Finding an equation for a plane

  • Thread starter Thread starter pvpkillerx
  • Start date Start date
  • Tags Tags
    Plane
pvpkillerx
Messages
22
Reaction score
0
Find an equation for the plane that contains the line x = 1+t, y = 3t, z = 2t and is parallel to the line of intersection of the planes -x+2y+z = 0 and x + z + 1 = 0.

On the attachment, the answer is there. My question is, (on the first page) right after he wrote, x,y, leading, z free, he wrote down the matrix form of the equation. Where did he get the values of the matrix right after z. The values are -1, -1 and 1. How did he find them, or where did he get them? Please explain, thanks!
 

Attachments

Physics news on Phys.org
If you write out the equations that step represents they are:

x + z = -1
y + z = -1/2 - z
z = z

Solve for (x,y,z) written as a column vector.
 
Wait, how exactly do you get that though? I see how you get x + z = -1, but how do you get y + z = -1/2 -z? I also understand z = z. Can you explain the second line? thanks.
 
LCKurtz said:
If you write out the equations that step represents they are:

x + z = -1
y + z = -1/2 - z
z = z

Solve for (x,y,z) written as a column vector.

pvpkillerx said:
Wait, how exactly do you get that though? I see how you get x + z = -1, but how do you get y + z = -1/2 -z? I also understand z = z. Can you explain the second line? thanks.

Isn't it obvious to you that it is just a typo? Did you try fixing it to see if it answered your question?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top