CAIE P3 2005 June — Question 10 11 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2005
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Cross Product & Distances
TypeShow lines are skew (non-intersecting)
DifficultyStandard +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.
Spec1.10b 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

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 \(\mathbf { r } = 4 \mathbf { i } - 2 \mathbf { j } + 2 \mathbf { k } + s ( \mathbf { i } + 2 \mathbf { j } + \mathbf { k } )\).
  1. Prove that the line \(I\) does not intersect the line through \(A\) and \(B\).
  2. Find the equation of the plane containing \(l\) and the point \(A\), giving your answer in the form \(a x + b y + c z = d\).

AnswerMarks Guidance
(i) State or imply a direction vector for \(AB\) is \(-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\), or equivalentB1
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 equivalentB1√
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 intersectA1 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 unknownM1
Obtain one unknown, e.g. \(x = 4\) or \(y = -2\)A1
Obtain corresponding remaining values, e.g. of \(z\), and show lines do not intersectA1 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 evaluationM1
Obtain at least two correct simplified terms of the three terms of the complete expansion of the triple product or of the corresponding determinantA1
Obtain correct non-zero value, e.g. \(-20\), and state that the lines do not intersectA1 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 equivalentA1 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 productA1
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 equivalentA1 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 formA1
Obtain answer \(6x + y - 8z = 6\), or equivalentA1 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]}}