Download Vectors Review_Ans

1.
(a)
2y + 8x = 4
–3x + 2y = –7
2x + 6 – 2x = 6
M1
A1
Note: Award M1 for attempt at components, A1 for two correct equations.
No penalty for not checking the third equation.
solving : x = 1, y = –2
(b)
A1
  4  4 
   
│a + 2b│=   3   2  2 
 2   2 
   
 4 
 
=   7
 6 
 
 a  2b  4 2  (7) 2  6 2
=
(M1)
101
A1
[5]
2.
(a)
(b)
AB = b – a
A1
CB = a + b
A1
AB CB = (b – a)•(b + a)
= │b│2–│a│2
= 0 since │b│=│a│
M1
A1
R1
Note: Only award the A1 and R1 if working indicates that they
understand that they are working with vectors.
so AB is perpendicular to CB i.e. AB̂C is a right angle
AG
[5]
3.
(a)
4
 
Finding correct vectors AB   3 
  1
 
  3
 
AC   1 
 1 
 
Substituting correctly in scalar product AB AC = 4(–3) + 3(1) – 1(1)
= –10
IB Questionbank Mathematics Higher Level 3rd edition
A1A1
A1
AG
N0
1
(b)
AB  26
AC  11
Attempting to use scalar product formula, cos BÂC 
(A1)(A1)
 10
M1
26 11
= –0.591 (to 3 s.f.)
BÂC = 126°
A1
A1
N3
[8]
4.
(a)
OP = i + 2j – k
the coordinates of P are (1, 2, –1)
(b)
EITHER
x = 1 + t, y = 2 – 2t, z = 3t – 1
y2
z 1
x – 1 = t,
 t,
t
2
3
y  2 z 1
x–1=

2
3
(M1)
A1
M1
A1
AG
N0
OR
 x  1   1 
     
 y    2   t  2 
 z    1  3 
     
x–1=
(c)
y  2 z 1

2
3
(i)
2(1 + t) + (2 – 2t) + (3t – 1) = 6  t = 1
(ii)
coordinates are (2, 0, 2)
M1A1
AG
M1A1
N1
A1
Note: Award A0 for position vector.
(iii)
distance travelled is the distance between the two points
(2  1) 2  (0  2) 2  (2  1) 2  14 (= 3.74)
IB Questionbank Mathematics Higher Level 3rd edition
(M1)
(M1)A1
2
(d)
(i)
distance from Q to the origin is given by
d(t) =
t 4  (1  t ) 2  (1  t 2 ) 2 (or equivalent)
2
e.g. for labelled sketch of graph of d or d
(ii)
M1A1
(M1)(A1)
the minimum value is obtained for t = 0.761
A1
the coordinates are (0.579, 0.239, 0.421)
A1
N3
Note: Accept answers given as a position vector.
(e)
(i)
(ii)
 0
1
 4 
 
 
 
a   1 , b   0  and c    1 
1
 0
  3
 
 
 
substituting in the equation a – b = k(b – c), we have
1  4 
 0  1
  1
  3
   
   
 
 
 1    0   k  0   1    1   k 1 
  0    3 
1  0
1
 3
   
 
 
   
1
 k  1 and k  which is impossible
3
so there is no solution for k
BA and CB are not parallel
(hence A, B, and C cannot be collinear)
(M1)A1
(M1)
A1
R1
R2
Note: Only accept answers that follow from part (i).
[23]
IB Questionbank Mathematics Higher Level 3rd edition
3
5.
(a)
Use of cos =




OA  AB
(M1)
OA AB

AB = i  j + k
A1


AB =

3 and OA = 3 2
A1

OA AB = 6
A1
substituting gives cos =

(b)


  6  or equivalent
6  3 
2
M1
N1

L1: r = OA + s AB
or equivalent
L1: r = i  j + 4k + s(i  j + k)
or equivalent
(M1)
A1
Note: Award (M1)A0 for omitting “r =” in the final answer.
(c)
Equating components and forming equations involving s and t
1 + s = 2 + 2t, 1  s = 4 + t, 4 + s = 7 + 3t
Having two of the above three equations
Attempting to solve for s or t
Finding either s = 3 or t = 2
Explicitly showing that these values satisfy the third equation
Point of intersection is (2, 2, 1)
(M1)
A1A1
(M1)
A1
R1
A1
N1
Note: Position vector is not acceptable for final A1.
(d)
METHOD 1
 1   2
  3
   
 
r =   1    1     3 
 4   3
  3
   
 
x = 1 + 2  3, y = 1 +  + 3 and z = 4 + 3  3
Elimination of the parameters
x + y = 3 so 4(x + y) = 12 and y + z = 4 + 3
so 3(y + z) = 12 + 9
(A1)
M1A1
M1
3(y + z) = 4(x + y) + 9
A1
Cartesian equation of plane is 4x + y  3z = 9 (or equivalent)
A1
IB Questionbank Mathematics Higher Level 3rd edition
N1
4
METHOD 2
EITHER
The point (2, 4, 7) lies on the plane.
The vector joining (2, 4, 7) and (1,  1, 4) and 2i + j + 3k
are parallel to the plane. So they are perpendicular to the
normal to the plane.
(i  j + 4k)  (2i + 4j + 7k) =  i  5j  3k
i
j
k
n  1  5  3
2
(A1)
1
M1
3
=  12i  3j + 9k
or equivalent parallel vector
A1
OR
L1 and L2 intersect at D (2, 2,1)

