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A-Level · Further Pure Mathematics (Further Vectors)

FP.CP5 Further Vectors

EDEXCEL 9FM0 · Calculator allowed · about 130 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Freya is programming a drone's flight path, which she models as the line l with vector equation r = (4, -2, 1) + λ(3, -1, 2), where λ is a real parameter and distances are in metres.

(a)Write down the cartesian equations of l in the form (x-a)/p = (y-b)/q = (z-c)/r.(2)

(b)Find the coordinates of the point where l crosses the plane z = 7.(3)

(c)Find the coordinates of the point where l crosses the plane y = 0.(2)

(Total for Question 1 is 7 marks)

Two coastal survey markers used by Jamal's team are at A(2, -1, 3) and B(5, 4, -2), with coordinates in kilometres relative to a harbour office. Jamal models the straight path between them as the line l1 through A and B.

(a)Find the vector AB.(2)

(b)Write down a vector equation for l1.(2)

(c)A third marker C has coordinates (11, 14, -12). Show that C lies on l1.(3)

(Total for Question 2 is 7 marks)

Two straight paths through a country park are modelled by the lines l1: r = (1, 2, -1) + s(2, -1, 2) and l2: r = (4, 0, 3) + t(1, 2, 2), where s and t are real parameters.

(a)Find the acute angle between l1 and l2, giving your answer in degrees to 1 decimal place.(4)

(b)State, with a reason, whether l1 and l2 are parallel, perpendicular or neither.(2)

(Total for Question 3 is 6 marks)

A triangular solar panel has corners at the points P(1, 0, 2), Q(3, 4, 1) and R(0, 2, 5), with coordinates measured in metres.

(a)Find the vectors PQ and PR.(2)

(b)Calculate the vector product PQ x PR.(3)

(c)Hence find the area of triangle PQR, giving your answer in m² to 3 significant figures.(3)

(Total for Question 4 is 8 marks)

A triangular plot of land for a proposed footbridge has corners at A(1, 2, -1), B(3, 0, 2) and C(-1, 4, 0), with coordinates measured in metres.

(a)Find the vectors AB and AC.(2)

(b)Find a vector normal to the plane containing A, B and C.(3)

(c)Show that the cartesian equation of the plane ABC can be written as x + y = 3.(3)

(Total for Question 5 is 8 marks)

A plane Pi has equation 2x - y + 2z = 5. A sensor is positioned at D(4, 3, -1), with coordinates in metres.

(a)Show that D does not lie in the plane Pi.(1)

(b)Find the shortest distance from D to Pi, giving your answer as an exact fraction.(3)

(c)Find the coordinates of the foot of the perpendicular from D to Pi.(4)

(Total for Question 6 is 8 marks)

A line l has equation r = (3, -1, 2) + t(1, 2, -2). A plane Pi has equation 3x - 4y + z = 6.

(a)Find the angle between l and Pi, giving your answer in degrees to 1 decimal place.(4)

(b)Show that l intersects Pi, and find the coordinates of the point of intersection.(4)

(Total for Question 7 is 8 marks)

Two planes are given by Pi1: x + 2y - 2z = 4 and Pi2: 3x + 4z = 10.

(a)Find the acute angle between Pi1 and Pi2, giving your answer in degrees to 1 decimal place.(4)

(b)Find a vector equation for the line of intersection of Pi1 and Pi2.(5)

(Total for Question 8 is 9 marks)

A plane Pi has equation 2x - y + kz = 8, where k is a constant. A line l has equation r = (1, 3, -2) + t(4, 1, -3).

(a)Given that l is parallel to Pi, find the value of k.(3)

(b)Show that, for every value of k, the line l does not lie in the plane Pi.(4)

(Total for Question 9 is 7 marks)

Two straight underground cable routes are modelled by the lines l1: r = (1, 2, 0) + s(1, -1, 2) and l2: r = (0, 3, 1) + t(2, 1, -1), where distances are in metres.

(a)Show that l1 and l2 are skew lines.(5)

(b)Find the shortest distance between l1 and l2, giving your answer to 3 significant figures.(5)

(Total for Question 10 is 10 marks)

A line l has equation r = (1, 0, 2) + t(2, 1, -2). A navigation beacon is located at E(5, 4, -1).

(a)Find the position vector of the foot of the perpendicular from E to l.(5)

(b)Hence find the position vector of the reflection of E in the line l.(2)

(Total for Question 11 is 7 marks)

Three edges of a tetrahedral bracket meet at a common vertex and are represented by the vectors a = (2, 1, -1), b = (1, 3, 2) and c = (-1, 2, 4), with lengths in centimetres.

(a)Calculate the scalar triple product a . (b x c).(4)

(b)Hence find the volume of the tetrahedron with edges a, b and c meeting at the vertex, giving your answer to 3 significant figures.(2)

(Total for Question 12 is 6 marks)

A line l has equation r = (2, -1, 3) + t(1, 2, -1). A point F has coordinates (0, 4, 1).

