2010 prelim revision - vectors - additional questions

8/6/2019 2010 Prelim Revision - Vectors - Additional Questions

1/11

1 The line 1l has equation 1 2 x y z = = .The line 2l passes through the point ( 2,1,7)A and is parallel to the vector 3 5 + +i j k.

(i) Write down the vector equations of lines 1l and 2l . [2]

(ii) Show that the lines 1l and 2l intersect and find the coordinates ofE, the point of

intersection of 1l and 2l . [3]

(iii) The acute angle between the lines 1l and 2l is denoted by . By finding cos , show

that2 2

sin35

= . Hence find the shortest distance fromA to the line 1l , leaving your

answer in exact form. [4]

2 The plane 1 has equation

1

2 4

1

=

r . The line lpasses through two points A and B

whose position vectors are

2

0

2

and

7

18

1

respectively.

(a) Show that the position vector of the pointNon 1 such thatBNis perpendicular

to 1 is

1

2

7

. Hence find the perpendicular distance fromB to 1 . [5]

(b) Verify that point A lies on 1 . Hence show that the reflection of l in 1 is

parallel to

7

14

13

. [3]

(c) The plane 2 is perpendicular to 1 and contains l. Find the equation of 2 inthe form p =r n . [3]

(d) 3 is perpendicular to both 1 and 2 and contains the lineBN. By consideringthe triangleABN, or otherwise, determine


2/11

(i) the distance of A from 3 ;

(ii) the acute angle between land 3 . [4]

3 A line l has equation

4 3

0 2

5 1

= +

r , .

(i) Find the position vector ofP, the foot of the perpendicular from the origin O to l.

[3]

(ii) Find a cartesian equation of the plane 1 containing O and l. [3]

(iii) It is given that l also lies in a plane 2 with equation

2

3

1

k =

r , k . Show

that7

2k= . [2]

(iv) Find the angle between the planes 1 and 2 , giving your answer in degrees.

[3]

(v) A third plane 3 has cartesian equation 2 7 x y z + + = . Determine the nature ofthe intersection of the three planes 1 , 2 and 3 . [3]

4 Verify that the point (2, 4, 6) lies on the plane 1 and 2 , such that 1

6

: . 5 32

4

=

r and

2

5

: . 1 24

3

=

r . Given that 1 and 2 meet in the line l, find a vector equation of the line

lin the form = +r a b . [4]

The plane 3 contains the line land passes through the point with position vector 4i + 3j +

2k. Find the vector equation of the plane 3 and express it in the form ax + by + cz= d.


3/11

[4]

Deduce, or prove otherwise, that the system of equations6 5 4 32

5 3 24

9 2 5 40

x y z

x y z

x y z

= + =

+ =has an infinite number of solutions. [2]

The plane 4 is parallel to 3 and has equation 9 2 5 50 x y z + = . Comment on thegeometrical representation of the 3 planes 1 , 2 and 4 . [2]

5 As part of a sculpture, an artist erects a flat triangular sheet ABC in his garden. The vertices

are attached to vertical poles DA, EB and FC. The coordinate axes Ox and Oy are horizontal,

and Oz is vertical. The coordinates of triangle are A(2, 0, 2), B(2, 0, 1) and C(0, 4, 3), withunits in metres.

i) Find the length of the side AC. [2]

ii) Find the scalar product ABAC, and the angle BAC. [4]

iii) Show that 2i + 3j 8kis perpendicular to the lines AB and AC. Hence find the Cartesianequation of the plane ABC. [4]

iv) The artist decides to erect another vertical pole GH based at the point G(1, 1, 0).

Calculate the height of the pole if H is to lie in the plane ABC. [5]

C(0, 4, 3)

B(2, 0,1)

A(2, 0, 2)

F

D

E

O

y

x

z


4/11

Solution

1(i) Equation of 1l is

1 1

2 1 ,

0 1

r = +

Equation of 2l is2 3

1 1 ,

7 5

r = +

1(ii)

1

If 1 2 3 ----(1)

and 2 1 ----(2)

and 7 5 ----(3),

we have from (1) & (2), 3 and 2 which satisfies (3) since

R.H.S of (3) = 7 + 5( 2) = 3 = L.H.S of (3) .andl

+ = = +

= += =

2 intersect and the coordinates of is (4, 1, 3).l E

(iii)

2

1

3 1

1 1

5 1 9 3 3cos

35 3 35 3 35

27 2 2

sin 1 cos 1 35 35

6

sin , where is the shortest dist from to and 2

10

2 2140 4 2

35

.

p p A l AE

AE

p

= = =

= = =

= =

= =


5/11

2. (a)

2. (b)

2. (c)

1

1

: 2 4

1

=

r

7 1

: 18 2

1 1

BNl

= +

r ,

7

18 2

1

ON

+ = +

uuur

for some

Thus,

7 1

18 2 2 4

1 1

+ = +

( )7 2 18 2 1 4 8 + + = =

( )

7 8 1

18 2 8 2

1 8 7

ON

+ = = +

uuur(shown)

Perpendicular distance

=

1 7 1

2 18 8 2 8 1 4 1 8 6

7 1 1

BN ON OB

= = = = + + =

uuur uuur uuur.

