chapter 2: vectors in three dimensions test a name:

© John Wiley & Sons Australia, Ltd 1

Chapter 2: Vectors in three dimensions Test A Name: _____________________ Some pretty ordinary working out here … it will take years to fix it all up. So just do the best you can with it and use common sense when the textbook solutions are silly! Simple familiar 1 Given the point , calculate the unit

vector parallel to .

3

2 Determine the position vector joining the points and .

2

3 If , calculate the value of .

3

4 Convert to Cartesian form.

4

( )2, 3,5P -

OP!!!"

( ) ( ) ( )

( )

2 2 2

ˆˆ ˆ2 3 5

2 3 5

38

let

1 ˆˆ ˆˆ 2 3 538

OP i j k

OP

OP p

p i j k

= - +

= + - +

=

=

= - +

!!!"

!!!"

!!!"

#

#

u AB=!!!"

#( )7, 2,5A - ( )3,9,12B - ( ) ( )ˆ ˆˆ ˆ ˆ ˆ3 9 12 7 2 5

ˆˆ ˆ10 11 7

AB u OB OA

i j k i j k

i j k

= = -

= - + + - - +

= - + +

!!!" !!!" !!!"

#

ˆ ˆˆ ˆ ˆ ˆ6 10 and 3 2a i j k b i j k= - + - = - + -! !

a b×! !

( ) ( )ˆ ˆˆ ˆ ˆ ˆ6 10 3 2

6 3 10 1 1 218 10 230

a b i j k i j k× = - + - × - + -

= - ´- + ´ + - ´-= + +=

! !

20, 45 ,60u é ù= ë û! !

"20, 45 ,60

20, 45 , 60

u

r q f

é ù= ë û= = =

! !

! !

"

( ) ( )cos cos

20cos 60 cos 45

7.071

x r f q=

=

=

! !

( ) ( )cos sin

20cos 60 sin 45

7.071

y r f q=

=

=

! !

( )sin

20sin 60

17.321

z r f=

=

=

!

ˆˆ ˆ7.071 7.071 17.321u i j k= + +!

Maths Quest 12 Specialist Mathematics Units 3 & 4 for Queensland Chapter 2: Vectors in three dimensions Test A


5 A line segment is formed between two points, and .

Calculate the distance between A and B.

Ahhh, Oh dear … how about

𝐴𝐵#### = 𝑏& − 𝑎) = ⟨3, −6,−7⟩

|𝐴𝐵####| = 13! + 6! + 7! = √94 Yes, my 2 lines are FULL marks!

3

6 Determine the Cartesian equation for the sphere with centre and radius,

.

3

7 Pay attention … ! Determine the Vector equation for the sphere with centre and radius, .

That last question is a Methods question! … Specialist Maths is about Vectors!

|�̃� − ⟨−4, 2, 5⟩| = 121

8 Pay attention … ! Determine the Vector equation for the sphere; 𝑥! + 𝑦! + 𝑧! + 8𝑥 − 4𝑦 − 10𝑧 − 76 = 0

𝑥! + 2 × 4𝑥 + 16 + 𝑦! − 2 × 2𝑦 + 4 + 𝑧! − 2 × 5𝑧 + 25

= 76 + 16 + 4 + 25

(𝑥 + 4)! + (𝑦 − 2)! + (𝑧 − 5)! = 121

|�̃� − ⟨−42, 5⟩| = 121

Yes, those three lines of working is FULL marks!

( )9, 5, 9A - - ( )6,1, 2B -( ) ( ) ( )

( ) ( ) ( )

2 2 22 1 2 1 2 1

2 2 26 9 1 5 2 9

9 36 49

94

d x x y y z z= - + - + -

= - + - - + - - -

= + +

=

( )4,2,5C -11r =

( ) ( ) C , , 4,2,54 , 2, 5

c c c

c c c

x y z Cx y z

= -

\ = - = =

2 2

Radius 1111 121r

r= =

\ = =

( ) ( ) ( )( ) ( ) ( )

2 2 22

2 2 2121 4 2 5c c cr x x y y z z

x y z

= - + - + -

= + + - + -

( )4,2,5C - 11r =



9 Determine the equivalent Cartesian equation for whose parametric equations are given by:

a)

b)

a)

b)

7

10 Determine the vector equation for the line passing through the points and

.

