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Date May 2009 Marks available 23 Reference code 09M.2.hl.TZ1.11
Level HL only Paper 2 Time zone TZ1
Command term Find, Hence, Show that, and State Question number 11 Adapted from N/A

Question

The position vector at time tt of a point PP is given byOP=(1+t)i+(22t)j+(3t1)k, t0.

(a)     Find the coordinates of P when t=0 .

(b)     Show that P moves along the line L with Cartesian equationsx1=y22=z+13

(c)     (i)     Find the value of t when P lies on the plane with equation 2x+y+z=6 .

  (ii)     State the coordinates of P at this time.

  (iii)     Hence find the total distance travelled by P before it meets the plane.

 

The position vector at time t of another point, Q, is given byOQ=(t21t1t2), t0.

(d)     (i)     Find the value of t for which the distance from Q to the origin is minimum.

  (ii)     Find the coordinates of Q at this time.

(e)     Let a , b and c be the position vectors of Q at times t=0, t=1 and t=2 respectively.

  (i)     Show that the equation ab=k(bc) has no solution for k .

  (ii)     Hence show that the path of Q is not a straight line.

Markscheme

(a)     OP=i+2jk     (M1)

the coordinates of P are (1, 2, –1)     A1

[2 marks]

 

(b)     EITHER

x=1+t , y=22t , z=3t1     M1

x1=t , y22=t , z+13=t     A1

x1=y22=z+13     AG     N0

OR

(xyz)=(121)+t(123)     M1A1

x1=y22=z+13     AG

[2 marks]

 

(c)     (i)     2(1+t)+(22t)+(3t1)=6t=1     M1A1     N1

(ii)     coordinates are (2, 0, 2)     A1

Note: Award A0 for position vector.

(iii)     distance travelled is the distance between the two points     (M1)

(21)2+(02)2+(2+1)2=14   (=3.74)     (M1)A1

[6 marks]

 

(d)     (i)     distance from Q to the origin is given by

d(t)=t4+(1t)2+(1t2)2   (or equivalent)     M1A1

e.g. for labelled sketch of graph of d or d2     (M1)(A1)

     or     

the minimum value is obtained for t=0.761     A1     N3

(ii)     the coordinates are (0.579, 0.239, 0.421)     A1

Note: Accept answers given as a position vector.

[6 marks]

 

(e)     (i)     a=(011), b=(100) and c=(413)     (M1)A1

substituting in the equation ab=k(bc) , we have     (M1)

(011)(100)=k((100)(413))(111)=k(313)     A1

k=1 and k=13 which is impossible

so there is no solution for k     R1

(ii)     BAand CBare not parallel     R2

(hence A, B, and C cannot be collinear)

Note: Only accept answers that follow from part (i).

[7 marks]

 

Total [23 marks]

Examiners report

Generally this question was answered well by those students who attempted it. It was common to see confusion between coordinates and position vectors. Part (d) was most easily answered with the use of a GDC, but fewer candidates took advantage of this. In part (e) many students had difficulties expressing their reasoning well to obtain the marks.

Syllabus sections

Topic 4 - Core: Vectors » 4.3 » Simple applications to kinematics.

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