Equation a plane through 3 points.

  • Thread starter Thread starter ScullyX51
  • Start date Start date
  • Tags Tags
    Plane Points
ScullyX51
Messages
35
Reaction score
0

Homework Statement


Find an equation of the plane through the points: A=(1,2,3), B=(0,1,3) and C=(2,14)


Homework Equations


r= r[0]+tv
vector equation of a plane: a(x-x[/0)+ b(y-y[/0]) + c(z-z[/0])=0


The Attempt at a Solution


Since I am given three points and no vector, I am a bit confused on what to do. I tried to compute the normal vector, and got:
(1,2,3)-(0,1,3)
n=<1,1,3>

Any corrections/suggestions on what to do next?
 
Physics news on Phys.org
(1,2,3)-(0,1,3) is (1,1,0) not (1,1,3). And that's a tangent vector, not a normal vector. Hint: you can find a normal vector if you can find two different tangent vectors and take the cross product.
 
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