4

11 The vector equation of a line is given by .

a) Convert the vector equation to parametric form.

b) Express the parametric form of the line as a Cartesian equation.

a)

b)

4

25 , 4x t y t= - =

( ) ( )2sec , 3tanx yq q= =

( )( )

2

2

2

2

5.....[1]5 [1 ]

4 [2]sub [1b] into [2]

4 5

4 10 25

4 40 100

x tt x by t

y x

y x x

y x x

= -= +

=

= +

= + +

= + +

( ) ( )( ) ( )

( ) ( )

( ) ( )

2 2 2 2

2 22 2

2 22 2

2 2

2sec [1] 3tan [2]

4sec 9 tan

sec [3] tan [4]4 9[3] [4]

sec tan4 9

14 9

x y

x y

x y

x y

x y

q q

q q

q q

q q

= =

= =

= =

-

- = -

- =

( )9,7,5A

( )6,6,1B -

ˆˆ ˆ9 7 5ˆˆ ˆ6 6

OA a i j k

OB b i j k

= = + +

= =- + +

!!!"

#!!!"

#

( ) ( )ˆ ˆˆ ˆ ˆ ˆ6 6 9 7 5

ˆˆ ˆ15 4

d b a

d i j k i j k

d i j k

= -

= - + + - + +

= - - -

! ! !

!

!

( ) ( )ˆ ˆˆ ˆ ˆ ˆ9 7 5 15 4

r a k d

r i j k k i j k

= +

= + + + - - -! ! !

!

( ) ( )ˆ ˆˆ ˆ ˆ ˆ5 4 6 2 8 11r i j k k i j k= - + + + + -! ( ) ( ), , 5 2 ,4 8 ,6 11

5 2 , 4 8 , 6 11x y z k k kx k y k z k

= - + + -

= - + = + = -

5 2 , 4 8 , 6 115 4 6

2 8 115 4 6

2 8 11The cartesian equation is,

5 4 6 2 8 11

x k y k z kx y zk k k

x y zk

x y z

= - + = + = -+ - -

= = =-

+ - -= = =

-

+ - -= =

-



12 Find the point of intersection between the following lines;

𝑟"C = 2𝚤̂ + 3𝚥̂ + 𝜆(3𝚤̂ + 2𝚥̂)

𝑟#C = 𝚤̂ + 2𝚥̂ + 𝜆(�̂� + 𝚥̂)

Clearly these two lines are in 2D and they have different direction vectors, they MUST intersect!

These are NOT vector functions of time, but we find the point of intersection the same way (not collision, but crossing paths).

Set,

𝑟"C = 2𝚤̂ + 3𝚥̂ + 𝑡(3𝚤̂ + 2𝚥̂)

𝑟"C = 𝚤̂ + 2𝚥̂ + 𝑢(𝚤̂ + 𝚥̂) From 𝑟"C ;

𝑥 = 2 + 3𝑡

𝑦 = 3 + 2𝑡 From 𝑟#C ;

𝑥 = 1 + 𝑢

𝑦 = 2 + 𝑢 Equate components,

2 + 3𝑡 = 1 + 𝑢

3 + 2𝑡 = 2 + 𝑢 Solve simiultaneously,

𝑡 = 0𝑎𝑛𝑑𝑢 = 1 Hence POI is,

⟨2,3⟩ Evaluate the reasonableness, by setting 𝑡 = 0𝑎𝑛𝑑𝑢 = 1; 𝑟"C = 2𝚤̂ + 3𝚥̂ + 0 × (3𝚤̂ + 2𝚥̂) = ⟨2, 3⟩

𝑟"C = 𝚤̂ + 2𝚥̂ + 1 × (𝚤̂ + 𝚥̂) = ⟨2, 3⟩

Hence, both lines go through the position ⟨2, 3⟩.



