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P1 June 2010 Solutions:
A1
Cross product – result is a vector [1 mark], perpendicular to both given vectors [1
mark] whose magnitude is equal to the area of the parallelogram defined by the two
given vectors (when sketched such to start from the same point) [1 mark].
5 1 11 5 1
1 5 5 1 25 1 4 4 24 [2 marks]
| | 4 4 24 24.66
5 1
1
1 1
5
x
y z
A2
32
12
1 1
00
[1 mark]
For m=1 and k=2,
3 12 2
00
3 12 2
00
[1 mark]
3 12 2 0
[1 mark]
3 2 2 0
1; 4
[1 mark]
1; 2
[1 mark]
A3
[5 marks]
F
c
k
m
x
A4 (Calculus II)
a)
⁄ ⁄ ⁄
⁄ ⁄ ⁄
⁄ ⁄ ⁄
[1 mark]
⁄ ⁄ ⁄
32
⁄ ⁄ ⁄ ⁄ ⁄
32
⁄ ⁄ / 23
⁄
⁄ ⁄ /
⁄
⁄ /
⁄
⁄
⁄
⁄
[1 mark]
b)
1
[1 mark1]
⁄
⁄
1⁄
⁄ 1⁄
⁄
⁄
⁄
32
[2 marks]
B1
a) Determinant of the matrix is
[1 mark]
Solving
[2 marks]
b)
1 0 0
[1 mark]
Determinant
1 0 0
[3 marks]
c)
1 1 10 2 30 4 9
1 1 10 2 30 4 9
0
1 11 6 0
[2 marks]
eigenvalues
11 √97
2; 1;
11 √972
0.575571; 1; 10.42443
[2 marks]
corresponding eigenvectors (normalised)
1 1 10 2 30 4 9
000
For 0.575571
1 0.575571 1 10 2 0.575571 30 4 9 0.575571
1 000
15 √97
611 √97
3
10.80814
0.383714
0.745280.60229
0.28597
For 1
1 1 1 10 2 1 30 4 9 1
1 000
100
100
For 10.42443
1 10.42443 1 10 2 10.42443 30 4 9 10.42443
1 000
15 √97
611 √97
3
12.47481
6.949619
0.134330.332430.93351
[4 marks]
B2 (Vectors)
a) At intersection point the expressions for the plane is
·
while for the line is
[2 marks]
Substituting into the equation of the plane we get
·
· ·
··
Hence
··
[5 marks]
b) Firstly, find the plane perpendicular to the line through point denoted by .
Plane perpendicular to the line is that for which
[1 mark]
Hence the equation for the plane
·
becomes
·
[1 mark]
when point is on that plane
·
[1 mark]
The plane is defined by the following equation
· ·
[1 mark]
Then, intersection between that plane and the line
is
· ·
such that
· ··
· · · ·1
· · ·
and
· · · ·
Hence, the distance between the points and
| | | · · · |
B3 (Calculus II)
a)
b) Substitution
, ,
reveals unit sphere in u‐v‐w coordinate system
1
c) Using
, ,
Jacobian can be obtained as follows
, ,, ,
0 00 00 0
d) The volume of the ellipsoid in x‐y‐z space is replaced by the unit sphere is u‐v‐w space. The
value of the integral once the constant has been taken out of the integral is (i.e. the
volume of the unit sphere)
43
x
y
z
a
b b
e) The mass is calculated as follows
Quick solution
43
, 1
43
x
y
z (out of plane) b
a
x
u
v
w (out of plane) 1
1
x
1
Unit sphere
Ellipsoid
Long solution
41
123 9
· 01
121 9 1 9
· 01
121 9 1 9
43
1
43
w (out of plane)
Unit sphere
u
v
1
1
r
u
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