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### Transcript of Vectors Qn Solutiona- H2 Revision: Vectors 2010 Mathematics Department Page 3 of 23 (ii) 10 0 6 3 0

• H2 Revision: Vectors 2010 Mathematics Department

Page 1 of 23

Vectors

Qn Solution 1 AJC/2009 Prelim/I/13 (i) 0 2

1 , 0 0 0 2

1 1

x z y r λ

� � � � � � � �= − − = � = +� � � � � � � �− −� � � �

Vector parallel to 2

1 0 1 1 0 1

0 1 1 π

� � � � � � � � � � � �= − − = −� � � � � � � � � � � �−� � � � � �

1

1 2 1 1 0 3

1 1 2

1 0 1 1 : 3 0 3 3 2

2 1 2 2

n

r rπ

� � � � � � � � � � � �= − × =� � � � � � � � � � � �−� � � � � � � � � � � � � � � � � � � � � �• = • � • = −� � � � � � � � � � � � � � � �−� � � � � � � �

(ii) Let foot of perpendicular from Q to 2π be N.

3 1 13 3 6 2

ON λ � � � � � � � �= +� � � � � � � � � � � �

����

3 1 13 3 3 2 6 2 2

4

λ λ λ

λ

+� � � � � � � �+ • =� � � � � � � �+� � � � � = −

1 1 2

ON

−� � � �= � � � �−� �

����

(iii) 1 3 5 '

' 2 1 13 11 2

2 6 10

OQ OQ ON OQ

− −� � � � � � + � � � � � �= � = − = −� � � � � �

� � � � � �− −� � � � � �

���� ����� ���� �����

2 2

5 1 6 3 ' 11 1 10 2 5

10 0 10 5

1 3 5 3 : 1 5 OR : 11 5

0 5 10 5

PQ

l r l rλ λ

− −� � � � � � � � � � � � � � � �= − − − = − = −� � � � � � � � � � � � � � � �− −� � � � � � � �

−� � � � � � � � � � � � � � � �∴ = − + ∴ = − +� � � � � � � � � � � � � � � �−� � � � � � � �

�����

� �

• H2 Revision: Vectors 2010 Mathematics Department

Page 2 of 23

(iv) 1 3 2

� � � � � � � � � �

= 6 2 4

a

k a � � � �

� =� � � � � �

Method 1:

1 2

1 1 : 3 2 : 3

2 2 2

b r rπ π � � � � � � � �• = − • =� � � � � � � � � � � �

Distance between the 2 planes = ( )2

2 224 1 9 4

b − − =

+ + 108 or -116b� =

Method 2:

Distance QN = ( )( ) ( ) ( )( )2 223 1 13 1 6 2 224− − + − + − − = 2 3 2 2 5 2 6 13 6 108 OR ' 6 11 6 116 4 6 4 4 10 4

b OQ b OQ

−� � � � � � � � � � � � � � � � � � � � � � � �= • = • = = • = − • = −� � � � � � � � � � � � � � � � � � � � � � � �−� � � � � � � � � � � �

���� �����

Method 3

1

1 1 1 2

: 3 2 3 14 14

2 2 r rπ � � � � � � � �• = − � • = −� � � � � � � � � � � �

2

1 1 2

: 3 224 14 14

2

1 3 2 56 54 or 58 108 or 116 2

r

r b

π � � � �• = − ±� � � � � �

� � � �� • = − ± = − � = −� � � � � �

(v) 2,a b≠ ∈�

Qn Solution 2 AJC/2009 Prelim/II/2 (i)

' 4 tan ' 3 4 ' 4

3 3 EE

EE E E kθ= � = × = ∴ = �����

' ' 3 4 10OF OE E E EF j k i= + + = + + ���� ����� ����� ����

� ��

(Shown)

• H2 Revision: Vectors 2010 Mathematics Department

Page 3 of 23

(ii) 10 0 6 3

0 4 DB DE

� � � � � � � �= − = −� � � � � � � � � � � �

���� ����

10 0 24 12 6 3 40 2 20

0 4 30 15 n

−� � � � � � � � � � � � � � � �= − × − = − = −� � � � � � � � � � � � � � � �−� � � � � � � �

12 10 20 3 15 4

sin 769 125

50.7228.. 50.7

θ

θ

� � � � � � � �•� � � � � � � � � � � �=

= =� �

Qn Solution 3 ACJC/Prelim 2009/I/12 (i) Let N be the foot of perpendicular from A to the plane 1p .