13 How does it work in 3D?

Find the point of intersection between the following lines;

𝑟"C = 𝚥̂ + 3𝑘N + 𝑠(𝚤̂ + 𝚥̂ − 2𝚥̂)

𝑟#C = 7�̂� + 5𝚥̂ + 4𝑘N + 𝑡P2𝚤̂ + 𝚥̂ + 𝑘NQ

From 𝑟"C ; 𝑥 = 0 + 𝑠

𝑦 = 1 + 𝑠

From 𝑟#C ;

𝑥 = 7 + 2𝑡

𝑦 = 5 + 𝑡 Equate components,

0 + 𝑠 = 7 + 2𝑡

1 + 𝑠 = 5 + 𝑡 Solve simiultaneously,

𝑡 = −3𝑎𝑛𝑑𝑠 = 1 𝑟"C = 𝚥̂ + 3𝑘N + 1 × (�̂� + 𝚥̂ − 2𝚥̂) = ⟨1, 2, 1⟩

Evaluate the reasonableness, by checking the second equation; 𝑟#C = 7𝚤̂ + 5𝚥̂ + 4𝑘N − 3 × P2𝚤̂ + 𝚥̂ + 𝑘NQ

= ⟨1, 2, 1⟩ Hence, both lines go through the position ⟨1, 2, 1⟩.

*** Always ccheck if the quiestion asks for an evaluaetion of your solution!

14 Describe the relationship between the line;

𝑟C = ⟨0, 2, 4⟩ + 𝜆⟨1, −1, 2⟩

and the Plane;

𝑥 − 𝑦 + 2𝑧 = 1

The line has a direction vector of

𝑑R = ⟨1,−1, 2⟩ And the plane

𝑟C ∙ ⟨1, −1, 2⟩ = 1 has a normal

𝑛C = ⟨1,−1, 2⟩

Hence, because 𝑛C is parallel to 𝑑R ,the line is perpendicular to the plane.



15 Describe the relationship between the line;

𝑟C = ⟨0, 2, 4⟩ + 𝜆⟨−2, 2, 4⟩

and the Plane;

3𝑥 − 3𝑦 + 6𝑧 = 2


𝑑R = ⟨−2, 2, 4⟩ And the plane

𝑟C ∙ ⟨3, −3, 6⟩ = 2 has a normal

𝑛C = ⟨3,−3, 6⟩

Because 𝑑R = − !$𝑛C then 𝑛C is parallel

to 𝑑R ,the line is perpendicular to the plane.

16 Find 𝛼 and 𝛽 , given the line;

𝑟C = 𝚤̂ + 𝑘N + 𝜆P6𝚤V + 𝛼𝚥W + 𝛽𝑘XQ

is perpendicular to the Plane;

2𝑥 − 3𝑦 + 6𝑧 = 5


𝑑R = ⟨6, 𝛼, 𝛽⟩ And the plane

𝑟C ∙ ⟨2, −3, 6⟩ = 2 has a normal

𝑛C = ⟨2,−3, 6⟩

Because the line and the plane are perpendicular, then 𝑛C = 𝑘𝑑R

Here, clearly 𝑘 = 3 , hence;

𝛼 = −9 and 𝛽 = 18

** can you see those Brief statements that JUSTIFY the processes … yes, that small amount is sufficient for FULL marks!



17 a) Calculate given and

using the distributive law

of vectors. b) Calculate using determinants, given

and .

a)

b)

8

18 Calculate the scalar triple product for the vectors , and

.

4

19 The parametric equations defines the

path of a particle.

Determine the point, P of intersection between the particle and the plane given by

.

8

a b´! !

ˆˆ ˆ7 2 5a i j k= - +! ! !ˆˆ ˆ3 2b i j k= + +

! ! !

a b´! !ˆˆ ˆ3 2a i j k= - + -

!

ˆˆ ˆ2 2 4b i j k= - +!

