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 P, Q 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.
(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.
Show that the line segment [CP] also includes the point G.
The coordinates of the points A, B 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.
Markscheme
→BR=→BA+→AR(=→BA+12→AC) (M1)
=(a−b)+12(c−a)
=12a−b+12c A1
[2 marks]
(i) rBR=b+λ(12a−b+12c)(=λ2a+(1−λ)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)(=(1−μ)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+(1−λ)b+λ2c=(1−μ)a+μ2b+μ2c
attempt to equate the coefficients of the vectors a, b and c M1
λ2=1−μ1−λ=μ2λ2=μ2} (A1)
λ=23 or μ=23 A1
substituting parameters back into one of the equations M1
→OG=12∙23a+(1−23)b+12∙23c=13(a+b+c) AG
Note: Accept solution by verification.
[9 marks]
→CP=12a+12b−c (M1)A1
so we have that rCP=c+β(12a+12b−c) 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]
→OG=13((131)+(37−5)+(221))=(24−1) A1
Note: This independent mark for the vector may be awarded wherever the vector is calculated.
→AB×→AC=(24−6)×(1−10)=(−6−6−6) 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)
=166√3|α|√3=12
⇒|α|=4 A1
Note: Condone absence of absolute value.
this gives us the position of X as (24−1)±(444)
X(6, 8, 3) or (−2, 0, −5) A1
Note: Award A1 for either result.
[9 marks]
Total [23 marks]