Few Geometry tasks.

[xRazor]

Hello, there's an exam coming soon and i need some help with few of my pre exam preparation tasks.

First of all i would like to ask if this solution is correct.

So the task number one.

"Write an equation for a plane that goes through a point A(1, -1, 9) and is parallel to plane x = z."

Heres my solution:
1. The normal vector of the plane is n = (1,0,-1) because if i transform the equation of the 2nd plane i get x - z = 0 or 1x - 0y -1z = 0
2. Knowing that the equation of a plane is n < (x-x0,y-y0,z-z0) = 0 i wrote an equation like this: (1,0,-1) * ( x-1, y+1, z-9) = 0
so
1(x-1)-0(y+1)-(z-9) = 0
x-1-0+0-z+9=0
x-z = -8

Is this correct?

Task number 2.

Find an equation of a plane that is parallel to a plane 3x - 4y -9z +3 = 0. The distance between those planes is 10.

Task number 3.
Write parametric equations of y = 5x + 8
I know that the form of parametric equation is a system of equations:
x=x0 + axt
y=y0 + ayt

But i realy have no idea how to turn the fgiven equation into parametric equation...

Would be great if anyone could help me out with these.
Thank you.
Maris L.

 

[stapel]

Task number 3.
Write parametric equations of y = 5x + 8
I know that the form of parametric equation is a system of equations:
x=x0 + axt
y=y0 + ayt

It doesn't have to be so complicated, in this particular case. You have y(x) = 5x + 8. So let x = t. Then y(x) = y(t) =...

Yes, it's as simple as that!

 

[xRazor]

Oh, sorry. I translated the task wrong. I had to write parametric equations of a line y=5x+8.
Anyways, i think i solved that one. What i did there was

°1. i found a point on the line. x = 1, y = 5+8, so the point on the line is A(1,13).
2. Found the direction vector (as much as i understood, parametrs at y and x represents the direction vector) so it must be n = (5,1),
3. Started to make the equation. So the formula is (x-x1)/a = (y-y1)/b = t where x1,y1 are points on the line and a,b are the coordinates of direction vector?
So i ended up with something like 1(y-13) = 5(x-1) = t
so 1(y-13) = t and 5(x-1) = t
Which means the parametric equation of line y = 5x+8 is
x = 1+t
y = 13+t

Is this correct?

Also, question about the 2nd task.
To write the equation of a plane that is parallel to plane 3x-4y-9z+3 = 0 and distance between those planes is 10, i used the distance formula (it is written in my 2nd task description in the main post). So parameters A,B,C are going to be the same, because the planes are parallel. To write the equation of that plane i need to find D2.

I am stuck at this part:
|D2-3| = 10*sqrt(106)

How do i sepparate the module in the left side? So that i can find the value of D2 and make the equation? Also, that square root makes me think im doing something wrong...
Thanks again.

 

[xRazor]

Sorry.

Sorry for disturbing you guys. Solved those tasks. This topic can be locked.