| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Point on line satisfying condition |
| Difficulty | Standard +0.8 This is a multi-part vectors question requiring: (i) showing parallelism via dot product of direction vector and normal (straightforward), (ii) finding line-plane intersection (standard substitution), and (iii) finding points equidistant from two planes using the given distance formula, requiring solving |expression1|/√9 = |expression2|/√9, handling absolute values, and finding two solutions. Part (iii) elevates this above routine as it requires careful algebraic manipulation of the distance formula and consideration of cases, though the formula is provided. Slightly above average difficulty for Further Maths Pure. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04f Line-plane intersection: find point |
| Answer | Marks |
|---|---|
| Substitute coordinates of a general point of \(l\) in given equation of plane \(m\) | M1 |
| Obtain equation in \(\lambda\) in any correct form | A1 |
| Verify that the equation is not satisfied for any value of \(\lambda\) | A1 |
| Answer | Marks |
|---|---|
| Substitute for \(\mathbf{r}\) in the vector equation of plane \(m\) and expand scalar product | M1 |
| Obtain equation in \(\lambda\) in any correct form | A1 |
| Verify that the equation is not satisfied for any value of \(\lambda\) | A1 |
| Answer | Marks |
|---|---|
| Expand scalar product of a normal to \(m\) and a direction vector of \(l\) | M1 |
| Verify scalar product is zero | A1 |
| Verify that one point of \(l\) does not lie in the plane | A1 |
| Answer | Marks |
|---|---|
| Use correct method to find perpendicular distance of a general point of \(l\) from \(m\) | M1 |
| Obtain a correct unsimplified expression in terms of \(\lambda\) | A1 |
| Show that the perpendicular distance is 4/3, or equivalent, for all \(\lambda\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Use correct method to find the perpendicular distance of a particular point of \(l\) from \(m\) | M1 | |
| Obtain answer 4/3, or equivalent | A1 | |
| Show that the perpendicular distance of a second point is also 4/3, or equivalent | A1 | [3] |
| Answer | Marks |
|---|---|
| Express general point of \(l\) in component form, e.g. \((1 + 2\lambda, 1 + \lambda, -1 + 2\lambda)\) | B1 |
| Substitute in given equation of \(n\) and solve for \(\lambda\) | M1 |
| Obtain position vector \(5\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}\) from \(\lambda = 2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| State or imply plane \(n\) has vector equation \(\mathbf{r}(2\mathbf{i} - 2\mathbf{j} + \mathbf{k}) = 7\), or equivalent | B1 | |
| Substitute for \(\mathbf{r}\), expand scalar product and solve for \(\lambda\) | M1 | |
| Obtain position vector \(5\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}\) from \(\lambda = 2\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Form an equation in \(\lambda\) by equating perpendicular distances of a general point of \(l\) from \(m\) and \(n\) | M1* | |
| Obtain a correct modular or non-modular equation in \(\lambda\) in any form | A1⬇ | |
| Solve for \(\lambda\) and obtain a point, e.g. \(7\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}\) from \(\lambda = 3\) | A1 | |
| Obtain a second point, e.g. \(3\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) from \(\lambda = 1\) | A1 | |
| Use a correct method to find the distance between the two points | M1(dep*) | |
| Obtain answer 6 | A1 | [6] |
**(i)**
**Main scheme (EITHER):**
| Substitute coordinates of a general point of $l$ in given equation of plane $m$ | M1 | |
| Obtain equation in $\lambda$ in any correct form | A1 | |
| Verify that the equation is not satisfied for any value of $\lambda$ | A1 | |
**OR1:**
| Substitute for $\mathbf{r}$ in the vector equation of plane $m$ and expand scalar product | M1 | |
| Obtain equation in $\lambda$ in any correct form | A1 | |
| Verify that the equation is not satisfied for any value of $\lambda$ | A1 | |
**OR2:**
| Expand scalar product of a normal to $m$ and a direction vector of $l$ | M1 | |
| Verify scalar product is zero | A1 | |
| Verify that one point of $l$ does not lie in the plane | A1 | |
**OR3:**
| Use correct method to find perpendicular distance of a general point of $l$ from $m$ | M1 | |
| Obtain a correct unsimplified expression in terms of $\lambda$ | A1 | |
| Show that the perpendicular distance is 4/3, or equivalent, for all $\lambda$ | A1 | |
**OR4:**
| Use correct method to find the perpendicular distance of a particular point of $l$ from $m$ | M1 | |
| Obtain answer 4/3, or equivalent | A1 | |
| Show that the perpendicular distance of a second point is also 4/3, or equivalent | A1 | [3] |
**(ii)**
**Main scheme (EITHER):**
| Express general point of $l$ in component form, e.g. $(1 + 2\lambda, 1 + \lambda, -1 + 2\lambda)$ | B1 | |
| Substitute in given equation of $n$ and solve for $\lambda$ | M1 | |
| Obtain position vector $5\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}$ from $\lambda = 2$ | A1 | |
**OR:**
| State or imply plane $n$ has vector equation $\mathbf{r}(2\mathbf{i} - 2\mathbf{j} + \mathbf{k}) = 7$, or equivalent | B1 | |
| Substitute for $\mathbf{r}$, expand scalar product and solve for $\lambda$ | M1 | |
| Obtain position vector $5\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}$ from $\lambda = 2$ | A1 | [3] |
**(iii)**
| Form an equation in $\lambda$ by equating perpendicular distances of a general point of $l$ from $m$ and $n$ | M1* | |
| Obtain a correct modular or non-modular equation in $\lambda$ in any form | A1⬇ | |
| Solve for $\lambda$ and obtain a point, e.g. $7\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}$ from $\lambda = 3$ | A1 | |
| Obtain a second point, e.g. $3\mathbf{i} + 2\mathbf{j} + \mathbf{k}$ from $\lambda = 1$ | A1 | |
| Use a correct method to find the distance between the two points | M1(dep*) | |
| Obtain answer 6 | A1 | [6] |
**Note:** [The f.t. is on the components of $l$.]10 Two planes, $m$ and $n$, have equations $x + 2 y - 2 z = 1$ and $2 x - 2 y + z = 7$ respectively. The line $l$ has equation $\mathbf { r } = \mathbf { i } + \mathbf { j } - \mathbf { k } + \lambda ( 2 \mathbf { i } + \mathbf { j } + 2 \mathbf { k } )$.\\
(i) Show that $l$ is parallel to $m$.\\
(ii) Find the position vector of the point of intersection of $l$ and $n$.\\
(iii) A point $P$ lying on $l$ is such that its perpendicular distances from $m$ and $n$ are equal. Find the position vectors of the two possible positions for $P$ and calculate the distance between them.\\[0pt]
[The perpendicular distance of a point with position vector $x _ { 1 } \mathbf { i } + y _ { 1 } \mathbf { j } + z _ { 1 } \mathbf { k }$ from the plane $a x + b y + c z = d$ is $\frac { \left| a x _ { 1 } + b y _ { 1 } + c z _ { 1 } - d \right| } { \sqrt { } \left( a ^ { 2 } + b ^ { 2 } + c ^ { 2 } \right) }$.]
\hfill \mbox{\textit{CAIE P3 2012 Q10 [12]}}