( ) ( )ˆ ˆˆ ˆ ˆ ˆ7 2 5 3 2

ˆˆ ˆ ˆ ˆ ˆ7 7 2 2 3ˆ ˆ ˆˆ ˆ ˆ2 2 5 3 5

ˆ ˆˆ ˆ ˆ ˆ7 14 6 4 15 5ˆˆ ˆ9 13

a b i j k i j k

i j i k j i

j k k i k j

k j k i j i

i j k

´ = - + ´ + +

= ´ + ´ - ´

- ´ + ´ + ´

= - + - + -

= - + +

! !

( ) ( )( )

ˆˆ ˆ

3 1 22 2 4

1 2 3 2 3 1ˆˆ2 4 2 4 2 2

ˆ ˆ1 4 2 2 3 4 2 2ˆ 3 2 1 2ˆˆ8 4

i j ka b

I j k

i j

k

j k

´ = - --

- - - -= - +

- -

= ´ - - ´- - - ´ - - ´

+ - ´- - ´

= +

! !

ˆˆ ˆ2 6a i j k= + -!

ˆˆ ˆ2 5 5b i j k= + -!ˆˆ ˆ4c i j k= + +

!

( )2 1 62 5 51 4 1

5 5 2 5 2 52 1 64 1 1 1 1 4

2 25 1 7 6 350 7 1825

a b c-

× ´ = -

- -= - -

= ´ - ´ - ´= - -=

! ! !

5 , 1 2 , 4 2x t y t z t= - = + = +

( ), ,x y z

4 8 2 12x y z- + - =

( ) ( ) ( )4 5 8 1 2 2 4 2 1220 4 8 16 8 4 12

16 20 1216 12 20

32162

t t tt t t

tt

t

t

- - + + - + =

- + + + - - =- =

= +

=

=

( )

1 : 2,5 , 1 2 , 4 25 2 , 1 2 2, 4 2 23, 5, 8

The intersection occurs at the point P 3,5,8 .

l tx t y t z tx y zx y z

== - = + = += - = + ´ = + ´= = =


Chapter 2: Vectors in three dimensions Test A Name: _____________________ Complex familiar 20 A triangle is formed by the points

and .

Classify the triangle as either scalene, isosceles or equilateral and prove the triangle is right-angled.

Triangle ABC is a right-angled scalene triangle.

8 ( ) ( )3,5,6 , 2,7,9A B- - ( )2,1,7C

( ) ( )

( ) ( )

( ) ( )

( ) ( )

ˆ ˆˆ ˆ ˆ ˆ2 7 9 3 5 6

ˆˆ ˆ2 3

14

ˆ ˆˆ ˆ ˆ ˆ2 7 2 7 9

ˆˆ ˆ4 6 2

56

ˆ ˆˆ ˆ ˆ ˆ3 5 6 2 7

ˆˆ ˆ5 4

42

ˆ ˆˆ ˆ ˆ ˆ2 3 4 6 2

4 12 61

AB i j k i j k

i j k

AB

BC i j k i j k

i j k

BC

CA i j k i j k

i j k

CA

AB BC i j k i j k

= - + + - - + +

= + +

=

= + + - - + +

= - -

=

= - + + - + +

= - + -

=

× = + + × - -

= + - + -= -

!!!"

!!!"

!!!"

!!!"

!!"

!!"

!!!" !!!"

( ) ( )

( ) ( )

4ˆ ˆˆ ˆ ˆ ˆ4 6 2 5 4

20 24 242

ˆ ˆˆ ˆ ˆ ˆ2 3 5 4

5 8 3

0 are

BC CA i j k i j k

AB CA i j k i j k

AB CA

× = - - × - + -

= - + - += -

× = + + × - + -

= - + -

= \ × ^

!!!" !!"

!!!" !!"

!!!" !!"



21 Two lines are formed between the points . The first line joins

and The second line passes through and . Calculate the acute angle between two lines.

8

22 Calculate the equation of the plane perpendicular to containing the the point .

5

, , and D E F G ( )1,2,1D( )2,5,6 .E -

,0,1)( 3F - ( )3,2,5G

( ) ( )

1

2

1

1 1 1

The vector equation of both lines is in the form, .The directional vector is given by

.

