| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2005 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Show lines are skew (non-intersecting) |
| Difficulty | Standard +0.3 This is a standard Further Maths vectors question requiring routine techniques: finding the direction vector of line AB, setting up parametric equations for both lines, attempting to solve simultaneously to show no intersection exists, then using the cross product to find a normal vector for the plane. All steps are algorithmic with no novel insight required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| (i) State or imply a direction vector for \(AB\) is \(-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\), or equivalent | B1 | |
| Either: State equation of \(AB\) is \(\mathbf{r} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k} + t(-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})\), or equivalent | B1√ | |
| Equate at least two pairs of components of \(AB\) and \(l\) and solve for \(s\) or for \(t\) | M1 | |
| Obtain correct answer for \(s\) or for \(t\), e.g. \(s = 0\) or \(t = -2; s = -\frac{3}{5}\) or \(t = -\frac{1}{3}\) or \(s = 5\) or \(t = 3\) | A1 | |
| Verify that all three pairs of equations are not satisfied and that the lines fail to intersect | A1 | 5 |
| Or: State a Cartesian equation for \(AB\), e.g. \(\frac{x-2}{-1} = \frac{y-2}{2} = \frac{z-1}{2}\), and for \(l\), e.g. \(\frac{x-4}{-1} = \frac{y+2}{2} = \frac{z-2}{2}\) | B1√ | |
| Solve a pair of equations, e.g. in \(x\) and \(y\), for one unknown | M1 | |
| Obtain one unknown, e.g. \(x = 4\) or \(y = -2\) | A1 | |
| Obtain corresponding remaining values, e.g. of \(z\), and show lines do not intersect | A1 | 5 |
| Or: Form a relevant triple scalar product, e.g. \((2\mathbf{i} - 4\mathbf{j} + \mathbf{k}) \cdot ((-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) \times (\mathbf{i} + 2\mathbf{j} + \mathbf{k}))\) | B1√ | |
| Attempt to use correct method of evaluation | M1 | |
| Obtain at least two correct simplified terms of the three terms of the complete expansion of the triple product or of the corresponding determinant | A1 | |
| Obtain correct non-zero value, e.g. \(-20\), and state that the lines do not intersect | A1 | 5 |
| (ii) Either: Obtain a vector parallel to the plane and not parallel to \(l\), e.g. \(2\mathbf{i} - 4\mathbf{j} + \mathbf{k}\) | B1 | |
| Use scalar product to obtain an equation in \(a, b\) and \(c\), e.g. \(a + 2b + c = 0\) | B1 | |
| Form a second relevant equation, e.g. \(2a - 4b + c = 0\) and solve for one ratio, e.g. \(a : b\) | M1 | |
| Obtain final answer \(a : b : c = 6 : 1 : -8\) | A1 | |
| Use coordinates of a relevant point and values of \(a, b\) and \(c\) in general equation and find \(d\) | M1 | |
| Obtain answer \(6x + y - 8z = 6\), or equivalent | A1 | 6 |
| Or: Obtain a vector parallel to the plane and not parallel to \(l\), e.g. \(2\mathbf{i} - 4\mathbf{j} + \mathbf{k}\) | B1 | |
| Obtain a second relevant vector parallel to the plane and attempt to calculate their vector product, e.g. \((\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \times (2\mathbf{i} - 4\mathbf{j} + \mathbf{k})\) | M1 | |
| Obtain two correct components of the product | A1 | |
| Obtain correct answer, e.g. \(6\mathbf{i} + \mathbf{j} - 8\mathbf{k}\) | A1 | |
| Substitute coordinates of a relevant point in \(6x + y - 8z = d\), or equivalent, to find \(d\) | M1 | |
| Obtain answer \(6x + y - 8z = 6\), or equivalent | A1 | 6 |
| Or: Obtain a vector parallel to the plane and not parallel to \(l\), e.g. \(2\mathbf{i} - 4\mathbf{j} + \mathbf{k}\) | B1 | |
| Obtain a second relevant vector parallel to the plane and correctly form a 2-parameter equation for the plane, e.g. \(\mathbf{r} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k} + \lambda(2\mathbf{i} - 4\mathbf{j} + \mathbf{k}) + \mu(\mathbf{i} + 2\mathbf{j} + \mathbf{k})\) | M1 | |
| State 3 correct equations in \(x, y, z, \lambda\) and \(\mu\) | A1 | |
| Eliminate \(\lambda\) and \(\mu\) | M1 | |
| Obtain equation in any correct form | A1 | |
