vectors extra questions solutions

8/3/2019 Vectors Extra Questions Solutions

1/47

Vectors Extra Questions Solutions

AJC/I/1313 i) distance between point A and the plane 1

2 2 2 2 2 2

3 2

1 1

4 2 6 6 1 8 2 332 1 2 2 1 2

= = =+ + + +

unit

ii)

0

5

2

=

AB ,2 2 2

2 21 1

1 132 1 ( 2) 2 6

n

= = + +

length of projection

2 2 2

0 21 5 13

2 2

12 61 2 2 2

4 2 6 2 5 65 units3 3 3 3

10 5

= = = = = + + =

AB n

iii)1 2

Area of triangle = ( )( 65)2 3

ABC AC

21 2(3 2)( 65) 2 65 units2 3

= =

iv)

1 3

: 4 2 ,

5 1

= +

l r , Two vectors // to 2 are1

4

5

and

3

2

1

,

normal to 2 is // to1 3 14 1

4 2 14 14 1

5 1 14 1

= =

Equation of 2 :1

1 0

1

=

r , ie. 0 + + = x y z

To find line of intersection:0

2 2 6

+ + =+ =

x y z

x y z

The augmented matrix, M is1 1 1 0

2 1 2 6

, RREF (M) =1 0 1 2

0 1 0 2

Alternatively,use i AB n ,

followed byPythagoras thm


2/47

2

2

==

x z

y

2 2 1

2 2 0

0 1

+ = = +

x z

y z

z z

ie. Equation of line:

2 1

2 0 ,

0 1

= +

r

[Alternatively, Cartesian equation of line: 2 , 2 = = x z y ]

AJC/II/44i

4ii

4iii

a b =

cos cos 2sin sin 2

1 1 2

t t

t t1cos cos 2 sin sin 22

= t t t t 1 1cos( 2 ) cos32 2

= + = t t t

( )2

2 2 2 2

1cos3 2 12cos cos3

5 21cos sin 1 cos 2 sin 2 2

= = = + + + +

t a b AOB t

a b t t t t

For maximum AOB , since 0 AOB and cos is a decreasing function over [ ]0, , we aim

to minimize cos AOB , ie.2 1

cos3

5 2

t .

Thus, cos3 1= t , ie.3= t (since 0 t < ).

When2

t = ,

0 1

1 , 0

1 1 2

= =

a b

1 0 1

0 1 1

1 2 1 3 / 2

AB

= =

and

0

1 1

0 1 1

r r

AC s s

= =

Since A, B and C are collinear, AB k AC =

1

1 1

3 / 2 1

r

k s

=

Looking at the z component, k= - 3/2, so r = -2/3 and s = 1/3

ACJC/I/9

9(i)

(ii)

Length of the projection of OA on OB

=

3 51 20

1 4 2 250 50

3 3

= =

or 4

2

Method 1:

From (i), 2 2OC = 5 5

1 22 2 4 4

5503 3

OC

= =

Method 2:

Line OB:

0 5

0 4

0 3

= +

r

A

C BO


3/47

(iii)

5 3 5 3

4 1 4 1

3 3 3 3

AC OC OA

= = =

Since AC OB

,

5 3 5

4 1 . 4 03 3 3

25

=

=

52

45

3

OC

=

Since25

OC OB=

, : 2 : 3OC CB = 5

1 1' 4

5 53

OB OB

= =

5 3 201 1

' ' 4 1 15 5

3 3 18

AB OB OA

= = =

Vector equation of line ' AB is

3 20

1 1 ,

3 18

r = +

ACJC/II/5

5

(i)

(ii)

1

1

: 2 3

1

p r

=

1

1 2

: 0 1 ,

1 4

l r

= +

2 11 . 2 2 2 4 0

4 1

= + =

1 10 . 2 1 0 1 3

1 1

= +

l1 is parallel to p1. l1 is not contained in p1.

Alternative method:1 2 1

2 2 3

1 4 1

+ = +

i

Since no solution for , l1 is parallel and not contained to p1


4/47

(iii)

(iv)

(v)

1 2 3

2 1 3 2

1 4 1

=

2

3 1 3

: 2 0 2 3 0 1 4

1 1 1

p r

= = + + =

i 2 : 3 2 4 p x y z + + =

3

2 4 2

: 1 1 1 8 1 4 5

4 1 4

p r

= = + =

3

2

: 1 5

4

p r

=

2

1 2

: 0 0 ,

1 3

l r

= +

1

1

: 2 3

1

p r

=

( )1 2 1

0 . 2 3 2 5 3 1 point is 1,0, 2

1 3 1

+ = + = =

B

2 11 1 5

sin 0 213 6 78

3 1

= =

Length of the projection of AB on p1 = cos AB

=

225 53 53 1

0 1 13 31878 78 6 6

3

= = =

CJC/I/77. (i) AB = 2 BP

OP =12 (3 OB OA ) =

8

1

5

21

(ii) Equation of l AB: r =

+

2

1

1

1

2

1


5/47

Equation of l: r =

+

1

1

2

2

1

0

If they intersect,

+=

++

2

1

2

21

2

1

3,2 == Check by substituting into unused equation, 1 + 6 2 + 2Hence they do not intersect.