For line 1, DE

For line 2, ˆ ˆˆ ˆ ˆ ˆ2 5 6 2

ˆˆ ˆ3 3 5ˆˆ ˆ

r a k d

k

k

FG k

k i j k i j k

i j k

a i b j c k

= +

=

=

= - + + - + +

= - + +

= + +

! ! !

"""#

"""#

( ) ( )

( )( ) ( )( ) ( )( )( ) ( ) ( ) ( ) ( ) ( )

2

2 2 2

1 2 1 2 1 2

2 2 2 2 2 21 1 1 2 2 2

2 2 2 2 2 2

ˆ ˆˆ ˆ ˆ3 2 5 3

ˆˆ ˆ6 2 4ˆˆ ˆ

cos

cos

3 6 3 2 5 4

3 3 5 6 2 4

k i j k i k

i j k

a i b j c k

a a bb c c

a b c a b c

q

q

= + + - - +

= + +

= + +=

+ +

+ + + +

=

- + +

- + + + +

1

8cos43 56

8cos2 602

80.62

q

q

q

-

=

æ ö= ç ÷è ø

= !

ˆˆ ˆ3 5 2n i j k= - + +!

( )0 3,7,4P -

( ) ( )0, , and 3,7,4r x y z r= = -! !

( ) ( ) ( ) ( )0ˆˆ ˆ3 7 4

ˆˆ ˆ3 5 2

r r x i y j z k

n i j k

- = + + - + -

= - + +! !

!

( )( ) ( ) ( )( ) ( )

0

ˆ ˆˆ ˆ ˆ ˆ3 7 4 3 5 2 0

0

x i y j z k i j k

r r n

+ + - + - × - + + =

- × =! ! !

( ) ( ) ( )3 3 5 7 2 4 03 9 5 35 2 8 0

3 5 2 52

x y zx y z

x y z

- + + - + - =

- - + - + - =- + + =



23 a) Calculate the equation of the plane determined by the points

and .

b) Determine the axis intercepts for the plane.

a)

b)

8

( ) ( )5, 6,7 , 3, 8,1P Q- - ( )12,6, 13R - - ( ) ( ) ( )

( ) ( ) ( )

ˆˆ ˆ3 5 8 6 1 7ˆˆ ˆ2 2 6

ˆˆ ˆ12 5 6 6 13 7ˆˆ ˆ17 12 20

PQ i j k

i j k

PR i j k

i j k

= - + - - - + -

= - - -

= - - + - - + - -

= - + -

!!!"

!!!"

ˆˆ ˆ

2 2 617 12 20

n PQ PR

i j kn

= ´

= - - -- -

!!!" !!!"

#

#

1 2 3

ˆˆ ˆ112 62 58( ) ( ) ( ) 0n i j kn x a n y b n z c= + -- + - + - =

!

112( 5) 62( 6) 58( 7) 0112 560 62 372 58 406 0112 62 58 21856 31 29 109

x y zx y zx y zx y z

- + + - - =- + + - + =+ - = -+ - = -

109 109 109 , ,56 31 29

x y z- -= = =



24 a) Calculate the equation of the plane determined by the points

𝑃(1, 2, 3), 𝑄(2, 4, 6)𝑎𝑛𝑑𝑅(5, 3, 1)

c) Determine the axis intercepts for the plane.

Funny how the Textbook sucks with it’s setting out! Let’s try that last question again,

Find 2 vectors in the plane;

𝑄𝑃#### = 𝑝) − 𝑞) = ⟨−1,−2,−3⟩

𝑄𝑅#### = �̃� − 𝑞) = ⟨3, −1, 3⟩

𝑛) = 𝑄𝑃#### × 𝑄𝑅####

𝑛) = ^𝑖 𝑗 𝑘−1 −2 −33 −1 3

^ = ⟨−3,−6, 5⟩

Now, vector equation to the plane becomes;

𝑟 ∙ ⟨−3,−6, 5⟩ = ⟨1, 2, 3⟩⟨−3,−6, 5⟩

𝑟 ∙ ⟨−3,−6, 5⟩ = 0

In cartesian form,

−3𝑥 − 6𝑦 + 5𝑧 = 0

Hmmm … I wasn’t planning on that zero being there … please check this and let me know if it is wrong … I was going to find the axes intercepts, but this plane goes through the Origin!