(a)Show that F does not lie on l.(2)

(b)Find the cartesian equation of the plane containing l and F, giving your answer in the form ax + by + cz = d.(5)

(Total for Question 13 is 7 marks)

Mark scheme · FP.CP5 Further Vectors

Question 1

(a) B1 correct point used, e.g. (4,-2,1) appearing as a, b, c
(a) B1 (x-4)/3 = (y+2)/-1 = (z-1)/2, oe
(a) Answer: (x-4)/3 = (y+2)/(-1) = (z-1)/2
(b) M1 sets 1 + 2*λ = 7
(b) A1 λ = 3
(b) A1 point (13, -5, 7), cao
(b) Answer: (13, -5, 7)
(c) M1 sets -2 - λ = 0 and solves for λ = -2
(c) A1 point (-2, 0, -3), cao
(c) Answer: (-2, 0, -3)

Question 2

(a) M1 attempts B - A
(a) A1 AB = (3, 5, -5), cao
(a) Answer: AB = (3, 5, -5)
(b) B1 uses point A (or B) as the position vector
(b) B1 r = (2, -1, 3) + t(3, 5, -5), oe (e.g. using B and/or a scalar multiple of AB)
(b) Answer: r = (2, -1, 3) + t(3, 5, -5)
(c) M1 forms two equations from components, e.g. 2+3t=11 and -1+5t=14
(c) A1 solves to get t=3 from both equations
(c) A1 checks third component 3-5t = 3-15 = -12, consistent, so C lies on l1, cso
(c) Answer: C lies on l1 (t = 3 is consistent for all three components)

Question 3

(a) M1 finds the dot product of the direction vectors, (2,-1,2).(1,2,2) = 4
(a) M1 finds both magnitudes, √4+1+4=3 and √1+4+4=3
(a) M1 uses cos(θ) = (d1.d2)/(|d1||d2|)
(a) A1 θ = awrt 63.6 degrees
(a) Answer: 63.6 degrees (1 dp)
(b) B1 neither
(b) B1 valid reason, e.g. dot product = 4 not equal to 0 so not perpendicular, and (2,-1,2) is not a scalar multiple of (1,2,2) so not parallel
(b) Answer: Neither parallel nor perpendicular

Question 4

(a) M1 attempts Q-P and R-P
(a) A1 PQ = (2,4,-1) and PR = (-1,2,3), both cao
(a) Answer: PQ = (2, 4, -1), PR = (-1, 2, 3)
(b) M1 sets up the determinant/component method for the vector product
(b) A1 at least two components correct
(b) A1 PQ x PR = (14, -5, 8), fully correct, cao
(b) Answer: PQ x PR = (14, -5, 8)
(c) M1 finds |PQ x PR| = √14²+(-5)²+8²
(c) M1 uses area = (1/2)|PQ x PR|
(c) A1 awrt 8.44 m²
(c) Answer: 8.44 m² (3 sf)

Question 5

(a) M1 attempts B-A and C-A
(a) A1 AB = (2,-2,3) and AC = (-2,2,1), both cao
(a) Answer: AB = (2, -2, 3), AC = (-2, 2, 1)
(b) M1 attempts AB x AC
(b) A1 unsimplified normal (-8, -8, 0), oe
(b) A1 simplified normal, e.g. (1, 1, 0), oe (any nonzero scalar multiple accepted)
(b) Answer: Normal direction (1, 1, 0) (or any scalar multiple, e.g. (-8,-8,0))
(c) M1 uses normal (1,1,0) with point A(1,2,-1) in n.(r-a)=0
(c) M1 expands to (x-1)+(y-2)+0(z+1)=0
(c) A1 correctly simplifies to x+y=3 (printed result), cso
(c) Answer: x + y = 3 (printed result)

Question 6

(a) B1 substitutes D into 2x-y+2z to get 3, and states 3 is not equal to 5 so D is not on Pi, cso
(a) Answer: 2(4) - 3 + 2(-1) = 3, and 3 ≠ 5, so D does not lie in Pi
(b) M1 uses distance = |2x0-y0+2z0-5| / √2²+(-1)²+2²
(b) M1 substitutes D correctly to get |3-5|/3
(b) A1 distance = 2/3, cao
(b) Answer: 2/3 m
(c) M1 sets up point on the normal line through D: (4+2t, 3-t, -1+2t)
(c) M1 substitutes into the plane equation to form a linear equation in t
(c) M1 solves to get t = 2/9
(c) A1 coordinates (40/9, 25/9, -5/9), oe (equivalent decimals awrt (4.44, 2.78, -0.56) accepted), ft
(c) Answer: (40/9, 25/9, -5/9), i.e. awrt (4.44, 2.78, -0.56)