(iii) Shortest distance from C to AB =3

22

6

2

1

1

x

2

0

2

=

Area of triangle ABC = 113

22

2

1

1

2

1 =

CJC/II/22.

(a)(i) Normal of p1 =

=

3

1

2

1

1

2

x

1

1

1

Direction of l1 =

1

1

1

Angle between l1 and normal of p1 = o

.

1.128)111)(914(

1

1

1

3

1

2

cos 1 =++++

Hence angle between l1 and p1 = 128.1 o 90 o = 38.1 o

(ii)

5.2

0

75.3

and

5.2

5.0

4

are two points on the line25

;2154 ==

z y x

Substituting each point into equation of plane,3.75 + 2.5 = 1 . (1)3.25 + 2.5 = 0 . (2) = 2, = 2.6


6/47

=

1

2

of Normal 3 b p

=

0

1

5.0

of vector Direction 2l

1;0

0

1

5.0

1

2==

bb .

islinecontaining planeof Eqn 24 l p

r. 10

1

1

2

.

5.2

0

75.3

1

1

2=

=

2

63

6

1

6

10 and betweenDistance 32 == pl

DH/I/55(i) If A, B and C are collinear, then

2 3

7 3 5 7

2 1 2

i.e. 2, 0, 4

AB BC

b b

a

a b

=

=

= = =

5(ii)If OA

is perpendicular to ,OB

then

0

2

3 7 0

2

i.e. 2 21 2 0

OA OB

b

a

b a

=

=

+ + =

i

i

2

2

2

2 33 5

1cos60

13 35

2(6 15 ) 13 35

31 168 1309 0

10 (nearest int.) or 4 (nearest int.)

20 (nearest int.) 6 (nearest int.)

a

a

a a

a a

a a

b b

=++ + = + =

= = = =

i


7/47

DH/II/44(i) 2 1

4 3

1 1 15cos| || | 21 11 21 11

9.3 .

1 1

1 2

n nn n

= = =

=

i

i

(ii) 1

1

2

= =

1 2d n n

Set z=0,2 4 10

3 8

1, 3

x y

x y

x y

+ =+ =

= =

1

1 1

3 1 , .

0 2

l

= +

1 1: r a + d =

Alternative

1

2 4 10

3 8

Let ,

2 4 10

3 8

1 , 3 ,2 2

1 1

: r 3 1 ,2

0 2

x y z

x y z

z t

x y t

x y t

t t x y

t l

+ + =+ + =

= + =

+ =

= + =

= + =

(iii) Since the point with co-ordinates (6,m.5) lies on the first plane,

1 1

6 2

4 10

5 1

12 4 5 10

D

m

m

= =

+ + =

a di

i

His claim is not necessarily true since points O , A, B and C may not be coplanar.


8/47

7.

4m =

(iv)

2

2 2

0 , .

7 1

l m = +

2 2: r a + d =

1 21 0 2 2 0 ( )

2 1

independent of thevalueof m = = = 1 2d di i

Therefore lines l1 and l2 are perpendicular for all real values of m.

HCI/I/1111i OA = 14ii Plane ABD

5 1 16

4 0 200 4 4

4 14 4

. 5 0 . 5 56

1 0 1

r

= = =

4 5 56

5 6 36

x y z

x y z

+ = + =

Using GC to solve:

4 1

8 1 ,

0 1

r = +

OR 14 5 1

4 . 5 36

4 6

14 5 20 24 36

25 25 50

2

+ =

+ + == +

= +


9/47

14 5 1

0 4 ( 2) 0

0 0 4

12 4

0 4

8 4

12 1

0 1 ,

8 1

r

= + + +

= + = +

iii 4 12

8 8 0

8

(Reason: j is zero.)

48

0

12 14 2

0 0 0

8 0 8

12 4 8

0 8 8

8 0 8

+ = + = =

=

= = = =

OD OD

OB

AD

BD

1Area =

28 2 1 1 4

18 0 8 1 0 8 5

28 8 1 4 1

8 42 51.8 (3 s.f.)

= = =

= =

ABD BD AD

iv 2(4) 7( 8) (0)

2(12) 7(0) 864, 5

+ =

+ = = =

OR

The 3 planes intersect at the line4 1

8 1 ,

0 1

r = +


10/47

2 1

7 . 1 0

1

2 7 0

5

= + ==

4 2

8 . 7 8 56 640 5

64

= + = =

v Their partitions are parallel to each other.

There is no intersection point.

