nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 1

TUTORIAL 3

Question 1

a) 1,3, 4u v

2 2 2

1 3 4

26

u v

b) 2 3 12, 19,12u v

2 2 2

2 3 12 19 12

649

u v

c) 40 34 83 3 3

24 , ,

3v u

2 2 2

40 34 83 3 3

24

3

2705 /17.7012

3

v u

d) 16v u

16v u

Question 2

a) u v b) u v w

c) v w d) w u

Question 3

2

u av bw

i j a i j a i j

Compare LHS and RHS

: 2

: 1

i a b

j a b

Therefore,

3 1,

2 2a b

u +v v

u

v-w

-w

v

w w - u

-u

w

u -v+w

-v

u

nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 2

Question 4

Finding angles: .

cosa b

a b

a) 2 2 , 3 4u i j k v i k

2 2 2 2 2 2

1

2, 2,1 3,0,4cos

2 2 1 3 0 4

10

15

10cos

15

48.19

b) 3 7 , 3 2u i j v i j k

2 22 2 2 2

1

3, 7,0 3,1, 2cos

3 7 0 3 1 2

4

4 26

4cos

4 26

101.31

c) 2 2 ,u i j k v i j k

2 22 2 2 2

1

1, 2, 2 1,1,1cos

1 2 2 1 1 1

1

15

1cos

15

104.96

Question 5

a) / /u w since

15 3 3

3 5

3

w i j k

i j k

u

b) None is perpendicular to each other.

0u w and 0v w

nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 3

Question 6

a) Line equation: AC tv

Since we know that v is parallel to AB , find v.

3, 2,5

v AB

(Using point A) Substitute into the formula: Let C(x,y,z) be a point on line L

1, , 2 3, 2,5

1 3 , 2 , 2 5

AC tv

x y z t

x t y t z t

(Using point B) Substitute into the formula: Let C(x,y,z) be a point on line L

4, 2, 3 3, 2,5

4 3 , 2 2 , 3 5

BC tv

x y z t

x t y t z t

b) Parallel lines:

Since L1 is parallel to L2, then both lines are parallel to the same vector v.

2 : 3 , 6 , 2 7

3 1

0 6

2 7

1,6,7

L x t y t z t

x

y t

z

v

B(4,-2,3)

A(1,0,-2)

L

L2: x=3-t, y=6t, z=2+7t P(0,1,2)

L1

nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 4

Equation of line L1 that passes through point P(0,1,2):

Let Q(x,y,z) be a point on line L1.

, 1, 2 1,6,7

, 1 6 , 2 7

PQ tv

x y z t

x t y t z t

c) Line orthogonal to the plane

Since L is orthogonal to the plane, then L is parallel to the normal vector of the plane, n.

Given the equation of the plane is

3 2 10x y z ,

Therefore, the normal vector is 3, 1, 2n .

Equation of line L that passes through point P(1,4,-2):

Let Q(x,y,z) be a point on line L.

1, 4, 2 3, 1,2

1 3 , 4 , 2 2

PQ tv

x y z t

x t y t z t

d) Line of intersection

Since plane M and N are intersect, vector v is parallel to the line of intersection.

P(1,4,-2)

L n

M

N nM

nN

v = nM x nN

nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 5

: 5 4 5,1,1

: 10 6 10,1, 1

M

N

M x y z n

N x y z n

Find v:

5 1 1 2 15 5

10 1 1

M N

i j k

v n n i j k

Find P(x,y,z) which lies on the line of intersection.

: 5 4 1

: 10 6 2

M x y z

N x y z

Assume 0 :z

5 4 3

10 6 4

2, 2

5

x y

x y

x y

Therefore, the point of intersection is 2

, 2,05

P

.

Equation of intersection line that passes through point 2

, 2,05

P

:

Let Q(x,y,z) be a point on the line.

2, 2, 2,15, 5

5

22 , 2 15 , 5

5

PQ tv

x y z t

x t y t z t

Question 7

a) Equation of plane

Find two intersecting vectors:

1, 1, 4 1,3,0AB AC

Find the normal vector, n:

1 1 4 12 4 2

1 3 0

i j k

n AB AC i j k

A C

B

n

nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 6

Let Q(x,y,z) be a point on the plane:

0

1, 2, 12,4,2 0

12 12 4 8 0

12 4 20

AQ n

x y z

x y z

x y z

b) Plane orthogonal to the line

Equation of line:

4 1

1 2

0 8

x

y t

z

Since line L is orthogonal to the plane, then we can say

that L is parallel to the normal vector, n of the plane.