If you had to find the axes intercepts, please justtify your process, just like we have since grade 8;

𝑥 − 𝑖𝑛𝑡 → 𝑦 = 0𝑎𝑛𝑑𝑧 = 0

Etc


Chapter 2: Vectors in three dimensions Test A Name: _____________________ Complex unfamiliar 25 The flight path of two hand gliders,

measured in metres, is given by the vector functions of time below.

Determine if the helicopters will ever cross paths. If possible, calculate the time and vector location of the collision.

Equate the components to determine a value for the parameter .

Test

The hand gliders collide when , at .

LOL, a helicopter and a hang glider collide … Messy!

Also LOL that the Textbook thinks this is Complex Unfamiliar … nowhere near an “A” level question!

9

( ) ( ) ( )( )

( ) ( ) ( )

( )

2 3

3

2

ˆ ˆ12 3 151 5

ˆ2 77 , where 0 minutesˆ ˆ2 337 5 26

ˆ4 13 where 0 minutes

a

b

r t t i t t j

t k t

r t t i t j

t k t

= + + - -

+ + ³

= - + + - +

+ - ³

!

!

k̂t

2

2

2 77 4 130 4 2 90(2 9)( 5) 0

95,2

5, since 0

t tt t

t t

t

t t

+ = -

= - -+ - =-

=

= ³

5t =

( ) ( ) ( ) ( )

( )( )

( ) ( ) ( )( )( )

2 3

3

2

ˆ ˆ5 12 3 5 151 5 5 5

ˆ2 5 77

ˆˆ ˆ87 87

ˆ ˆ5 2 5 337 5 5 26

ˆ4 5 13

ˆˆ ˆ87 87

a

b

r i j

k

i j k

r i j

k

i j k

æ ö æ öç ÷ ç ÷è ø è ø

æ öç ÷è øæ öç ÷è ø

= + + - -

+ +

= + +

= - + + - +

+ -

= + +

!

!

( ) ( ) ˆˆ ˆ5 5 87 87a br r i j k= = + +! !

5t =ˆˆ ˆ87 87i j k+ +



26 A tetrahedron is a pyramid consisting of four triangular faces. The volume of a tetrahedron is given by the rule,

.

Show that the volume of a tetrahedron with vertices

and is 9 units3.

Three vectors are to be created, with , as the shared vertex.

7

27 Determine the parametric equations of the line containing the point that is perpendicular to the line

.

7

( )16

V a b c= × ´! ! !

( ) ( )11,2,0 , 8,5,0 ,A B

( )4,9,2C ( )D 14,8,3

( )16

V a b c= × ´! ! !

( ) ( ) ( ) ( )11,2,0 , 8,5,0 , 4,9,2 and D 14,8,3A B C

( )D 14,8,3

( )

( )

ˆ ˆˆ ˆ ˆ ˆ3 6 3 6 3 3ˆˆ ˆ10

16

The vector product, ,ˆˆ ˆ

6 3 310 1 1

ˆˆ ˆ6 24 361 ˆ ˆˆ ˆ ˆ ˆ3 6 3 6 24 366

AD i j k BD i j k

CD i j k

Volume AD BD CD

BD CD

i j kBD CD

BD CD i j k

V i j k i j

= + + = + +

= - +

= × ´

´

´ =-

´ = + -

= + + × + -

!!!" !!!"

!!!"

!!!" !!!" !!!"

!!!" !!!"

!!!" !!!"

!!!" !!!"