AD = (2i + 2j + k)  (i  j + 4k) = 3i + 3j  3k
i
j
k
n 2
1
3
(A1)
M1
3 3 3
= 12i  3j + 9k
or equivalent parallel vector
A1
THEN
r • n = (i  j + 4k) • (12i  3j + 9k)
= 27
Cartesian equation of plane is 4x + y  3z = 9 (or equivalent)
M1
A1
A1
N1
[20]
IB Questionbank Mathematics Higher Level 3rd edition
5
6.
METHOD 1
for finding two of the following three vectors (or their negatives)
0
  2
  2
 
 
 
AB   2 , AC   2 , BC   0 
  1
  2
 1
 
 
 
and calculating
(A1)(A1)
EITHER
  2
 
AB  AC  0 2  1   2 
 2 2  2  4 
i
j
k
1
AB AC
2
area ∆ABC =
M1A1
M1
OR
 2 
 
BA  BC  0  2 1    2 
 2 0  1   4 
i
area ∆ABC =
j
k
1
BA  BC
2
M1A1
M1
OR
  2
 
CA  CB  2  2 2   2 
2 0 1  4 
i
area ∆ABC =
j
k
1
CA  CB
2
M1A1
M1
THEN
area ∆ABC =
=
24
2
6
IB Questionbank Mathematics Higher Level 3rd edition
A1
AG
N0
6
METHOD 2
for finding two of the following three vectors (or their negatives)
0
  2
  2
 
 
 
AB   2 , AC   2 , BC   0 
  1
  2
 1
 
 
 
(A1)(A1)
EITHER
cos A =
AB AC
M1
AB AC
=
6

5 12
6 
3 
 or

60 
15 
2
5
sin A =
A1
1
AB AC sin A
2
1
2
=
5 12
2
5
1
=
24
2
= 6
area ∆ABC =
M1
A1
AG
N0
OR
cos B =
BA  BC
M1
BA BC
1
= 
5 5

1
5
24 
24 
or

25 
5 
1
area ∆ABC =
BA BC sin B
2
1
24
=
5 5
2
25
1
=
24
2
= 6
sin B =
IB Questionbank Mathematics Higher Level 3rd edition
A1
M1
A1
AG
N0
7
OR
cos C =
CA  CB
M1
CA CB
6
=

12 5
sin C =
6 
3 
 or

60 
15 
2
5
1
CA CB sin C
2
1
2
=
12 5
2
5
1
=
24
2
= 6
area ∆ABC =
A1
M1
A1
AG
N0
METHOD 3
for finding two of the following three vectors (or their negatives)
0
  2
  2
 
 
 
AB   2 , AC   2 , BC   0 
  1
  2
 1
 
 
 
AB =
5  c, AC  12  2 3  b, BC  5  a
52 3 5
 3 5
2
area ∆ABC = s( s  a)(s  b)(s  c)
s=
(A1)(A1)
M1A1
M1
=
( 3  5 )( 3 )( 5  3 )( 3 )
=
3(5  3)
A1
=
6
AG
IB Questionbank Mathematics Higher Level 3rd edition
N0
8
METHOD 4
for finding two of the following three vectors (or their negatives)
0
  2
  2
 
 
 
AB   2 , AC   2 , BC   0 
  1
  2
 1
 
 
 
AB = BC = 5 and AC =
∆ABC is isosceles
12  2 3
let M be the midpoint of [AC], the height BM =
area ∆ABC =
=
2 3 2
2
6
(A1)(A1)
M1A1
53  2
M1
A1
AG
N0
[6]
IB Questionbank Mathematics Higher Level 3rd edition
9
7.
(a)
EITHER
normal to plane given by
i
j k
2 3 2
6 3 2
= 12i + 8j – 24k
equation of π is 3x + 2y – 6z = d
as goes through (–2, 3, –2) so d = 12
π :3x + 2y – 6z = 12
M1A1
A1
(M1)
M1A1
AG
OR
x = –2 + 2λ + 6μ
y = 3 + 3λ – 3μ
z = –2 + 2λ + 2μ
(b)
eliminating μ
x + 2y = 4 + 8λ
2y + 3z = 12λ
eliminating λ
3(x + 2y) – 2(2y + 3z) = 12
π : 3x + 2y – 6z = 12
M1A1A1
therefore A(4, 0, 0), B(0, 6, 0) and C(0, 0, –2)
A1A1A1
M1A1A1
AG
Note: Award A1A1A0 if position vectors given instead of coordinates.
(c)
area of base OAB =
V=
(d)
(e)
1
 4  6 = 12
2
1
 12  2 = 8
3
 3  1
   