1, 2,8n

Let Q(x,y,z) be a point on the plane:

0

3, 2, 1 1, 2,8 0

3 2 4 8 8 0

2 8 7

PQ n

x y z

x y z

x y z

c) Parallel planes

Since the planes are parallel, then 5, 3,2M Nn n .

Plane M passes through point P.

Let Q(x,y,z) be a point on plane M.

0

4, 1, 2 5, 3,2 0

5 20 3 3 2 4 0

5 3 2 27

MPQ n

x y z

x y z

x y z

d) Equation plane that contains 2 lines

Find the normal vector, n:

1 2 1 2 3 13 7 9

1 3 2

i j k

n v v i j k

L: x = 4+t, y = 1-2t, z = 8t

n

n

L1 L2

N: 3x - y + 2z = 10

nN

M

nM

P

P

nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 7

P(4,0,1) is the intersection point and let Q(x,y,z) be a point on the plane:

0

4, , 1 13,7,9 0

13 52 7 9 9 0

13 7 9 61

PQ n

x y z

x y z

x y z

e) Two planes orthogonal.

Find the normal vector of plane N

2 1 2 3 8 7

3 2 1

N M

i j k

n PQ n i j k

Let R(x,y,z) be a point on plane N

0

1, 2, 4 3, 8, 7 0

3 3 8 16 7 28 0

13 7 9 41

PR n

x y z

x y z

x y z

Question 8

a) Intersection point between line and plane

2 3 4

2 3 3 1 2 2 4

13 3

3

13

x y z

t t t

t

t

Therefore, the intersection point is 9 19 23

, ,13 13 13

P

n N

P(1,2,4)

Q(-1,3,2)

M : 3x + 2y - z = 4

N

n M = <3,2,1>

L

P

nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 8

b) Intersection between lines

1 1 1 1

2 2 2 2

: 1 2 , 3 , 5

: 6 , 2 4 , 3 7

L x t y t z t

L x t y t z t

Equate x and y to find t1 and t2

1 2 1 2

1 2 1 2

1 2

: 1 2 6 2 5 1

: 3 2 4 3 4 2 2

2, 1

x t t t t

y t t t t

t t

Therefore the point of intersection is P(5,6,10).

Question 9

a) Line through two points

2, 2,2

v PQ

Let R(x,y,z) be a point on the line

(i) P(1,2,-1)

1, 2, 1 2, 2,2

1 2 , 2 2 , 1 2

PR tv

x y z t

x t y t z t

Therefore the parametric for the line by using P(1,2,-1) is

1 2 , 2 2 , 1 2x t y t z t .

(ii) Q(-1,0,1)

1, , 1 2, 2,2

1 2 , 2 , 1 2

QR tv

x y z t

x t y t z t

Therefore the parametric for the line by using Q(-1,0,1) is

1 2 , 2 , 1 2x t y t z t .

b) Parallel lines

Since L1 is parallel to L2, then both lines are parallel to the same vector, v.

Let Q(x,y,z) be a point on line L1.

3, 2, 1 2, 1,3

3 2 , 2 , 1 3

PQ tv

x y z t

x t y t z t

Therefore, the parametric for is 3 2 , 2 , 1 3x t y t z t .

nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 9

c) Line perpendicular to the plane

Since line L is perpendicular to the plane, then

we can say that L is parallel to the normal vector, n of the plane.

3,7, 5n

Let Q(x,y,z) be a point on the line:

2, 4, 5 3,7, 5

3 3 , 4 7 , 5 5

PQ tn

x y z t

x t y t z t

Therefore, the parametric for is 2 3 , 4 7 , 5 5x t y t z t .

Question 10

a) Parallel planes

Since these two planes are parallel, then 1 2n n .

Let Q(x,y,z) be point on the plane.

0

1, 1, 3 3,1,1 0

3 3 1 3 0

3 1

PQ n

x y z

x y z

x y z

b) Line perpendicular to the plane

Since the line is perpendicular to the plane, then v n

Let P(x,y,z) be a point on the plane.

0

2, 4, 5 1,3,4 0

2 3 12 4 20 0

3 4 34

QP n

x y z

x y z

x y z

Question 11

a) Intersection point between lines

1

2

: 2 , 3 2, 4

: 2, 2 4, 4 1

L x t y t z t

L x s y s z s

L n

P(2,4,5)

n

2

n

1

P(1,-1,-3)

3x + y + z = 7

L: x = 5+t, y = 1+3t, z = 4t n

Q(2,4,5)

nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 10

Equate x and y to find t and s

: 2 1 2 2 1 1

: 3 2 2 4 3 2 2 2

0, 1

x t s t s

y t s t s

t s

Therefore the point of intersection is P(1,2,3).