2 1

0 2 2 2 4

2 1

= + =

. Therefore pointA lies in 1. (Verified)

7 2 8 7 7

2 18 0 2 16 14 // 14

1 2 8 13 13

AB AB BN

= + = + =

uuuur uuur uuur(shown)

7 2 9 3

18 0 18 3 6

1 2 3 1

AB

= = =

uuur

Two vectors // to 2 are

3

6

1

and

1

2

1

.

1

B (7, 18, 1)

A (2, 0, 2)

N

B

n1

2

l


6/11

2. (d)(i)

2.(d)(ii)

3 1 4 2

6 2 2 2 1

1 1 0 0

= =

Vector perpendicular to 2, 2

2

1

0

=

n . Also, pointA lies in 2.

Therefore,

2 2 2

1 0 1 4

0 2 0

= =

r .

Equation of 2

2

: 1 4

0

=

r

DistanceA from 3,

1 2 1

2 0 2

7 2 5

1 4 25 30

AN ON OA=

= =

= + + =

Acute angle between land 3

=

1 1 30tan tan 15.68 6

ANABN

BN

= = = R (to 1 dec. pl.)

Qn Suggested Solutions

3 Vectors

(i)

SincePlies on l,

4 3

2

5

OP

+ = +

for some R .

Since

3

, 2 0.

1

OP l OP

=

B

A

N

3

1

2

l


7/11

Now

4 3 3

2 2 0

5 1

+ = + 12 9 4 5 0

14 7

+ + + + =

=

1

2=

14 3

25

1 12 2

2 211

15

2

OP

+ = = +

(ii) 3 5 3 20 101 1

2 2 2 38 192 2

1 11 1 16 8

OP

= = = 10

19 0

8

=

r

The cartesian equation is 10 19 8 0 x y z + + = .

Remark: Better to use

4

0

5

instead of OP .

(iii) 4 3 2

2 3

5 1

k

+ = +

8 6 2 5 3k + + + =

3 (7 2 ) 3k + = 7 2 0k =

7

2k =

OR Use pointP(or any point on l).


8/11

5

22

1 3

11 1

2

k

=

Alt. Since llies in 2,

2

1

l k

.

2 3

2 0

1 1

k

=

6 2 1 0k + =

7

2k=

(iv) Let be the angle between planes 1 and 2 .

7

2

10 2

19

8 1cos

525 17.25

=

6.8 = o

(v) The system of equations is10 19 8 0

72 3

2

2 7

x y z

x y z

x y z

+ + =

+ + =

+ + =

Hence the augmented matrix is7

2

10 19 8 0

2 1 3

1 2 1 7


9/11

Using GC, the RREF is

1 0 3 0

0 1 2 0

0 0 0 1

The last row 0 1 = impossible!

Therefore, there is no solution. The system is inconsistent.

Hence the three planes 1 , 2 and 3 do not intersect.

Alt. lis the line of intersection of1 and 2.

Check to see iflintersects with 3.

2 x y z + +

= ( ) ( ) ( )4 3 2 2 5 + + + += 1

7

ldoes not intersect with 3.Hence the 3 planes do not intersect at any common point.

4

Since

2 6

4 5 32

6 4

=

g and

2 5

4 1 24

6 3

=

g , thus the point (2, 4, 6) lies on 1 and 2 .

Direction vector of line l=

6 5 19 1

5 1 38 19 2

4 3 19 1

= =

.

Thus

2 1

4 2: , .

6 1

rl

= +

3

2 1 4 2

: 4 2 3 46 1 2 6

r

= + + =

2 1 2

4 2 1 , ,6 1 4

+ +

Normal vector of 3 =

1 2 9

2 1 2 2

1 4 5

9

5

= =

.


10/11

Thus 3

9 2 9

: 2 4 2 40

5 6 5

r

= =

g g .

Thus Cartesian equation of 3 is 9 2 5 40 x y z + = .

The system of linear equations represents the intersection between 1 2 3, a .nd

Since lis a common line to 1 2 3, a ,nd thus the system has an infinite number of

solutions.

Since 4 is parallel to 3 , the planes 1 2 4, and will form an infinite triangular

prism.

5. (i) AC = OC OA = 4j + 3k (2i + 2k) = 2i + 4j + kLength of side AC = |AC| = 211164 =++

(ii) AB = OB OA = 2i + k (2i + 2k) = 4i i

AB.AC = 718

1

4

2

1

0

4

==

Let BAC = .

cos =2117

7

1164116

7=

+++=|AC||AB|

AC.AB

0368 .=

(iii) Vector perpendicular to AB and AC

= AB x AC =

=

=

8

3

2

2

16

6

4

1

4

2

1

0

4

x

83

2

is perpendicular to the lines AB and AC.

Vector equation of plane ABC is

12

8

3

2

2

0

2

8

3

2

=

=

..r

Cartesian equation of plane ABC is

2x + 3y 8z = 12

(iv) OG = i +j OH = i +j + k

GH =

0

0

H lies on plane ABC. Then


11/11

8

17

178

12)(8)1(3)1(2

12

8

3

2

1

1

12

8

3

2

OH

=

=

=+

=

=

GH =

8

17

0

0

|GH| =8

17

Height of pole GH =8

17or 2.125m

2010 prelim revision - vectors - additional questions

Documents