HCI/II/4

4i 0 1 12 (1 ) 0 2

0

0 0 0

0 (1 ) 2 2 2

2

0 1 1

2 2 2 2 4

2 3

= + = = + = +

= =

+ +

OX

t t

OY

t t t t

XY

t t t t t

OR 1

2 AB

t

=

0

2

2

BC

t

=

( )1 1

1 2 2 2 XB

t t t

= =

0 02 2

2 2

BY

t t

= =

1 0 1

2 2 2 2 4

2 3

XY XB BY

t t t t t

= + = + =

ii Suppose O, X , Y are collinear.Then


11/47

1 0

2 2 2

2

1 0 1 (Out of range)

= = +

= =

OX kOY

k

t t t

Thus O, X , Y are not collinear.

iii 1 0

2 2 2

2

OX OY

t t t

= +

i i

= (4 4 + t 2 2 t 2)= 0

2

2 2

4 1 10 (reject) or = +

4 2 2 2t t t

+ = =+ +

For all t \{0}, 0 < < 1.

Thus XOY can be 90 when 0t .iv 12 4

3

projection vector

1 4 4

2 4 . 1 . 1

3 0 0

17

44 4 2 4

117

0

42

117

0

XY

t t

t t

= +

+ =

+ = =


12/47

IJC/I/44a 1

2

OA =

1 1 2

0

2 1 1

BA

= =

cosOA BA OA BA OAB =

2 21 2

1 4 4 1 cos

2 1

OAB = + + + +

2

24cos5

OAB

+=+

4b Area of quadrilateral OABC = 2 Area of AOC = 2 1

2OA OC

= OA OC

Area of AOB =

12

OA OB

Area of BOC = 12

OB MC

= ( )1 42 OB AM

since 4 AM MC =

=1

42

OB AM

= ( )4 area of AOB

=1

42

OA OB

= 2 OA OB

Thus, area of quadrilateral OABC = Area of AOB + Area of BOC =

12

OA OB

+ 2 OA OB

=52

OA OB

OA OC

=52

OA OB

(Shown)

B

O

A

M

C

B

O

A

1

4


13/47

IJC/I/1212i

1 :2

r 1 9

3

=

.

1 :

3 4

r 0 1

0 2

= +

For point of intersection,3 4 2

0 1 1 9

0 2 3

+ =

.

3 2 4 2

0 1 1 1 9

0 3 2 3

+ =

. .

9 6

38 1 6

= =

OP =

3 4 15

0 3 1 3

0 2 6

+ =

Thus, coordinates of the P are ( )15, 3, 6 .

15 3 12

3 0 3

6 0 6

AP

= =

Shortest distance from A to 1

=

2

1

3

2

1

3

AP

.

12 2

3 16 3

4 1 9

=+ +

.

=24 3 18 3 3 14

1414 14

= = =

1

A (3, 0, 0)

P (15, 3, 6)

2

1

3

1


14/47

Alternative solution for shortest distance :

Line AN :3 2

r 0 1

0 3

= +

N is the point of intersectionof line AN and plane 1 .

3 2 2

0 1 1 9

0 3 3

+ =

.

3 2 2 2

0 1 1 1 9

0 3 3 3

+ =

. .

6 (4 1 9) 9 + + + =

314

=

Thus,

3 2 483 1

0 1 314 14

0 3 9

ON

= + =

48 3 21 3

3 0 114 14

9 0 3

AN

= =

Thus, shortest distance AN

=3 3 14

4 1 914 14

AN = + + =

1

A (3, 0, 0)

P

2

1

3

1

N


15/47

ii

1 :2

r 1 9

3

=

.

2 :1 2 2

r 1 0 2

1 3 1

s t

= + +

A normal to plane 2 =2 2 6 3

0 2 (8) 2 4

3 1 4 2

= =

2 :3 1 3

r 4 1 4 3 4 2 1

2 1 2

= = + =

. .

3

r 4 1

2

=

.

Since

3 2

4 1

2 3

k

for any k ,

the normal of planes 1 and 2 are not parallel to each other The planes are not parallel to each other.The planes will intersect in a line.


16/47

iii

2 :

3 2

3 1

2 1

+

r =

3 : x y z + =

1

1

=

r.

1 , 2 and 3 do not have any points in common 2 is parallel to 3 and does not lie on 3

2 is perpendicular to 1

1

and

3

3 1

2 1

.

Thus,

2

1 1 0

1 1

=

.

2 1 1 0 + + = 1 =

and

3

3 1

2 1

. 3( 1) 3 2 +

2

JJC/I/75 1

14 2

11 3

6 3

12 2 6

8 4

AB

= = =

3 2

6 3 6 18 24

4 6

0

= +

=

l is parallel to p.

1 2

2 3 2 6 18

3 6

14 7

= +

=

l is parallel but not contained on the plane p .

(i) Let m be the line perpendicular to p and passing through A.

3

1

1

2

1

1

(3, 3, 2)

2


17/47

Vector equation of line

1 2

: 2 3 ,

3 6

m

= +

r

Let F be the foot of perpendicular of A to p .

