| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 12 |
| 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: (i) setting equal and showing inconsistency (standard skew lines proof), (ii) using dot product formula for angle between direction vectors (direct application), and (iii) finding a plane containing a line and parallel to another (requires cross product for normal vector). All parts follow textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks |
|---|---|
| Express general point of \(l\) or \(m\) in component form, e.g. \((2 + \lambda, -\lambda, 1 + 2\lambda)\) or \((\mu, 2 + 2\mu, 6 - 2\mu)\) | B1 |
| Equate at least two pairs of components and solve for \(\lambda\) or for \(\mu\) | M1 |
| Obtain correct answer for \(\lambda\) or \(\mu\) (possible answers for \(\lambda\) are \(-2\), \(\frac{1}{4}\), \(7\) and for \(\mu\) are \(0, 2\frac{1}{4}, -4\frac{1}{2}\)) | A1 |
| Verify that all three component equations are not satisfied | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| State a relevant scalar triple product, e.g. \((2\mathbf{i} - 2\mathbf{j} - 5\mathbf{k}) \cdot ((\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \times (\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}))\) | B1 | |
| Attempt to use the correct method of evaluation | M1 | |
| Obtain at least two correct simplified terms of the expansion of the triple product or of the corresponding determinant, e.g. \(-4\), \(-8\), \(-15\) | A1 | |
| Obtain correct non-zero value, e.g. \(-27\), and state that the lines do not intersect | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Carry out the correct process for evaluating scalar product of direction vectors for \(l\) and \(m\) | M1 | |
| Using the correct process for the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine of the result | M1 | |
| Obtain answer \(47.1°\) or 0.822 radians | A1 | [3] |
| Answer | Marks |
|---|---|
| Use scalar product to obtain \(a - b + 2c = 0\) | B1 |
| Obtain \(a + 2b - 2c = 0\), or equivalent, from a scalar product, or by subtracting two point equations obtained from points on \(m\), and solve for one ratio, e.g. \(a : b\) | M1* |
| Obtain \(a : b : c = -2 : 4 : 3\), or equivalent | A1 |
| Substitute coordinates of a point on \(m\) and values for \(a\), \(b\) and \(c\) in general equation and evaluate \(d\) | M1(dep*) |
| Obtain answer \(-2x + 4y + 3z = 26\), or equivalent | A1 |
| Answer | Marks |
|---|---|
| Attempt to calculate vector product of direction vectors of \(l\) and \(m\) | M1* |
| Obtain two correct component | A1 |
| Obtain \(-2\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}\), or equivalent | A1 |
| Form a plane equation and use coordinates of a relevant point to evaluate \(d\) | M1(dep*) |
| Obtain answer \(-2x + 4y + 3z = 26\), or equivalent | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Form a two-parameter plane equation using relevant vectors | M1* | |
| State a correct equation e.g. \(\mathbf{r} = 2\mathbf{j} + 6\mathbf{k} + s(\mathbf{i} - \mathbf{j} + 2\mathbf{k}) + t(\mathbf{i} + 2\mathbf{j} - 2\mathbf{k})\) | A1 | |
| State three correct equations in \(x\), \(y\), \(z\), \(s\) and \(t\) | A1 | |
| Eliminate \(s\) and \(t\) | M1(dep*) | |
| Obtain answer \(-2x + 4y + 3z = 26\), or equivalent | A1 | [5] |
**(i)**
**EITHER:**
Express general point of $l$ or $m$ in component form, e.g. $(2 + \lambda, -\lambda, 1 + 2\lambda)$ or $(\mu, 2 + 2\mu, 6 - 2\mu)$ | B1 |
Equate at least two pairs of components and solve for $\lambda$ or for $\mu$ | M1 |
Obtain correct answer for $\lambda$ or $\mu$ (possible answers for $\lambda$ are $-2$, $\frac{1}{4}$, $7$ and for $\mu$ are $0, 2\frac{1}{4}, -4\frac{1}{2}$) | A1 |
Verify that all three component equations are not satisfied | A1 |
**OR:**
State a relevant scalar triple product, e.g. $(2\mathbf{i} - 2\mathbf{j} - 5\mathbf{k}) \cdot ((\mathbf{i} + \mathbf{j} + 2\mathbf{k}) \times (\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}))$ | B1 |
Attempt to use the correct method of evaluation | M1 |
Obtain at least two correct simplified terms of the expansion of the triple product or of the corresponding determinant, e.g. $-4$, $-8$, $-15$ | A1 |
Obtain correct non-zero value, e.g. $-27$, and state that the lines do not intersect | A1 | [4]
**(ii)**
Carry out the correct process for evaluating scalar product of direction vectors for $l$ and $m$ | M1 |
Using the correct process for the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine of the result | M1 |
Obtain answer $47.1°$ or 0.822 radians | A1 | [3]
**(iii)**
**EITHER:**
Use scalar product to obtain $a - b + 2c = 0$ | B1 |
Obtain $a + 2b - 2c = 0$, or equivalent, from a scalar product, or by subtracting two point equations obtained from points on $m$, and solve for one ratio, e.g. $a : b$ | M1* |
Obtain $a : b : c = -2 : 4 : 3$, or equivalent | A1 |
Substitute coordinates of a point on $m$ and values for $a$, $b$ and $c$ in general equation and evaluate $d$ | M1(dep*) |
Obtain answer $-2x + 4y + 3z = 26$, or equivalent | A1 |
**OR1:**
Attempt to calculate vector product of direction vectors of $l$ and $m$ | M1* |
Obtain two correct component | A1 |
Obtain $-2\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}$, or equivalent | A1 |
Form a plane equation and use coordinates of a relevant point to evaluate $d$ | M1(dep*) |
Obtain answer $-2x + 4y + 3z = 26$, or equivalent | A1 |
**OR2:**
Form a two-parameter plane equation using relevant vectors | M1* |
State a correct equation e.g. $\mathbf{r} = 2\mathbf{j} + 6\mathbf{k} + s(\mathbf{i} - \mathbf{j} + 2\mathbf{k}) + t(\mathbf{i} + 2\mathbf{j} - 2\mathbf{k})$ | A1 |
State three correct equations in $x$, $y$, $z$, $s$ and $t$ | A1 |
Eliminate $s$ and $t$ | M1(dep*) |
Obtain answer $-2x + 4y + 3z = 26$, or equivalent | A1 | [5]
10 With respect to the origin $O$, the lines $l$ and $m$ have vector equations $\mathbf { r } = 2 \mathbf { i } + \mathbf { k } + \lambda ( \mathbf { i } - \mathbf { j } + 2 \mathbf { k } )$ and $\mathbf { r } = 2 \mathbf { j } + 6 \mathbf { k } + \mu ( \mathbf { i } + 2 \mathbf { j } - 2 \mathbf { k } )$ respectively.\\
(i) Prove that $l$ and $m$ do not intersect.\\
(ii) Calculate the acute angle between the directions of $l$ and $m$.\\
(iii) Find the equation of the plane which is parallel to $l$ and contains $m$, giving your answer in the form $a x + b y + c z = d$.
\hfill \mbox{\textit{CAIE P3 2011 Q10 [12]}}