| Obtain answer \(6x + y - 8z = 6\), or equivalent | A1 | 6 |
**(i)** State or imply a direction vector for $AB$ is $-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$, or equivalent | B1 |
**Either:** State equation of $AB$ is $\mathbf{r} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k} + t(-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})$, or equivalent | B1√ |
Equate at least two pairs of components of $AB$ and $l$ and solve for $s$ or for $t$ | M1 |
Obtain correct answer for $s$ or for $t$, e.g. $s = 0$ or $t = -2; s = -\frac{3}{5}$ or $t = -\frac{1}{3}$ or $s = 5$ or $t = 3$ | A1 |
Verify that all three pairs of equations are not satisfied and that the lines fail to intersect | A1 | 5
**Or:** State a Cartesian equation for $AB$, e.g. $\frac{x-2}{-1} = \frac{y-2}{2} = \frac{z-1}{2}$, and for $l$, e.g. $\frac{x-4}{-1} = \frac{y+2}{2} = \frac{z-2}{2}$ | B1√ |
Solve a pair of equations, e.g. in $x$ and $y$, for one unknown | M1 |
Obtain one unknown, e.g. $x = 4$ or $y = -2$ | A1 |
Obtain corresponding remaining values, e.g. of $z$, and show lines do not intersect | A1 | 5
**Or:** Form a relevant triple scalar product, e.g. $(2\mathbf{i} - 4\mathbf{j} + \mathbf{k}) \cdot ((-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) \times (\mathbf{i} + 2\mathbf{j} + \mathbf{k}))$ | B1√ |
Attempt to use correct method of evaluation | M1 |
Obtain at least two correct simplified terms of the three terms of the complete expansion of the triple product or of the corresponding determinant | A1 |
Obtain correct non-zero value, e.g. $-20$, and state that the lines do not intersect | A1 | 5
**(ii)** **Either:** Obtain a vector parallel to the plane and not parallel to $l$, e.g. $2\mathbf{i} - 4\mathbf{j} + \mathbf{k}$ | B1 |
Use scalar product to obtain an equation in $a, b$ and $c$, e.g. $a + 2b + c = 0$ | B1 |
Form a second relevant equation, e.g. $2a - 4b + c = 0$ and solve for one ratio, e.g. $a : b$ | M1 |
Obtain final answer $a : b : c = 6 : 1 : -8$ | A1 |
Use coordinates of a relevant point and values of $a, b$ and $c$ in general equation and find $d$ | M1 |
Obtain answer $6x + y - 8z = 6$, or equivalent | A1 | 6
**Or:** Obtain a vector parallel to the plane and not parallel to $l$, e.g. $2\mathbf{i} - 4\mathbf{j} + \mathbf{k}$ | B1 |
Obtain a second relevant vector parallel to the plane and attempt to calculate their vector product, e.g. $(\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \times (2\mathbf{i} - 4\mathbf{j} + \mathbf{k})$ | M1 |
Obtain two correct components of the product | A1 |
Obtain correct answer, e.g. $6\mathbf{i} + \mathbf{j} - 8\mathbf{k}$ | A1 |
Substitute coordinates of a relevant point in $6x + y - 8z = d$, or equivalent, to find $d$ | M1 |
Obtain answer $6x + y - 8z = 6$, or equivalent | A1 | 6
**Or:** Obtain a vector parallel to the plane and not parallel to $l$, e.g. $2\mathbf{i} - 4\mathbf{j} + \mathbf{k}$ | B1 |
Obtain a second relevant vector parallel to the plane and correctly form a 2-parameter equation for the plane, e.g. $\mathbf{r} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k} + \lambda(2\mathbf{i} - 4\mathbf{j} + \mathbf{k}) + \mu(\mathbf{i} + 2\mathbf{j} + \mathbf{k})$ | M1 |
State 3 correct equations in $x, y, z, \lambda$ and $\mu$ | A1 |
Eliminate $\lambda$ and $\mu$ | M1 |
Obtain equation in any correct form | A1 |
Obtain answer $6x + y - 8z = 6$, or equivalent | A1 | 6
10 With respect to the origin $O$, the points $A$ and $B$ have position vectors given by
$$\overrightarrow { O A } = 2 \mathbf { i } + 2 \mathbf { j } + \mathbf { k } \quad \text { and } \quad \overrightarrow { O B } = \mathbf { i } + 4 \mathbf { j } + 3 \mathbf { k }$$
The line $l$ has vector equation $\mathbf { r } = 4 \mathbf { i } - 2 \mathbf { j } + 2 \mathbf { k } + s ( \mathbf { i } + 2 \mathbf { j } + \mathbf { k } )$.\\
(i) Prove that the line $I$ does not intersect the line through $A$ and $B$.\\
(ii) Find the equation of the plane containing $l$ and the point $A$, giving your answer in the form $a x + b y + c z = d$.
\hfill \mbox{\textit{CAIE P3 2005 Q10 [11]}}