 2    0  = 3 = 7 × 1 × cos 
  6  0
   
3
  arccos
7
3
so θ = 90 – arccos = 25.4° (accept 0.443 radians)
7
d = 4 sin θ =
12
(= 1.71)
7
IB Questionbank Mathematics Higher Level 3rd edition
M1
M1A1
M1A1
M1A1
(M1)A1
10
(f)
8=
1 12
  area  area = 14
3 7
M1A1
Note: If answer to part (f) is found in an earlier part, award M1A1,
regardless of the fact that it has not come from their answers
to part (c) and part (e).
[20]
8.
EITHER
l goes through the point (1, 3, 6), and the plane contains A(4, –2, 5)
the vector containing these two points is on the plane, i.e.
 1   4    3
     
 3    2   5 
 6  5   1 
     
i
j k
  1   3 
   
 2    5    1 2  1 = 7i + 4j + k
  1  1   3 5 1
   
 4  7
   
  2    4   25
 5  1
   
hence, Cartesian equation of the plane is 7x + 4y + z = 25
(M1)A1
M1A1
(M1)
A1
OR
finding a third point
e.g. (0, 5, 5)
three points are (1, 3, 6), (4, –2, 5), (0, 5, 5)
equation is ax + by + cz = 1
system of equations
a + 3b + 6c = 1
4a – 2b + 5c = 1
5b + 5c = 1
7
4
1
a=
, from GDC
,b 
,c 
25
25
25
7
4
1
so
x
y
z 1
25
25
25
or 7x + 4y + z = 25
M1
A1
M1
M1A1
A1
[6]
IB Questionbank Mathematics Higher Level 3rd edition
11
9.
(a)
(i)
METHOD 1

 1  1  0
AB = b  a =  2    1    1 
 3  2 1
     
(A1)

 3  1   2 
AC = c  a =  0    1     1
 1   2    1
     
(A1)

i
j
k
1
1
2 1 1

AB  AC = 0
M1
= i (1 + 1)  j(0  2) + k (0  2)
(A1)
= 2j  2k
Area of triangle ABC =
A1


1
1
2 j  2k 
8  2 sq. units
2
2
M1A1
Note: Allow FT on final A1.
METHOD 2
AB  2 , BC  12 , AC  6
A1A1A1
Using cosine rule, e.g. on Ĉ
6  12  2
cos C =
2 72
Area ABC 

M1
2 2
3
A1
1
absin C
2
=
1
12 6 sin
2
M1


 arccos 2 2 

3 



2 2 
 2
= 3 2 sin  arccos
3 


A1
Note: Allow FT on final A1.
(ii)
AB =
2 =
2
1
1
AB h 
2  h , h equals the shortest distance
2
2
h=2
IB Questionbank Mathematics Higher Level 3rd edition
A1
(M1)
A1
12
(iii)
METHOD 1
 0 
 
 has form r •  2   d
  2
 
(M1)
Since (1, 1, 2) is on the plane
1  1 
   
d = 1 • 2 24 2
 2    2
   
M1A1
 0 
 
Hence r •  2  = 2
  2
 
2y  2z = 2 (or y  z = 1)
A1
METHOD 2
 1  0
2
   
 
r =  1   λ  1   μ   1
 2 1
  1
   
 
x = 1 + 2
(i)
y=1+
(ii)
z=2+
(iii)
(M1)
A1
Note: Award A1 for all three correct, A0 otherwise.
From (i)  =
x 1
2
 x 1 

 2 
substitute in (ii) y = 1 +   
 x 1 

 2 
=y1+ 
substitute  and  in (iii)
M1
 x 1   x 1 


 2   2 
z=2+y1+ 
 y  z = 1
IB Questionbank Mathematics Higher Level 3rd edition
A1
13
(b)
(i)
The equation of OD is
 0 
 
r =  2 ,
  2
 

 0 

 
 or r  λ  1  

  1 
 

M1
This meets  where
2 + 2 = 1
= 
(M1)
1
4
A1
1 1

Coordinates of D are  0 ,  , 
2 2

(ii)
2
2
1
 1 1
OD  0        
2
 2 2

A1
(M1)A1
[20]
10.
consider a vector parallel to each line,
 4 
 3
 
 
e.g. u =   2  and v   3 
 1 
1
 
 
let θ be the angle between the lines
u  v 12  6  1
cos θ =

uv
21 19
7
=
= 0.350...
21 19



7
 
so 0 = 69.5°  or 1.21 rad or arccos

21
19



A1A1
M1A1
(A1)
A1
N4
Note: Allow FT from incorrect reasonable vectors.
[6]
IB Questionbank Mathematics Higher Level 3rd edition
14
11.
METHOD 1
(from GDC)

1
1
1 0
 
6
12 

2 1

0 1  3  6 


0 0 0 0 




(M1)
1
1
x λ
6
12
A1
2
1
y λ
3
6
A1
1
 1
r =  i 
6
 12
2

 1

j    i  j  k 
3

 6

A1A1A1
N3
M1A1
(M1)A1
M1
A1
N3
METHOD 2
(Elimination method either for equations or row reduction of matrix)
Eliminating one of the variables
Finding a point on the line
Finding the direction of the line
The vector equation of the line
[6]
IB Questionbank Mathematics Higher Level 3rd edition
15