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Date November 2015 Marks available 2 Reference code 15N.1.hl.TZ0.13
Level HL only Paper 1 Time zone TZ0
Command term Find Question number 13 Adapted from N/A

Question

Consider the triangle ABC. The points PQ and R are the midpoints of the line segments [AB], [BC] and [AC] respectively.

Let OA=a, OB=b and OC=c.

Find BR in terms of a, b and c.

[2]
a.

(i)     Find a vector equation of the line that passes through B and R in terms of a, b and c and a parameter λ.

(ii)     Find a vector equation of the line that passes through A and Q in terms of a, b and c and a parameter μ.

(iii)     Hence show that OG=13(a+b+c) given that G is the point where [BR] and [AQ] intersect.

[9]
b.

Show that the line segment [CP] also includes the point G.

[3]
c.

The coordinates of the points AB and C are (1, 3, 1), (3, 7, 5) and (2, 2, 1) respectively.

A point X is such that [GX] is perpendicular to the plane ABC.

Given that the tetrahedron ABCX has volume 12 units3, find possible coordinates

of X.

[9]
d.

Markscheme

BR=BA+AR(=BA+12AC)     (M1)

=(ab)+12(ca)

=12ab+12c     A1

[2 marks]

a.

(i)     rBR=b+λ(12ab+12c)(=λ2a+(1λ)b+λ2c)     A1A1

 

Note:     Award A1A0 if the r= is omitted in an otherwise correct expression/equation.

Do not penalise such an omission more than once.

 

(ii)     AQ=a+12b+12c     (A1)

rAQ=a+μ(a+12b+12c)(=(1μ)a+μ2b+μ2c)     A1

 

Note:     Accept the use of the same parameter in (i) and (ii).

 

(iii)     when AQ and BP intersect we will have rBR=rAQ     (M1)

 

Note:     If the same parameters are used for both equations, award at most M1M1A0A0M1.

 

λ2a+(1λ)b+λ2c=(1μ)a+μ2b+μ2c

attempt to equate the coefficients of the vectors a, b and c     M1

λ2=1μ1λ=μ2λ2=μ2}     (A1)

λ=23 or μ=23     A1

substituting parameters back into one of the equations     M1

OG=1223a+(123)b+1223c=13(a+b+c)     AG

 

Note:     Accept solution by verification.

[9 marks]

b.

CP=12a+12bc     (M1)A1

so we have that rCP=c+β(12a+12bc) and when β=23 the line passes through

the point G (ie, with position vector 13(a+b+c))     R1

hence [AQ], [BR] and [CP] all intersect in G     AG

[3 marks]

c.

OG=13((131)+(375)+(221))=(241)     A1

 

Note:     This independent mark for the vector may be awarded wherever the vector is calculated.

AB×AC=(246)×(110)=(666)     M1A1

GX=α(111)     (M1)

volume of Tetrahedron given by 13×Area ABC×GX

=13(12|AB×AC|)×GX=12     (M1)(A1)

 

Note:     Accept alternative methods, for example the use of a scalar triple product.

 

=16(6)2+(6)2+(6)2×α2+α2+α2=12     (A1)

=1663|α|3=12

|α|=4     A1

 

Note:     Condone absence of absolute value.

 

this gives us the position of X as (241)±(444)

X(6, 8, 3) or (2, 0, 5)     A1

 

Note:     Award A1 for either result.

[9 marks]

Total [23 marks]

d.

Examiners report

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a.
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b.
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c.
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d.

Syllabus sections

Topic 4 - Core: Vectors » 4.1 » Algebraic and geometric approaches to the sum and difference of two vectors.

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