| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2009 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Parallel and perpendicular planes |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on parallel planes and lines. Part (i) requires using the fact that parallel planes share the same normal vector and substituting a point. Part (ii) uses the standard distance formula between parallel planes. Part (iii) requires finding a direction vector perpendicular to two normal vectors via cross product. All techniques are standard A-level further maths procedures with no novel insight required, making it slightly easier than average. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04d Angles: between planes and between line and plane4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Substitute coordinates (1, 4, 2) in \(2x - 3y + 6z = d\) | M1 | |
| Obtain plane equation \(2x - 3y + 6z = 2\), or equivalent | A1 | [2] |
| (ii) EITHER: Attempt to use plane perpendicular formula to find perpendicular from (1, 4, 2) to p | M1 | |
| Obtain a correct unsimplified expression, e.g. \(\frac{ | 2-3(4)+6(2)-16 | }{\sqrt{(2^2+(-3)^2+6^2)}}\) |
| Obtain answer 2 | A1 | |
| OR1: State or imply perpendicular from O to p is \(\frac{16}{7}\), or from O to q is \(\frac{2}{7}\), or equivalent | B1 | |
| Find difference in perpendiculars | M1 | |
| Obtain answer 2 | A1 | |
| OR2: Obtain correct parameter value, or position vector or coordinates of foot of perpendicular from (1, 4, 2) to p (μ = ±\(\frac{2}{7}\); (\(\frac{11}{7}, \frac{22}{7}, \frac{26}{7}\))) | B1 | |
| Calculate the length of the perpendicular | M1 | |
| Obtain answer 2 | A1 | |
| OR3: Carry out correct method for finding the projection onto a normal vector of a line segment joining a point on p, e.g. (8, 0, 0) and a point on q, e.g. (1, 4, 2) | M1 | |
| Obtain a correct unsimplified expression, e.g. \(\frac{ | 2(8-1)-3(-4)+6(-2) | }{\sqrt{(2^2+(-3)^2+6^2)}}\) |
| Obtain answer 2 | A1 | [3] |
| (iii) EITHER: Calling the direction vector \(ai + bj + ck\), use scalar product to obtain a relevant equation in a, b and c | M1* | |
| Obtain two correct equations, e.g. \(2a - 3b + 6c = 0, a - 2b + 2c = 0\) | A1 | |
| Solve for one ratio, e.g. \(a : b\) | M1(dep*) | |
| Obtain \(a : b : c = 6 : 2 : -1\), or equivalent | A1 | |
| State answer \(r = \lambda(6i + 2j - k)\) or equivalent | A1√ | |
| OR: Attempt to calculate vector product of two normals, e.g. \((i - 2j + 2k) \times (2i - 3j + 6k)\) | M2 | |
| Obtain two correct components | A1 | |
| Obtain \(-6i - 2j + k\), or equivalent | A1 | |
| State answer \(r = \lambda(-6i - 2j + k)\), or equivalent | A1√ | [5] |
**(i)** Substitute coordinates (1, 4, 2) in $2x - 3y + 6z = d$ | M1 |
Obtain plane equation $2x - 3y + 6z = 2$, or equivalent | A1 | [2]
**(ii)** EITHER: Attempt to use plane perpendicular formula to find perpendicular from (1, 4, 2) to p | M1 |
Obtain a correct unsimplified expression, e.g. $\frac{|2-3(4)+6(2)-16|}{\sqrt{(2^2+(-3)^2+6^2)}}$ | A1 |
Obtain answer 2 | A1 |
OR1: State or imply perpendicular from O to p is $\frac{16}{7}$, or from O to q is $\frac{2}{7}$, or equivalent | B1 |
Find difference in perpendiculars | M1 |
Obtain answer 2 | A1 |
OR2: Obtain correct parameter value, or position vector or coordinates of foot of perpendicular from (1, 4, 2) to p (μ = ±$\frac{2}{7}$; ($\frac{11}{7}, \frac{22}{7}, \frac{26}{7}$)) | B1 |
Calculate the length of the perpendicular | M1 |
Obtain answer 2 | A1 |
OR3: Carry out correct method for finding the projection onto a normal vector of a line segment joining a point on p, e.g. (8, 0, 0) and a point on q, e.g. (1, 4, 2) | M1 |
Obtain a correct unsimplified expression, e.g. $\frac{|2(8-1)-3(-4)+6(-2)|}{\sqrt{(2^2+(-3)^2+6^2)}}$ | A1 |
Obtain answer 2 | A1 | [3]
**(iii)** EITHER: Calling the direction vector $ai + bj + ck$, use scalar product to obtain a relevant equation in a, b and c | M1* |
Obtain two correct equations, e.g. $2a - 3b + 6c = 0, a - 2b + 2c = 0$ | A1 |
Solve for one ratio, e.g. $a : b$ | M1(dep*) |
Obtain $a : b : c = 6 : 2 : -1$, or equivalent | A1 |
State answer $r = \lambda(6i + 2j - k)$ or equivalent | A1√ |
OR: Attempt to calculate vector product of two normals, e.g. $(i - 2j + 2k) \times (2i - 3j + 6k)$ | M2 |
Obtain two correct components | A1 |
Obtain $-6i - 2j + k$, or equivalent | A1 |
State answer $r = \lambda(-6i - 2j + k)$, or equivalent | A1√ | [5]10 The plane $p$ has equation $2 x - 3 y + 6 z = 16$. The plane $q$ is parallel to $p$ and contains the point with position vector $\mathbf { i } + 4 \mathbf { j } + 2 \mathbf { k }$.\\
(i) Find the equation of $q$, giving your answer in the form $a x + b y + c z = d$.\\
(ii) Calculate the perpendicular distance between $p$ and $q$.\\
(iii) The line $l$ is parallel to the plane $p$ and also parallel to the plane with equation $x - 2 y + 2 z = 5$. Given that $l$ passes through the origin, find a vector equation for $l$.
\hfill \mbox{\textit{CAIE P3 2009 Q10 [10]}}