• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

3D Vectors: Parametric equation 🥲 (1 Viewer)

Hello_World2

Member
Joined
May 3, 2022
Messages
38
Gender
Male
HSC
2023
Screen Shot 2023-03-06 at 9.23.01 pm.png

Can anyone help, I am just confused about how to start on these two thanks!
 

cossine

Well-Known Member
Joined
Jul 24, 2020
Messages
628
Gender
Male
HSC
2017
9.

AP = (x-1)^2 + (y+2)^2 + (z-3)^2

PB = (x+3)^2 + (y-1)^2 + (z-2)^2

AP/PB = 1/sqrt(2)

I will leave you to finish it of.

10.

z^2 = x^2 + y^2 (note z>=0)

=> 2x^2 + 2y^2 = 1

=> x^2 + y^2 = 1/2 coordinates are such that (x, y) are on the point of the circumference of the circle with radius 1/sqrt(2)

If we know that x^2 +y^2 = 1/2 then this means z = sqrt(1/2) = 1 / sqrt(2)

I will leave you to finish it off.


Minor notes:

9) I just gave it a go.

10) Generally for parametric questions your making some variable the subject or performing substitution.
 

Hello_World2

Member
Joined
May 3, 2022
Messages
38
Gender
Male
HSC
2023
Hmmm for question 9 I still don't understand, how can I solve for x when there is a √2 ?
 

cossine

Well-Known Member
Joined
Jul 24, 2020
Messages
628
Gender
Male
HSC
2017
Hmmm for question 9 I still don't understand, how can I solve for x when there is a √2 ?
You don't solve for x. x is a variable or any other variable. You need to perform simplification.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top