| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2014 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Line lies in or parallel to plane |
| Difficulty | Standard +0.8 This is a two-part Further Maths vectors question requiring (i) perpendicular distance calculation using cross product or projection methods, and (ii) finding plane coefficients given that a line lies in it. Part (i) is a standard technique but requires careful vector manipulation. Part (ii) requires understanding that both the direction vector must be perpendicular to the normal and a point on the line must satisfy the plane equation—this involves simultaneous equations and is less routine than typical A-level questions. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| (i) EITHER: Find \(\overrightarrow{AP}\) (or \(\overrightarrow{PA}\)) for a point \(P\) on \(l\) with parameter \(\lambda\), e.g. \(\mathbf{i} - 17\mathbf{j} + 4\mathbf{k} + \lambda(-2\mathbf{i} + \mathbf{j} - 2\mathbf{k})\) | B1 | |
| Calculate scalar product of \(\overrightarrow{AP}\) and a direction vector for \(l\) and equate to zero | M1 | |
| Solve and obtain \(\lambda = 3\) | A1 | |
| Carry out a complete method for finding the length of \(\overrightarrow{AP}\) | M1 | |
| Obtain the given answer 15 correctly | A1 | |
| OR1: Calling \((4, -9, 9)\) \(B\), state \(\overrightarrow{BA}\) (or \(\overrightarrow{AB}\)) in component form, e.g. \(-\mathbf{i} + 17\mathbf{j} - 4\mathbf{k}\) | B1 | |
| Calculate vector product of \(\overrightarrow{BA}\) and a direction vector for \(l\), e.g. \((-\mathbf{i} + 17\mathbf{j} - 4\mathbf{k}) \times (-2\mathbf{i} + \mathbf{j} - 2\mathbf{k})\) | M1 | |
| Obtain correct answer, e.g. \(-30\mathbf{i} + 6\mathbf{j} + 33\mathbf{k}\) | A1 | |
| Divide the modulus of the product by that of the direction vector | M1 | |
| Obtain the given answer correctly | A1 | |
| OR2: State \(\overrightarrow{BA}\) (or \(\overrightarrow{AB}\)) in component form | B1 | |
| Use a scalar product to find the projection of \(BA\) (or \(AB\)) on \(l\) | M1 | |
| Obtain correct answer in any form, e.g. \(\frac{27}{\sqrt{9}}\) | A1 | |
| Use Pythagoras to find the perpendicular | M1 | |
| OR3: State \(\overrightarrow{BA}\) (or \(\overrightarrow{AB}\)) in component form | B1 | |
| Use a scalar product to find the cosine of \(ABP\) | M1 | |
| Obtain correct answer in any form, e.g. \(\frac{27}{\sqrt{9\sqrt{306}}}\) | A1 | |
| Use trig. to find the perpendicular | M1 | |
| Obtain the given answer correctly | A1 | |
| OR4: State \(\overrightarrow{BA}\) (or \(\overrightarrow{AB}\)) in component form | B1 | |
| Find a second point \(C\) on \(l\) and use the cosine rule in triangle \(ABC\) to find the cosine of angle \(A\), \(B\) or \(C\), or use a vector product to find the area of \(ABC\) | M1 | |
| Obtain correct answer in any form | A1 | |
| Use trig. or area formula to find the perpendicular | M1 | |
| Obtain the given answer correctly | A1 | |
| OR5: State correct \(\overrightarrow{AP}\) (or \(\overrightarrow{PA}\)) for a point \(P\) on \(l\) with parameter \(\lambda\) in any form | B1 | |
| Use correct method to express \(AP^2\) (or \(AP\)) in terms of \(\lambda\) | M1 | |
| Obtain a correct expression in any form, e.g. \((1 - 2\lambda)^2 + (-17 + \lambda)^2 + (4 - 2\lambda)^2\) | A1 | |
| Carry out a method for finding its minimum (using calculus, algebra or Pythagoras) | M1 | |
| Obtain the given answer correctly | A1 | [5] |
| (ii) EITHER: Substitute coordinates of a general point of \(l\) in equation of plane and either equate constant terms or equate the coefficient of \(\lambda\) to zero, obtaining an equation in \(a\) and \(b\) | M1* | |
| Obtain a correct equation, e.g. \(4a - 9b - 27 + 1 = 0\) | A1 | |
| Obtain a second correct equation, e.g. \(-2a + b + 6 = 0\) | A1 | |
| Solve for \(a\) or for \(b\) | M1(dep*) | |
| Obtain \(a = 2\) and \(b = -2\) | A1 | |
| OR: Substitute coordinates of a point of \(l\) and obtain a correct equation, e.g. \(4a - 9b = 26\) | B1 | |
| EITHER: Find a second point on \(l\) and obtain an equation in \(a\) and \(b\) | M1* | |
| Obtain a correct equation | A1 | |
