Recently, I found an equation r(s,t)= r0 + sv + tw ,
which is a plane with a point and two vectors lying on it.
I don't know why this is an equation of a plane.
can anyone teach me how to interpret r(s,t)= r0 + sv + tw ?
and
is there a method to convert r(s,t)= (-3,4,-9) +s(9,-5,9) + t(3,-2,3) into the form (r-r0).n=0?
Thank you.
Do you know that there exist a unique plane given two intersecting lines? If you are given point r0 and vector v, then r0+ s[/b]v[/b] gives all points on the line through r0 in the direction of vector v. r0+ tv gives all points on the line through r0 in the direction of w. Those two lines, given by t= 0 and s= 0 respectively, determine the plane. Given specific values of s and v, "r0+ sv+ tw" means "starting at ro, go for a distance s in the direction of vector v, the go distance t in the direction of vector w" to get to the corresponding point on the plane.
In order that, for r a point in the plane, (r- r0).n= 0, n must be perpendicular (or normal) to the plane. And that is given by the cross product of two vectors in the plane.
In your example, (9, -5, 9) and (3, -2, 3) are vectors in the plane. Their cross product, n= (3, 0, -3), is perpendicular to the plane. For any point in the plane, (x, y, z), we must have ((x, y, z)- (-3, 4, -9)).(3, 0, -3)= 0 or (x+ 3)(3)+ (y+ 4)(0)+ (z+ 9)(-3)= 3x+ 9- 3z- 27= 3x- 3z- 18= 0.
Now, you were given that x= -3+ 9s+ 3t, y= 4- 5s- 2t, z= -9+ 9s+ 3t. Putting those into 3x- 3z- 18, we have 3(-3+ 9s+ 3t)+ 0(4- 5x- 2t)- 3(-9+ 9s+ 3t)- 18= (-9+ 27s+ 9t)+ (27- 27s- 9t)- 18= (-9+ 27- 18)+ (27x- 27s)+ (9t- 9t)= 0 so it does, in fact, satisfy the equation.
I have a different answer. Let $\displaystyle r = \left\langle {x,y,z} \right\rangle \;\& \,r_0 = \left\langle {x_0 ,y_0 ,z_0 } \right\rangle $Originally Posted by happymatthematics
Then the plane is $\displaystyle (r-r_0)\cdot(v\times w)=0$.
  |
2Thanks
LinkBack URL
About LinkBacks