Question 7

(a) M1 finds d.n = (1,2,-2).(3,-4,1) = -7
(a) M1 finds |d|=3 and |n|=√26
(a) M1 uses sin(θ) = |d.n|/(|d||n|)
(a) A1 θ = awrt 27.2 degrees
(a) Answer: 27.2 degrees (1 dp)
(b) B1 since d.n = -7 is not equal to 0, l is not parallel to Pi so a unique intersection point exists
(b) M1 substitutes parametric coordinates (3+t, -1+2t, 2-2t) into the plane equation
(b) M1 solves the resulting equation to get t = 9/7
(b) A1 point (30/7, 11/7, -4/7), oe (equivalent decimals awrt (4.29, 1.57, -0.57) accepted)
(b) Answer: (30/7, 11/7, -4/7), i.e. awrt (4.29, 1.57, -0.57)

Question 8

(a) M1 finds n1.n2 = (1,2,-2).(3,0,4) = -5
(a) M1 finds |n1|=3 and |n2|=5
(a) M1 uses cos(θ) = |n1.n2|/(|n1||n2|) (taking the modulus for the acute angle)
(a) A1 θ = awrt 70.5 degrees
(a) Answer: 70.5 degrees (1 dp)
(b) M1 attempts n1 x n2 to find the direction of the line
(b) A1 direction vector (8,-10,-6), oe (simplified form (4,-5,-3) accepted)
(b) M1 sets y=0 (or another suitable substitution) and solves Pi1 and Pi2 simultaneously for a point
(b) A1 point (3.6, 0, -0.2), oe (equivalent fraction (18/5, 0, -1/5) accepted)
(b) B1 fully correct vector equation stated, e.g. r = (3.6,0,-0.2)+t(4,-5,-3), ft
(b) Answer: r = (3.6, 0, -0.2) + t(4, -5, -3)

Question 9

(a) M1 sets the direction vector dotted with the normal equal to zero: (4,1,-3).(2,-1,k)=0
(a) M1 forms and rearranges 7-3k=0
(a) A1 k = 7/3, oe (awrt 2.33 accepted)
(a) Answer: k = 7/3
(b) M1 recognises that for l to lie in Pi, the point (1,3,-2) on l must also satisfy the plane equation
(b) A1 substitutes the point into the plane equation to get -1-2k=8, giving k=-9/2
(b) B1 compares this with k=7/3 from part (a) and notes the two required values of k are different
(b) B1 concludes that l cannot lie in Pi for any single value of k, cso
(b) Answer: l cannot lie in Pi for any value of k, since containment would require k=7/3 (for the direction to be perpendicular to the normal) and k=-9/2 (for the point (1,3,-2) to satisfy the plane), which cannot both hold

Question 10

(a) B1 notes (1,-1,2) is not a scalar multiple of (2,1,-1) (e.g. ratios 1/2, -1/1, 2/-1 are not all equal), so the lines are not parallel
(a) M1 equates components to form simultaneous equations, e.g. 1+s=2t and 2-s=3+t
(a) M1 solves these two equations to get s=-1 and t=0
(a) A1 substitutes into the third equation, 2s=1-t, to get -2 = 1, a contradiction, cso
(a) B1 concludes l1 and l2 do not intersect and are not parallel, so they are skew
(a) Answer: l1 and l2 are skew (not parallel, and shown not to intersect)
(b) M1 finds the vector between points on the lines, (0,3,1)-(1,2,0) = (-1,1,1)
(b) M1 attempts d1 x d2
(b) A1 d1 x d2 = (-1, 5, 3), cao
(b) M1 applies distance = |(b-a).(d1xd2)| / |d1xd2|
(b) A1 awrt 1.52 m
(b) Answer: 1.52 m (3 sf)

Question 11

(a) M1 writes a general point on l as (1+2t, t, 2-2t)
(a) M1 forms the vector from this point to E
(a) M1 sets this vector's dot product with (2,1,-2) equal to zero
(a) M1 solves to get t = 2
(a) A1 foot of perpendicular = (5, 2, -2), cao
(a) Answer: (5, 2, -2)
(b) M1 uses E' = 2F - E where F is the foot of the perpendicular found in (a)
(b) A1 E' = (5, 0, -3), ft from (a)
(b) Answer: (5, 0, -3)

Question 12

(a) M1 attempts b x c
(a) A1 b x c = (8, -6, 5), cao
(a) M1 forms the dot product of a with their vector b x c
(a) A1 a.(bxc) = 5, cao
(a) Answer: a . (b x c) = 5
(b) M1 uses volume = (1/6)|a.(bxc)|
(b) A1 awrt 0.833 cm³
(b) Answer: 5/6 = 0.833 cm³ (3 sf)

Question 13

(a) M1 attempts to find t from two components, e.g. 2+t=0 giving t=-2, and -1+2t=4 giving t=2.5
(a) A1 shows the two values of t are inconsistent, hence F does not lie on l, cso
(a) Answer: F does not lie on l (x gives t=-2, y gives t=2.5, inconsistent)
(b) M1 finds the vector AF from a point on l (e.g. A=(2,-1,3)) to F, AF=(-2,5,-2)
(b) M1 attempts d x AF where d is the direction of l, as the normal to the plane
(b) A1 normal (1,4,9), oe
(b) M1 forms the plane equation using the normal and a point on the plane
(b) A1 x + 4y + 9z = 25, cao
(b) Answer: x + 4y + 9z = 25

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