| OR: Calculate scalar product of a direction vector for \(l\) and a vector normal to the plane and equate to zero | M1* | |
| Obtain a correct equation, e.g. \(-2a + b + 6 = 0\) | A1 | |
| Solve for \(a\) or for \(b\) | M1(dep*) | |
| Obtain \(a = 2\) and \(b = -2\) | A1 | [5] |
**(i)** **EITHER:** Find $\overrightarrow{AP}$ (or $\overrightarrow{PA}$) for a point $P$ on $l$ with parameter $\lambda$, e.g. $\mathbf{i} - 17\mathbf{j} + 4\mathbf{k} + \lambda(-2\mathbf{i} + \mathbf{j} - 2\mathbf{k})$ | B1 |
Calculate scalar product of $\overrightarrow{AP}$ and a direction vector for $l$ and equate to zero | M1 |
Solve and obtain $\lambda = 3$ | A1 |
Carry out a complete method for finding the length of $\overrightarrow{AP}$ | M1 |
Obtain the given answer 15 correctly | A1 |
**OR1:** Calling $(4, -9, 9)$ $B$, state $\overrightarrow{BA}$ (or $\overrightarrow{AB}$) in component form, e.g. $-\mathbf{i} + 17\mathbf{j} - 4\mathbf{k}$ | B1 |
Calculate vector product of $\overrightarrow{BA}$ and a direction vector for $l$, e.g. $(-\mathbf{i} + 17\mathbf{j} - 4\mathbf{k}) \times (-2\mathbf{i} + \mathbf{j} - 2\mathbf{k})$ | M1 |
Obtain correct answer, e.g. $-30\mathbf{i} + 6\mathbf{j} + 33\mathbf{k}$ | A1 |
Divide the modulus of the product by that of the direction vector | M1 |
Obtain the given answer correctly | A1 |
**OR2:** State $\overrightarrow{BA}$ (or $\overrightarrow{AB}$) in component form | B1 |
Use a scalar product to find the projection of $BA$ (or $AB$) on $l$ | M1 |
Obtain correct answer in any form, e.g. $\frac{27}{\sqrt{9}}$ | A1 |
Use Pythagoras to find the perpendicular | M1 |
**OR3:** State $\overrightarrow{BA}$ (or $\overrightarrow{AB}$) in component form | B1 |
Use a scalar product to find the cosine of $ABP$ | M1 |
Obtain correct answer in any form, e.g. $\frac{27}{\sqrt{9\sqrt{306}}}$ | A1 |
Use trig. to find the perpendicular | M1 |
Obtain the given answer correctly | A1 |
**OR4:** State $\overrightarrow{BA}$ (or $\overrightarrow{AB}$) in component form | B1 |
Find a second point $C$ on $l$ and use the cosine rule in triangle $ABC$ to find the cosine of angle $A$, $B$ or $C$, or use a vector product to find the area of $ABC$ | M1 |
Obtain correct answer in any form | A1 |
Use trig. or area formula to find the perpendicular | M1 |
Obtain the given answer correctly | A1 |
**OR5:** State correct $\overrightarrow{AP}$ (or $\overrightarrow{PA}$) for a point $P$ on $l$ with parameter $\lambda$ in any form | B1 |
Use correct method to express $AP^2$ (or $AP$) in terms of $\lambda$ | M1 |
Obtain a correct expression in any form, e.g. $(1 - 2\lambda)^2 + (-17 + \lambda)^2 + (4 - 2\lambda)^2$ | A1 |
Carry out a method for finding its minimum (using calculus, algebra or Pythagoras) | M1 |
Obtain the given answer correctly | A1 | [5]
**(ii)** **EITHER:** Substitute coordinates of a general point of $l$ in equation of plane and either equate constant terms or equate the coefficient of $\lambda$ to zero, obtaining an equation in $a$ and $b$ | M1* |
Obtain a correct equation, e.g. $4a - 9b - 27 + 1 = 0$ | A1 |
Obtain a second correct equation, e.g. $-2a + b + 6 = 0$ | A1 |
Solve for $a$ or for $b$ | M1(dep*) |
Obtain $a = 2$ and $b = -2$ | A1 |
**OR:** Substitute coordinates of a point of $l$ and obtain a correct equation, e.g. $4a - 9b = 26$ | B1 |
**EITHER:** Find a second point on $l$ and obtain an equation in $a$ and $b$ | M1* |
Obtain a correct equation | A1 |
**OR:** Calculate scalar product of a direction vector for $l$ and a vector normal to the plane and equate to zero | M1* |
Obtain a correct equation, e.g. $-2a + b + 6 = 0$ | A1 |
Solve for $a$ or for $b$ | M1(dep*) |
Obtain $a = 2$ and $b = -2$ | A1 | [5]10 The line $l$ has equation $\mathbf { r } = 4 \mathbf { i } - 9 \mathbf { j } + 9 \mathbf { k } + \lambda ( - 2 \mathbf { i } + \mathbf { j } - 2 \mathbf { k } )$. The point $A$ has position vector $3 \mathbf { i } + 8 \mathbf { j } + 5 \mathbf { k }$.\\
(i) Show that the length of the perpendicular from $A$ to $l$ is 15 .\\
(ii) The line $l$ lies in the plane with equation $a x + b y - 3 z + 1 = 0$, where $a$ and $b$ are constants. Find the values of $a$ and $b$.
\hfill \mbox{\textit{CAIE P3 2014 Q10 [10]}}