( )

1 2 2

2 3 3 73 6 6

2 1 2 3(2 3 ) 6(3 6 ) 7

17

+ + =

+ + + =

=

2 2 2

1 21

2 37

3 6

1 2 112 3 2

73 6 3

21

37

6

12 3 ( 6)

71

OF

AF

AF

= +

= + =

= + +

=

Distance of the line l from the plane p= 1 unit

(ii) Let the plane required be 1 p .

1

3 2 48

Normal of = 6 3 26

4 6 3

1 48

2 26 48 52 9 91

3 3

p

=

= + =

1

48

Equation of : 26 91 48 26 3 91

3

p r x y z

= + =

JJC/II/3


18/47

3(i) Using ratio theorem,

23

320 2

1 3 2 022 2

1

3

4

1

OA OBOM

OM OBOA

+=

=

= =

=

3(ii) 4 2

0

1 2

2

1

BC

=

=

Length of projection of BC onto the line OM =3

2

2 2

2

3

2

2 0

1

1 1 3

21 11 3

2 8 0

2 or 4 (rejected)

2

BC OM

OM

=

=

++ =

+ == =


19/47

3(iii) 2 1 1

0 3 3 1

2 4 2

4 1 5

2 3 1

1 4 51 5

Normal of plane 1 1

2 5

3

5

4

AB

AC

ABC

= =

= =

=

=

( ) ( )

2

2 2

2

3 3

0 5 0 5 sin 61 4 1 4

13 4 1 50

2

6 8 50 1

7 48 7 0

17 or - (reject)

7

7

a a

a a

a a

a a

a

a

= + = +

+ = +

=

=

=

MI/I/9d =

r n ------ (1)

r = a + b ------(2)

Substitute (2) into (1),( ) d

d

a + b n =

a n + b n =d

= a n

b n

(Shown)

(i) If l is in , a is on , d =a n . 0b . n = Hence, has infinitely many solutions.

(ii) When l and intersect at one point,a n

b nd

= , from

above. Substitute this into the equation of the line l, the position

vector required isa n

r a bb n

d = +


20/47

MI/II/33 (a)(i)

12 9 1

6 3

3 0

3 1

6 33 0

p

q

pq

= =

3

3

6 (3)(3) 15

q

p p

== = =

(a)(ii)2 3

52 53 3

9 12 142 5

6 15 213 3

3 3 3

OM OPON

OP OM ON

OP

+=

= + = + =

(b)(i)

Point (1, ,3) is on both planes:

1 2

1 43 1

2 3 4

1

b

b

b

b

= + ==

i

1 1

1 3 7

3

1 3 3 7

3

a

a

a

= + ==

i

(b)(ii)Line is perpendicular to the normals to both planes:

1 2 0

3 1 5

3 1 51 0 1 0

: 1 1 where or 1 5 where

3 1 3 5

l

l r r

=

= + = +


21/47

MJC/I/9

(i)

(ii)

Vector equation of the line l 1 1

= 0 1 ,

1 1

+

r

Angle between OA and the line l

( )o o

0 1

1 1

2 1 3 3cos

5 3 15 15

39.232 39.2 1 d.p.

= = =

=

i

Let the shortest distance from the origin to the line l be x.

( )

sin5

1.41 3 s.f.

x

x

=

=

(iii) Since

1

1

1 1

0 2 4 A point on lies on

1 3

1 1and 1 2 0 is parallel to

1 3

l

l

=

=

i

i

The line l lies on the plane (shown)

Alternative Solution (1):Since

1

1

1

0 1

-1 2 4 Point on lies on

2 3

1 1

0 2 4 Point on lies on

1 3

line lies on

A l

B l

l

= =

i

i

Alternative Solution (2):Since


22/47

1

1 1 1

0 1 2

1 1 3

1 1

2

1 3

1 2 3 3

4

line lies onl

+

+ =

= + + + =

i

i

(iv)2

1 1 5

normal of 1 2 4

1 3 1

= =

2

2

5 1 5

: 4 0 4

1 1 15

: 4 6

1

=

=

r

r

i i

i

(v)3

1 5 14 1

normal of 2 4 14 14 1 OR

3 1 14 1

= = =

3

1

normal of director vector of 11

l

= =

3

1

: 1

1

d =

r i

3 perpendicular distance from origin to 1 31

1

d d = =

3 33

d d = =

3

1

: 1 3 or

1

=

r i

3

1

: 1 3

1

=

r i


23/47

MJC/II/23 = p q p r

( ) =p q 3r 0 ( )// p q 3r

=q 3r p , where is a scalar.

( ) ( )3 . 3 . =q r q r p p 2 2 226 . 9 + =q q r r p

( ) ( )2 22 255 6 5 2 9 2 16

+ =

2 11

11

==

NYJC/I/55 (i) c = a + b

(ii) sin sinOB AOB OA OB AOB OA OBOA= = = =

Area a b

(iii) Since sin AOB has maximum value of 1, thus maximum area of OACB isa b .