1 1 1

AN λ � � � �= � � � � � �

����

1 1 1 2 1 2 4 1 4

ON OA AN

λ λ λ

λ

− − +� � � � � � � � � � � �= + = + = +� � � � � � � � � � � �+� � � � � �

���� ���� ����

Since N is a point on the plane 1p ,

1 . 1 7

1 ON

� � � � =� � � � � �

����

1 1 2 . 1 7 4 1

λ λ λ

− +� � � � � � � �+ =� � � � � � � �+� � � �

( 1 ) (2 ) (4 ) 7λ λ λ− + + + + + = 2λ =

2 1

31 1 2 1

2 2 8 3 3

4 142 4

3

ON

λ λ λ

� �− +� � − + −� � � �� � � � � �� �= + = + =� � � �� � � � � �+ � �� � � �

� �+� � � �

����

• H2 Revision: Vectors 2010 Mathematics Department

Page 4 of 23

(ii) 4 1 3 . 1 4 3 0 7 0 1

� � � � � � � � = + + =� � � � � � � � � � � �

(satisfies equation of plane 1p )

4 1 3 . 1 4 3 0 1 0 a

� � � � � � � �− = − + =� � � � � � � � � � � �

(satisfies equation of plane 2p )

Hence the point (4, 3, 0) is on both planes 1p and 2p . (4, 3, 0) is then a point on the line � where the 2 planes intersect.

1 1 1 1 1 1 1 2

a

a

a

+� � � � � � � � � � � �× − = −� � � � � � � � � � � �−� � � � � �

is the direction vector of the line � ,

Hence r 4 1 3 1 0 2

a

aλ +� � � �

� � � �= + −� � � � � � � �−� � � �

is the equation of line � .

(iii)

If the planes 3p with equation r 1

. 2 3

b � � � � =� � � � � �

intersects with 1p and 2p at line �

, then

1 1 2 . 1 0 3 2

a

a

+� � � � � � � �− =� � � � � � � �−� � � �

(1 ) 2(1 ) 6 0a a+ + − − = 3a = −

The line � lies in the plane 3p and thus

4 3 0

� � � � � � � � � �

lies in the plane 3p also.

4 1 3 . 2 4 6 0 10 0 3

b b � � � � � � � � = � = + + =� � � � � � � � � � � �

(iv) 0 1 1

2 2 4 5 4 1

AB OB OA

−� � � � � � � � � � � �= − = − − = −� � � � � � � � � � � � � � � � � �

���� .

Given the length of projection of line segment AB on the line � is 1 6

.

• H2 Revision: Vectors 2010 Mathematics Department

Page 5 of 23

1

. 1

2 1 1 6 1

2

a

AB a

a

a

+� � � �−� � � �−� � = +� �

� �−� � � �−� �

����

2 2 2

(1 ) 4(1 ) 2 1 6(1 ) (1 ) ( 2)

a a

a a

+ − − − =

+ + − + −

2

5( 1) 1 3( 3)

a

a

− = +

2

2

25( 1) 1 ( 3) 3

a a

− = +

2 275( 1) 3a a− = + 274 150 72 0a a− + = 150 1188 75 247

2(74) 74 a

± ±= =

0.781a = or 1.25a =

Qn Solution 4 ACJC/Prelim 2009/II/1

Equation of line AB: r 1 3 2 4 1 1

λ � � � � � � � �= + −� � � � � � � � � � � �

or r 4 3 2 4

2 1 λ

� � � � � � � �= − + −� � � � � � � � � � � �

C is on the line AB:

1 3 2 4 1

OC

λ λ

λ

+� � � �= −� � � �+� �

����

Let AOC BOCθ = =� � . .

cos OAOC OB OC

OA OC OB OC θ = =

�������� ��������

���� ���� ���� ����

Hence, . .OAOC OB OC

OA OB =

�������� ��������

���� ����

• H2 Revision: Vectors 2010 Mathematics Department

Page 6 of 23

1 1 3 4 1 3 2 . 2 4 2 . 2 4 1 1 2 1

1 1 2 2 2 1 1