar product of a direction vector for \(l\) and a vector normal for the plane to zero | M1 | |
| Obtain a correct equation, e.g. \(2a - b - 4 = 0\) | A1 | |
| Solve for \(a\) or for \(b\) | M1 | |
| Obtain \(a = 3\) and \(b = 2\) | A1 | |
| Total | 6 |
## Question 10(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Find $\overrightarrow{PQ}$ for a general point $Q$ on $l$, e.g. $-3\mathbf{i}+6\mathbf{k}+\mu(2\mathbf{i}-\mathbf{j}-2\mathbf{k})$ | B1 | |
| Calculate scalar product of $\overrightarrow{PQ}$ and a direction vector for $l$ and equate the result to zero | M1 | |
| Solve for $\mu$ and obtain $\mu = 2$ | A1 | |
| Carry out a complete method for finding the length of $\overrightarrow{PQ}$ | M1 | |
| Obtain answer 3 | A1 | |
| **Alternative:** Calling the point $(1,2,3)$ $A$, state $\overrightarrow{AP}$ (or $\overrightarrow{PA}$) in component form, e.g. $3\mathbf{i}-6\mathbf{k}$ | B1 | |
| Use scalar product with direction vector of $l$ to find projection of $\overrightarrow{AP}$ (or $\overrightarrow{PA}$) on $l$ | M1 | |
| Obtain correct answer in any form, e.g. $\frac{18}{\sqrt{9}}$ | A1 | |
| Use Pythagoras to find the perpendicular | M1 | |
| Obtain answer 3 | A1 | |
| **Alternative:** State $\overrightarrow{AP}$ (or $\overrightarrow{PA}$) in component form | B1 | |
| Calculate a vector product with a direction vector for $l$ | M1 | |
| Obtain correct answer, e.g. $6\mathbf{i}-6\mathbf{j}-3\mathbf{k}$ | A1 | |
| Divide modulus of the product by that of the direction vector | M1 | |
| Obtain answer 3 | A1 | |
| **Total** | **5** | |
## Question 10(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute coordinates of a general point of $l$ in the plane equation and equate constant terms | M1 | |
| Obtain a correct equation, e.g. $a + 2b + 6 = 13$ | A1 | |
| Equate the coefficient of $\mu$ to zero | M1 | |
| Obtain a correct equation, e.g. $2a - b - 4 = 0$ | A1 | |
| Substitute $(1, 2, 3)$ in the plane equation | M1 | |
| Obtain a correct equation, e.g. $a + 2b + 6 = 13$ | A1 | |
| **Alternative:** Find a second point on $l$ and obtain an equation in $a$ and/or $b$ | M1 | |
| Obtain a correct equation, e.g. $5a - 2 = 13$ | A1 | |
| Equate scalar product of a direction vector for $l$ and a vector normal for the plane to zero | M1 | |
| Obtain a correct equation, e.g. $2a - b - 4 = 0$ | A1 | |
| Solve for $a$ or for $b$ | M1 | |
| Obtain $a = 3$ and $b = 2$ | A1 | |
| **Total** | **6** | |10 The line $l$ has equation $\mathbf { r } = \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } + \mu ( 2 \mathbf { i } - \mathbf { j } - 2 \mathbf { k } )$.\\
(i) The point $P$ has position vector $4 \mathbf { i } + 2 \mathbf { j } - 3 \mathbf { k }$. Find the length of the perpendicular from $P$ to $l$.\\
(ii) It is given that $l$ lies in the plane with equation $a x + b y + 2 z = 13$, where $a$ and $b$ are constants. Find the values of $a$ and $b$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE P3 2019 Q10 [11]}}