(iv) Let a =a1a 2a3

and b =b1b2b3

. We have cos = ia b a b a b .

Thus2 2 2ia b a b . Since 1 1 2 2 3 3a b a b a b= + +ia b , we have

( )( )2 2 2 2 2 2 21 1 2 2 3 3 1 2 3 1 2 3)( b a b a b a a a b b ba + + + + + + .

NYJC/II/44 (a)(i) Equation of l is r = i j + k + (2i + j 2k ).

Substitute r =

++

21

1

21

into equation of p gives

++

21

121

1

11

= 3

1 + 2 1 + + 1 2 = 3 = 2.

Therefore

=

3

1

5

OB .


24/47

(ii) A

B p (side view)

Let be the angle between l and the normal vector of p . Takingscalar product of the direction vector of l and the normal vector of

p gives

21

2

1

1

1

= 33 cos cos =33

1.

[Note that is acute since cos > 0]Let be the acute angle between l and p .sin = sin(90 ) = cos =

33

1.

(iii) shortest distance from A to p

= AB sin =4

2

4

1

3 3

=6 2

3 3 3= .

(iv) AB = ( )22 24

2 4 2 4 6

4

= + + =

length of the projection of AB onto p

=2

2 2 266 233

= using Pythagoras Theorem

(b) By GC, the equation of the line of intersection of the planes given bythe last 2 equations is

r =

3 2

0 2

0 1

+

.

For the system to have infinite solutions, the above line must lie on the plane given by the first equation. So the line is perpendicular to thenormal vector of the plane. That is,

1

3

2

2

1

= 0 -2 + 2 + 3 = 0 = 12

Furthermore, any point of the line is also a point of the plane. Take(3,0,0) and substitute into the first equation gives 3 + (0) + 3(0) =

= 3


25/47

NJC/I/33(i) ( ) AB OP = b a

p

= b p a p = a p a p (since b p = a p )= 0

Hence, AB is perpendicular to OP. OR

b p = a pb p a p = 0( ) b a p = 0

0 AB OP =

Hence, AB is perpendicular to OP.

3(ii) Since =a b , then P must be the midpoint of AB.

Using ratio theorem, ( )12

OP = +a b

Thus, 2OD OP=

( )1

2 2

= + a b = +a b

3(iii) a b represents the

(1) area of rhombus OADB or OBDA . (or)

(2) magnitude of a vector which is perpendicular to a and b .

NJC/I/9

9 (i)r

13 3

2

= 3 2 3 x y z + + =

2 : 0 0 x y z + + =

Using GC: l:

3 14 23 1

,4 20 1

= +

r R

OR 34 13

1 ,4

20

= +

r


26/47

9 (ii) For the plane 2 : y x= 0 0 x y z + + =

Let 2n be normal vector to plane 2 , then 21

1

0

=

n

1 1

3 12 0 2 1 7

cos (shown)714 2 28 7

= = = =

i

9 (iii) Since point F lies on line l,

Let

3 1 34 2 4 23 1 3

4 2 4 20 1

OF

= + =

for some .

Then

3 34 2 4 203 1

14 2 4 2

0

AF

= =

Now, AF l AF

121

02

1

=

34 21

4 2

121

02

1

=

16

=


27/47

23 134 12

3 1 24 12 3

1 16 6

OF

= =

23 1 34 121 1 14 12 3

1 16 6

AF

= =

2 2 22 1 1 73 3 6 12

AF = + + =

Hence, exact length of projection from AF to the plane 2 cos AF =

7 712 7

=

1 1

12 2 3= = or 3

6

OR exact length of projection from AF to the plane 2

2

2

AF = nn

9 (iv) 3 14 2

3 1 , for some4 20 1

OP

= +

3 has equation 1 px qy+ = .

3 : r 10

p

q

=

2

1

Al

F


28/47

For the three planes to intersect exactly a point, l is not parallel to 3 , then:1212

1

0

0

p

q

1 10

2 2 p q

(ans) p q

PJC/I/14(i)

1 p :

2

2 1 4

0 1

=

2 2 4 = 1 =

position vector of A is i 2 j

(ii)3 1 4

3 2 1

2 0 2

AB OB OA=

= =

Shortest distance from B to 1 p4 2

1 91 1

6 62 1

= =

(iii)

2

2 1

1 2

1cos60

6 5

=

+

2

412 30 6

+=+

( )2 230 6 4 16 8 + = + + 2 16 17 0 =

( )( )17 1 0 + =

1 p A

B2

1

1

1 p

2

1

1

2 p

1

2

60

60

A


29/47

17 = (rejected) 1 =

2 p :

1 1 1

2 2 2

1 0 1

= =

r.