Find the normal vector of the plane:

1 2 2 3 4 20 12

1 2 4

i j k

n v v i j k

Let Q(x,y,z) be a point on the plane:

0

1, 2, 3 20,12,1 0

20 20 12 24 3 0

20 12 7

PQ n

x y z

x y z

x y z

Question 12

a) Distance from a point to the line

From the line equation we can find Q and v.

: 2 , 1 2 , 2

0,1,0 2,2,2

L x t y t z t

Q v

Distance from P to L is

QP vD

v

2 2 2

2 0 1 2 6 4

2 2 2

2 6 4 2 14

i j k

QP v i j k

QP v

2 2 2

2 14 42

32 2 2

D

Q

P(2,1,-1)

v

nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 11

b) Distance from a point to the plane

Given the equation of plane is 2 2 4 0x y z , then the

normal vector is 2,1, 2n and P(2,2,3).

2 2 2

2 2 2

2 2 4

2 1 2

2 2 2 2 3 4

2 1 2

8

3

PQ nD

n

x y z

c) Distance between planes

Find a point on plane M. Let x = 0 and y = 0, then .3

5z

2 2 2

53

2 2 2

2 6 1

1 4 6

0 2 0 6 1

1 4 6

9/1.4056

41

x y zD

d) Angles between planes

1 2

1 2

2 2 2 2 2 2

1

cos

1,2,0 2,1, 2

1 2 0 2 1 2

4

3 5

4cos

3 5

53.4

n n

n n

n

P

Q

n

2

nN = <1,2,6>

P(1,-1,-3)

N: x + 2y + 6z = 1

M: x + 2y + 6z = 10 D

n2 = <2,1,-2>

n1 = <1,2,0>

nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 12

Question 13

Intersection plane parallel to the line

: 2 2 5 : 5 2 0

1,2, 2 5, 2, 1M N

M x y z N x y z

n n

Find vector, v (intersection line)

1 2 2 6 9 12

5 2 1

M N

i j k

v n n i j k

Given the line equation is:

: 3 2 , 3 , 1 4

2,3,4

L x t y t z t

u

Therefore,

6, 9, 12

3 2,3,4

3

v

u

Since 3v u , therefore the intersection line is parallel to line L.

Question 14

Line intersects to a plane.

: 3 2 , 2 , : 3 4L x t y t z t M x y z

Find t by substitute x,y and z to M

3 2 3 2 4

1

t t t

t

Therefore, the point of intersection is P(1,-2,-1).

Find v:

2 2 1 5 3 4

1 3 1

i j k

u v n i j k

Therefore, the line equation on the plane that passes through P(1,-2,-1) and let Q(x,y,z) be a

point on the line:

L

P

nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 13

1, 2, 1 5,3,4

1 5 , 2 3 , 1 4

PQ tu

x y z t

x t y t z t

Therefore, the parametric for is 1 5 , 2 3 , 1 4x t y t z t

Question 15

26 263, 2,4 ,0,

5 5AB CD

Find the normal vector of the plane

26 265 5

52 182 523 2 4

5 5 50

i j k

n AB CD i j k

Let Q(x,y,z) be a point on the plane:

0

52 182 522, , 3 , , 0

5 5 5

52 104 182 52 1560

5 5 5 5 5

2 7 2 10

PQ n

x y z

x y z

x y z

Question 16

Intersection line

: 2 1 : 2 8

1,2,1 1, 1,2M N

M x y z N x y z

n n

Find vector v which is parallel to the line of intersection.

1 2 1 5 3

1 1 2

M N

i j k

v n n i j k

nhaa/EQT101/TUTORIAL3/CONFIDENTIAL pg. 14

Find the point of intersection. Let z =0, then 17 7

,3 3

x y . Therefore, the point of

intersection is 17 7

, ,03 3

P

.

Let Q(x,y,z) be a point the line.

17 7, , 5, 1, 3

3 3

17 75 , , 3

3 3

PQ tv

x y z t

x t y t z t

Therefore, the parametric for is 17 7

5 , , 33 3

x t y t z t .

Question 17

2 3 1 7 3 5

1 1 2

i j k

n u v i j k

Let Q(x,y,z) be a point on the plane:

0

1, 2, 3 7, 3, 5 0

7 7 3 6 5 15 0

7 3 5 14

AQ n

x y z

x y z

x y z