( )

( )1 18 144 1086

9The volume of the tetrahedron is 9.

k

V

V

= + -

=

( )0,2,4P

1 : 1 , 1 , 2l x t y t z t= + = - = ( )

( ) ( )

( )

( )

1Let : 1 , 1 , 2When 0,

1, 1, 01,1,0The direction of this line is given by the vector,

ˆˆ ˆ 2The vector joining 1,1,0 and 0,2, 4 is,

ˆˆ ˆ 4

ˆ ˆ

ˆˆ ˆ 4

1 ˆ ˆ 26

l x t y t z tt

x y z

a i j k

b i j k

b b a b a

b i j k

i j

^

^

= + = - ==

= = =

= - +

= - + +

= - ×

= - + +

- - +

!

!

! ! ! ! !

!

( ) ( ) ( )( ) ( )

( ) ( )

1ˆ ˆ ˆˆ ˆ ˆ ˆ4 26

ˆ ˆˆ ˆ ˆ ˆ4 2

ˆˆ ˆ2 2 2

This is the direction of the line to be developed. ˆ ˆˆ ˆ ˆ ˆ0 2 4 2 2 2

2 , 2 2 , 4 2

k i j k i j k

b i j k i j k

b i j k

r i j k t i j k

x t y t z t

^

^

é ùæ ö× - + + - +ê úç ÷è øë û

= - + + - - +

= - + +

\ = + + + - + +

= - = + = +

!

!

!



28 Let’s try that one again … Determine the parametric equations of the line containing the point that is perpendicular to the line

.

Convert line to symmetric form;

𝑡 = %&''

, 𝑡 = (&'&'

and 𝑡 = )!

𝑥 − 11

=𝑦 − 1−1

=𝑧2

So, 𝑑R = ⟨1,−1, 2⟩ and as given 𝑎C = ⟨0, 2, 4⟩

𝑟C = ⟨0, 2, 4⟩ + 𝜆⟨1, −1, 2⟩ Yes, that’s enough working for FULL marks! There’s more “crap” to fix in this worksheet than you can poke a stick at … just can’t get to it all L

( )0,2,4P

1 : 1 , 1 , 2l x t y t z t= + = - =



29 Calculate the distance between the two intersection points of the line

and sphere given by

.

Describe their geometric significance of this result.

a)

The two intersection points are, and .

b)

The radius of the sphere is . The diameter is therefore , which is the distance between the two intersection points. The intersection points represent the diameter of the sphere.

10

( ) ( )1ˆ ˆˆ ˆ ˆ ˆ34 28 49 2 2 3l i j k k i j k= - - - + - - -

( ) ( ) ( )2 2 268 2 8 5x y z= - + - + -

( ) ( )

( ) ( ) ( )( ) ( )( )

( ) ( ) ( )( ) ( )

1

1

2 2 2

2 2

2

2 2 2

2 2

2

2

ˆ ˆˆ ˆ ˆ ˆ34 28 49 2 2 3

34 2: 28 2

49 3

68 2 8 5

68 34 2 2 28 2 8

49 3 5

68 36 2 36 2 54 3

68 2 36 2 54 3

68 17 612 55080 17 612 5440

20,

l i j k k i j k

x kl y k

z k

x y z

k k

k

k k k

k k

k kk k

k k

= - - - + - - -

= - -ìï = - -íï = - -î

= - + - + -

= - - - + - - -

+ - - -

= - - + - - + - -

= - - + - -

= + +

= + += - = -16

( ) ( )

( ) ( )

( ) ( )

1

1

1

ˆ ˆˆ ˆ ˆ ˆ34 28 49 2 2 3

16ˆ ˆˆ ˆ ˆ ˆ34 28 49 16 2 2 3

ˆˆ ˆ2 420

ˆ ˆˆ ˆ ˆ ˆ34 28 49 20 2 2 3

ˆˆ ˆ6 12 11

l i j k k i j k

k

l i j k i j k

i j kk

l i j k i j k

i j k

= - - - + - - -

= -

= - - - - - - -

= - + -= -

= - - - - - - -

= + +

( )1 2,4, 1P - -

( )2 6,12,11P

( ) ( ) ( )2 2 21 2 6 2 12 4 11 1

64 64 144

272

4 17

PP = - - + - + - -

= + +

=

=

!!!!"

68 2 17=2 2 2 17 4 17r = ´ =

chapter 2: vectors in three dimensions test a name:

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