1

2 51

=

r

(iv) Since 3 p passes through the origin, 0 = 3 p 1 p and 2 p (i.e n 3 // line of intersection of 1 p and 2 p )

2 1 1

1 2 1

1 1 1

=

or use GC to obtain

1 1

2 1

0 1

s= +

r

equation of 3 p is 0 x y z + + = 1 =

(v) 3 p // 1 p

n 3

2

1

1

=

2 = and 4

PJC/II/3(i)

5 5 10

2 2 4

6 3 3

AC

= =

line AC :

5 10

r 2 4

6 3

= +

or

5 10

r 2 4

3 3

= +

, R

(ii)Let point R be the top of the pillar

15

6OR

h

=

lies on line AC

15 5 10

6 2 4

6 3h

= +

1 p

2 p

3 p

A

3n

O

1 p

2

1

1

3 p

All points on 3 p

equidistant to 1 p


30/47

15 5 10 1 = + = 6 2 4 1 = + =

6 3 9h = + = Height of pillar is 9m.

(iii)

5 10

2 4

6 3

OX = +

for some

5 10 5 10 10

2 4 2 4

6 3 6 3

DX

+ = + =

10 10 10

4 4 0

3 3

+ =

100 100 16 9 0 + + + = 45

=

5 10 54 3

2 4 25 5

6 3 6

OX

= =

RJC/I/11Q11[10](i)[3]

Since the two planes intersect in a line l, with a vector equation given by

r = 4 22 1 , ,5 3

s s +

R

The line l lies in p2.

So2 3

1 0 6 18 0 12

3 6

= + + = =

4

2

5

satisfies equation of p2.

So4 3

2 12 2 30 12 24 30 6

5 6

= = = + =

(ii)[2] Since p1 and p2 intersect in line l, and the 3 planes have no common point of intersection, l does not intersect p3.Hence l is parallel to p3, but l does not lie in p3.

A X

D ( )5,2,6

10

43


31/47

2 5

21 8 0 10 8 3 0

33

t t

t

= + + = =

Also

4 52

2 . 8 123

5 23

=

which means (4, 2, 5) is not in 3 p and so l does not lie in p3.

(iii)[5]

Since p4 contains l,2

1

3

is parallel to p4.

Since p4 contains the points (1, 1, 2) and (4, 2, 5).

Hence4 1 3

2 1 1

5 2 7

=

is parallel to p4.

A vector normal to p4 is3 2 4

1 1 5

7 3 1

=

.

Equation of p4 is given by :4 1 4

5 1 5 1 4 5 1

1 2 1

x y z

= = + + =

r

Let be the angle between p1 and p4.

cos =2 2 2 2 2 2

2 4

5 53 1 14

38 422 5 3 4 5 1

=

+ + + +

= 110.514

Acute angle between p1 and p4 = 180 = 69.5 (1 d.p.)

RJC/I/12

Q12[11](a) [3] By Ratio Theorem,22

12 14 112

4 82 2

10 6 12

28 11So 12 = 12 24 28 11 4

+2 2

16 16 4Also, 4 = 4 10

+2 2 6 6

OA OBOP

t

t t t

+=+

= + + +

+ + = + =+

+ = + = +


32/47

(b) [8](i)[4]

11 12 1

2 4 6

12 10 2

11 14 3

2 8 6

12 6 6

Length of projection of onto1 3

6 6

2 6

PB OB OP

AB OB OA

PB AB

= = =

= = =

=

2 2 2

3 36 12 21 79 3813 6 6

+ = = =+ +

Shortest distance of P from AB 2

2 7 49 320 8 5413 9 3 3

BP = = = =

(ii)[4] BAPQ forms a parallelogram.Hence

3 12 9

6 4 10

6 10 16

AB PQ

OQ AB OP

= = + = + =

Area of parallelogram BAPQ

2 2 2

3 26 12

6 4

48

0

24

48 0 24

AB AP=

=

=

= + +

= 53.67 (to 2 dec places)

RVH/I/11Let PB be b and F be the foot of perpendicular from B tol1.

P Bb

a F l1

l2

l3


33/47

Then ( )2 PF = =

a ba b a a

a

2 PB PF FB = =

a bb a

a

,

which is parallel to l3.

( )2 2 2 2

4 8

1 48 4

2 34 1

4 1 23 2

= + +

a bb a

a

=

8 4

4 2 1

3 2

=

0

2

1

Since P (3, 5, 2) is a common point,

3

3 0

: 5 2

2 1

l = +

r , .

Angle between 2l and 3l =1

8 0

4 2

3 1cos

64 16 9 0 4 1

+ + + +

= 76.3

1

4 0 5

1 2 4

2 1 8

= =

n

So,

5

2 44 8

k

c

=

1252

c

k

=

=


34/47

2

5

: 4 18

8

=

r i

Since

2 5

0 4 18

1 8

=

i , a point on 2 is Q ( )2,0, 1 .

