Question 10:
--- 10 (i) ---
10 (i) |  1   3   26  −1
       
d.v. =  7 ×−5=−26 ≡  1 
       
−6 8 −26 1
       
finding one point on line: e.g. (0, 8, 11), (8, 0, 3), (11, –3, 0)
8 −1
   
ft Line equation in vector form r = 0+λ 1  (must have r = …, l = …)
   
3 1
   
ALTERNATIVE 1
finding two points on line: e.g. (0, 8, 11), (1, 7, 10), (8, 0, 3), (11, –3, 0) , …
−1
 
ft d.v. r =  1  (e.g.)
 
1
 
8 −1
   
ft Line equation in vector form r = 0+λ 1  (must have r = …)
   
3 1
   
ALTERNATIVE 2
Eliminating x (say): z = y + 3
Setting y = λ (or equivalent) and finding z and x in terms of the parameter
z = λ + 3 and x = 8 – λ
8 −1
   
ft Line equation in vector form r = 0+λ 1  (must have r = …)
   
3 1
    | M1 A1
M1 A1
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M1 A1
A1
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[5]
M1 A1
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(ii) | 1 7 −6
3 −5 8 =26k−130
k 2 3
= 0 when k = 5 ft from a linear eqn.
x+7y−6z =−10
−33y+33z =66 − y+z =2
e.g. 3x−5y+8z =48 → →
−26y+26z =78 − y+z =3
5x+2y+3z =16
eliminating one variable (twice); correctly
Inconsistency noted/explained (allow valid ft)
ALTERNATIVE
Substituting x = 8 – λ, y = λ and z = 3 + λ into kx + 2y + 3z = 16
⇒ λ(5 – k) + (11k – 22) = 0 λ terms; other terms
k = 5 gives non-unique soln.
Then 33 = 0 ft and system inconsistent (allow valid ft) | M1 A1
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11 (a)
(b) (i)
(ii) | pqr pqr pqr
Roots {qr, rp, pq} =  , , 
 p q r 
4 4 4 4
=  , ,  from pqr = 4, so subst. y =
p q r x
4 64 16 4
Setting x = ⇒ +2. +3. −4=0
y y3 y2 y
⇒ 64+32y+12y2 −4y3 =0
⇒ y3 −3y2 −8y−16=0
ALTERNATIVE (for last four-mark section)
∑α '= qr + rp + pq = 3
∑α 'β '= pqr(p + q + r) = 4 × (–2) = –8
∑α 'β 'γ ' = (pqr) 2 = 16
New eqn. is x3 −3x2 −8x−16=0 (must have “= 0”)
β and γ are complex conjugates (since the coefficients are real)
4
αβγ = 4 ⇒ βγ =
α
4
and | β | = | γ | since conjugates ⇒ βγ = β . γ = β 2 =
α | M1
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(iii) | 2
and | β | = (since α > 0 given) Answer Given
α
p(2.695) = –0.003… < 0 and p(2.705) = 0.049… > 0
so 2.695 < α < 2.705 (by the “Change-of-Sign Rule”) ⇒ α = 2.70 to 3s.f.
(Allow numerical methods, e.g. Newton-Raphson – to find α = 2.70 to 3s.f.)
α + β + γ = 4 ⇒ β + γ = 4 – α
Then 2.695 < α < 2.705 ⇒ 1.295 < 4 – α < 1.305
Now, if β = u + iv, then γ = u – iv and β + γ = 2u, i.e. 2 Re(β )
Thus 1.295 < 2 Re(β ) < 1.305 ⇒ 0.6475 < Re(β ) < 0.6525
and Re(β ) = 0.65 to 2 s.f.
(A0 for failure to justify 2s.f. accuracy properly) | A1
[2]
B1
M1
M1
A1
[4]
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