CAIE P3 2011 June — Question 10 12 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2011
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeLine intersection with line
DifficultyStandard +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.
Spec4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting

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.
  1. Prove that \(l\) and \(m\) do not intersect.
  2. Calculate the acute angle between the directions of \(l\) and \(m\).
  3. Find the equation of the plane which is parallel to \(l\) and contains \(m\), giving your answer in the form \(a x + b y + c z = d\).

(i)
EITHER:
AnswerMarks
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 satisfiedA1
OR:
AnswerMarks 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 evaluationM1
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 intersectA1 [4]
(ii)
AnswerMarks 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 resultM1
Obtain answer \(47.1°\) or 0.822 radiansA1 [3]
(iii)
EITHER:
AnswerMarks
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 equivalentA1
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 equivalentA1
OR1:
AnswerMarks
Attempt to calculate vector product of direction vectors of \(l\) and \(m\)M1*
Obtain two correct componentA1
Obtain \(-2\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}\), or equivalentA1
Form a plane equation and use coordinates of a relevant point to evaluate \(d\)M1(dep*)
Obtain answer \(-2x + 4y + 3z = 26\), or equivalentA1
OR2:
AnswerMarks Guidance
Form a two-parameter plane equation using relevant vectorsM1*
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 equivalentA1 [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]}}