2 3 1

0 5 5

1 2 3

PQ

= =

Required distance = 1PQ n

=

1 51

5 425 16 64

3 8

+ +

=13 105

35

Alternative Method 1 to calculate distance:5

4

8 211 105 105

5

4

8 182 105 105

: .

: .

=

=

r

r

Thus, O is between the planes.So, distance = 21

105+ 18

105= 39

105

Alternative Method 2 to calculate distance:Equation of line through P and perpendicular to

1 2and : 3 5

5 4

2 8

= +

r

Let N be the foot of perpendicular from P to 2 .3 5 5

135 4 4 18

352 8 8

+ = =

i

3 513

5 435

2 8

ON

= +


35/47

Required distance =

513 13 105

435 35

8

PN

= =

RVH/II/2(i)

Area of triangle OAB =12 OA OB

=

1 41

2 12

3 0

=

31

122

9

=3 26

2 units2

(ii) ( )1 23OP OA OB= +

by ratio thm

=

1 41

2 2 13

3 0

+

=

3

0

1

Since //OQ OP

,30

1

OQ c =

for some constant c.

(iii) 0 AQ BQ =

i

3 1 3 4

2 1 0

3

c c

c c

=

i

2 29 12 3 4 2 3 0c c c c c + + = 25 9 1 0c c + =

Using GC, c = 1.68 or c = 0.119 (rej. since OQ > OP )

SAJC/I/77(i) 2 4 8

2 6

x y z

x z

+ =+ =

From G.C, x = 6 2 z, y = 1 + 1.25 z, z = z


36/47

vector equation of l:6 2

1 1.25 ,

0 1

+

r =

(ii)1

x

OF y

z

=

F 1 is on 1 2 x + 4 y z = 8 --- (1)

1

6 2

9 4

2 1

OF

=

--- (2)

Solving (1) and (2), 2 = x = 2 , y = 1 , z = 0The foot of the perpendicular is (2,1,0) .

(iii)

Direction vector of 1 2

26 162 85 5

29 1 8 20

52 0 2 15 5

F F

= = =

Vector perpendicular to 3 =8 2 18.75 15520 1.25 10 8

41 1 50 40

= =

15 2 15

8 1 8 22

40 0 40

= =

r

vector equation of 3 :

15

8 22

40

=

r (shown)

(iv) (6, 1, 0) is a point on l.Perpendicular distance from (6, 1, 0) to the plane 3

6 2

9 4

2 1

x

y

z

= +


37/47

6 2 15 4 151 1 8 2 8

0 0 40 0 401.75

15 225 64 16008

40

= = =

+ +

15

1.75 , 8

40

vm

= =

SAJC/II/1

(i)

Length of projection = AC = 2 2

2

1

22 1 2

AG

+ + = 2 2

1 2

2 1

4 22 1 2

+ +

=2 2 8

3+ +

= 4 units

(ii)2 2

1

2 2

AC

= =

4 AC = 2 2 24 4 16 + + =

43

= 2

41

32

AC

=

By ratio theorem, 2 35 AG AC AI + =

2 8

4 4

8 8

510 5

1 28 4

5 516 8

AI

+ =

= =


38/47

(iii) Angle between AG and GC

=1sin

AG AC

AG AC

units

1

1 24

2 134 2

sin 60.81 4 16 4

= =+ +

or 1.06 (in radians)

SRJC/I/9

(i)

1 1

1 , 1

4 3

AB BC

= =

7

1

2

AB BC

=

Equation of plane :7 1 7 7

1 3 1 10 1 102 0 2 2

= = =

r r

(ii) Let foot of perpendicular from D be F .

Now,4 7

5 2

OF

+ = + , Since F is on ,

4 7 7

1 10

5 2 2

+ = +

1427

=

10271427

10727

OF

=

, therefore

10 9827 2714 1427 27

107 2827 27

4

0

5

DF

= =

392 14 627 9

DF

= =

(iii) Equation of line l :

4 0

0 1

5 1

= +

r Thus since

4

5

ON

= + lies on plane ,

4 728

1 103

5 2

= = +

Therefore 283133

4

ON

=

1'

2OF OD OD = +

88272827

7927

'OD

=

Thus 2827

7' 8

7

ND =

283

133

4 7' : 8

7

l

= +

r


39/47

TPJC/I/9(i) a b+

is acting along the angle bisector of AOB

(ii)

3 2 1

4 3 1

12 6 6

AB

= =

Equation of line l:

2 1

3 1 ,

6 6

r

= +

(iii)

21 37

6

a

=

,

31 4

1312

b

=

2 3 47

1 1 1 3 4 677 13 91

6 12 162

a b

+ = + =

Equation of OD:

47

67 ,

162

r

=

At D,

2 47

3 67

6 6 162

+ + = +

2 47 ......(1)

3 67 ......(2)

6 6 162 ......(3)

+ = + = + =

(2) (1), 120

=

471

6720

162

OD

=

TPJC/II/4

(i)

3 1 2 2 1

12 3 4 ; 1 ; 0 2 0

3 1 2 2 1

OA OB OC

= = = = =

3 2

12 0

3 2

1

12

5

OX OA AX OA OC = + = +

= + =


40/47

(ii)Height of X above the ground = 5 units

Let be the required angle.

