Question

Can you please see my attachment and help me please?

Answer 1

I don't see an attachment.

Answer 2

the badminton quesion please

Answer 3

ok i will sendit again

Answer 4

which sub question are you stuck with?

Answer 5

badminton one please

Answer 6

a, b c or d? or all of them?

Answer 7

all please

Answer 8

are you familiar with vectors?

Answer 9

kind of

Answer 10

but dont know how to start

Answer 11

(a) deals with coordinate systems. I'll take Y as an example. since Y is above O by a unit of 3, but not left or in front of O, so Y is only on k-axis. You follow?

Answer 12

yes

Answer 13

so, since it's 0 units on i, 0 units on j and 3 units on k, then the position vector of Y is \((3k)\)

Answer 14

or IOW, \(Y=0i + 0j +3k\)

Answer 15

yes

Answer 16

what about position vector A is it 607056i+1.55k +0j

Answer 17

ad for Y POSITION VECTOR is i 607056 and not 0

Answer 18

note that A is 6.096/2 in front of O on the j axis.

Answer 19

So don't consider it

Answer 20

yes, you'll need to consider it. It's on the coordinate system. if a point is 4 units on top of O and 3 units right on the x-y axes, then the point is (3,4). same with the x-y-z system.

Answer 21

What about the other parts can you help me please

Answer 22

for the next question, are you familiar with parametric or symmetric equations of vectors?

Answer 23

for the parametric eqs, \(Y=<a,b,c>\) and \(A=<d,e,f>\)
the vector equation will be \(YA=<a-d,b-e,c-f>\)
and any point on YA will be characterised by:
\(X=P_x-(a-d)t\), \(Y=P_y-(b-e)t\), \(Z=P_z-(c-f)t\)
where P is one point on the vector YA. do you follow?

Answer 24

and t is a parameter

Answer 25

Try. To

Answer 26

hmm...how much of vectors are you really familiar with?

Answer 27

I'll try to adjust the tips

Answer 28

thank you

Answer 29

are you familiar with y=mx +c? the vector line is quite similar to in terms of vectors, it will be r= tv +a, where t is a parameter(like m), a is like c, and v is like x

Answer 30

yes i know this

Answer 31

the vector line eq is r=tv+a. r and v are vectors.
v is the vector from the origin.
t is a parameter denoting that r is a vector parallel to v. (kinda like the gradient)
a is another parameter denoting how much r is from O (kinda like c)
you follow?

Answer 32

like y=mx +c, every linear line is simply y=mx after translation, right?

Answer 33

yes follow

Answer 34

The line of YA will be \(\vec {YA}=\vec Y-\vec A\)

follow?

Answer 35

yes

Answer 36

Any point of this line will have it's x, y and z components.
the vector components of these points is parallel to the vector components of YA, right?

Answer 37

so, let any point on the line be P.
the x-component of P is \(P_x\)
since it is parallel to \({YA}_x\), and taking either Y or A as a reference point, the point will be \(P_x=t({YA}_x)+Y_x\)
taking Y or A simply changes the parameter.
you follow?

Answer 38

yes

Answer 39

it's the same for the y and z components.
What did you get ?

Answer 40

PY=T(YAy)+y

Answer 41

PZ=T(OA)+YZ

Answer 42

for PZ why is it OA instead? :)
But I think you've got it. Do the numbers match?

Answer 43

WHAT DO YOU MEAN BY DO THE NUMBERS MATCH

Answer 44

well, since Y =<0,0,3> and A=<6.7056,3.048,1.55>,
do your answers match the answers given by your teacher/etc?

Answer 45

yes

Answer 46

but the 3.048 dont match i think it should be 3

Answer 47

do you agree

Answer 48

hmm,how did you get three? i got that from dividing the breadth of the court by two since A is the midpoint.

Answer 49

Yes recognise it sorry

Answer 50

so i guess a and b is done?

Answer 51

Thank

Answer 52

to find X, just let k=0.
c deals with the distance formula. Are you familiar with the distance formula of x-y axes?

Answer 53

No

Answer 54

for two points \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) on the x-y coord system, the distance between them is \(d=\sqrt {(x_2-x_1 )^2+(y_2-y_1 )^2}\)

in the x-y-z coord system, the distance formula is \(d=\sqrt {(x_2-x_1 )^2+(y_2-y_1 )^2+(z_2-z_1 )^2}\)

Answer 55

thank you

Answer 56

you're welcome :)