Hence,

5 5sin

1 144 25

22.5 oOX

= =+ +

=

(iii)1

2 0

1

1 2 1

12 1 13

5 2 3

BD OC

BX

= =

= =

1 1 13

0 13 21 3 13

=

Therefore, equation of plane

13 2 13

: 2 1 2 54

13 2 13

BDX

= =

r

(iv) Normal vector to plane OBDC is

1

2 4

1

OA

=

Let be the required angle.13 1

2 4

13 1cos

342 18

=

76.7 o =


41/47

TJC/I/1Soln:

0 2 2 1

2 0 2 2 1

1 3 4 2

AB

= = =

( )

2 2 1

2 0 2 2 12 1 3 2 4 2

AC

= + = + =

Therefore AB is parallel to AC and since A is a common point, we have A, B and C collinear.

From above, we have2

2 AC AB

=

Since C divides the line segment AB in the ratio 2:1,23

AC AB=

.

2 22 3

= 23 =

Alternative 1for last part:Since C divides the line segment AB in the ratio 2:1, we have AB: BC = 2:1

23

OA OBOC

+ =

2 01

2 0 2 23

2 1 3 1

= +

23

=

TJC/I/10Soln:

The vector equation of p1 is

1

1 6

3

=

r i ---(1)

The vector equation of the line passing thru A and perpendicular to p1 is1 1

0 1 ,

2 3

= +

r ----(2)

Sub (2) into (1):

1 1

1 6

2 3 3

+ =

1 6 9 6 + + + = 1 =


42/47

Therefore

1 1 2

0 1 1

2 3 1

OB

= + =

1 0 1

0 1 1

1 1 1

=

and origin is in p2,

Therefore p2 has Cartesian equation 0 x y z = ---(3)

And p1 has Cartesion equation 3 6 x y z+ = ---(4)

Using GC to solve (3) and (4):

We have x = 3+2 , y = 3+ , z = .

Therefore the equation of l is3 23 1 ,

0 1

r = +

Shortest distance =2 2 2

3 1 2

3 0 1

0 2 1

2 1 1

+ +

2 2 5

3 1 6

2 1 4

6 6

= =

( ) ( )2 225 6 4 7766

+ + = =

Since p3 passes through the points A and B, therefore p3 //

1

13

.

Since p1, p2 and p3 do not have a common point, therefore p3 //

2

1

1

.

An equation of p3 is

1 1 2

0 1 1 , ,

2 3 1

r = + +

.


43/47

VJC/I/8


44/47

VJC/II/3


45/47

YJC/I/1212(i) Since tan = 4

3 AN =

43

=

=+=

3

4

0

41

1

0

43

43

~~k jPA //

(ii)

=

43

13

RA

cos PAR =

RAPA

RAPA=

169

169

43

43

1910

1

3

1

0

++++

=16

1691625

1691 +

=657 //

(iii) A vector equation of the line AR is

+

= ,1

3

0

2

3

43

~r or

+

= ,

3

4

12

0

2

3

~r

(iv) Let F be the foot of perpendicular from P to AR

F on AR

+

=

43

1

3

0

2

3

PF for some

PF AR 01

3

43

=

PF

01

3

1

3

0

2

3

43

43

=

+

9 + 9 2 + +169

= 0169176=

=

+

=

132

162

21

1691

1

3

169176

0

2

3

43

PF

=

169132

,169162

,169

21F //


46/47

YJC/II/4

4(i)

=

2

1

12

// 21 k

b

a p p

k = 2, a = 2, b = 4

2

2

1

1

.4

4

2

2

.:2 =

=

rr p

Perpendicular distance between p1 and p2

411

4

411

2

++

++=

6

6= 6=

(ii)

Vectors:

=

03

1

11

1

14

2

,3

3

2

=

=

=

1

1

3

3

3

3

9

0

3

1

3

3

2

3n

Equation of p3: 3

1

1

3

.

1

1

1

1

1

3

. =

=

r

33 =+ z y x

From GC, RREF

=

49

47

10

47

41

01

3113

4211

47

41

47

41 +== z x z x

49

47

49

47 +== z y z y

Let z = 4 t

t x += 47

t y 749 +=

Vector eqn of l is

+

= t t ,4

7

1

0

4947

r


47/47

4(iii)(a)

=++ z y x p 5:4

0,

49

,47

and ( )1,4,2 are 2 points on l so they are also on p4

Sub into the eqn,

11049

47

5 ==++

( ) ( ) 3111425 ===++

(b) If 3= , 11 , the 3 planes do not intersect and they form a triangular prism.

vectors extra questions solutions

Documents