Find the equation of the plane that contains the points, A(1,0,0), 5(0,1,0) and

Question

Find the equation of the plane that contains the points, A(1,0,0), 5(0,1,0) and C(0,0,1). Does this plane contain the points (-1, 1, 1) and (1/3, 2/3, 0)? What about the point (1,1,-2) ? Find the distance from the origin to the plane 3x + 2y + z = 3. Can you compute the intersection of the line x = y/2 = z/3 with the plane 3x + 2y + z = 3? If yes, do so. Consider the two parallel planes P1, given by 2x - Zy + 4z = 1 and P2 given by x - (3/2)y + 2z = 5. Can you explain why the line x-x0/3 =y-y0/2 =z-z0/6 will intersect both planes no matter what (x0,y0,z0) are? Consider the planar curve below, Show that x2(t) + y2(t) = 1. What is the curvature of this curve? Can you compute the length?

Explanation / Answer

2)

The plane equation is:

x+y+z = 1

Since:

-1+1+1 = 1 and 1/3 + 2/3 + 0 = 1

the two points (-1,1,1) and (1/3,2/3,0) are both on the plane.

(1,1,-2) is not on plane because 1+1-2 = 0

3)

for the line we have: y=2x , z=3x , thus:

3x+2y+z = 3x+4x+3x = 10x = 3 -> x = 0.3

So the intersection is: (0.3 , 0.6 , 0.9)

4)

The normal vector of both planes is (2,-3,4). The direction vector of the line is (3,2,6), since (3,2,6).(2,-3,4) = 24 != 0 -> the line is not perpendicular to the normal vector of the planes -> it will intersect both planes.

5)

((1-t^2)/(1+t^2))^2 + (2t/(1+t^2))^2 =

(1+t^4-2t^2+4t^2)/(1+t^2)^2 = (1+2t^2+t^4)/(1+t^2)^2 =

(1+t^2)^2/(1+t^2)^2 = 1

This is a circle of radius R = 1, so the curvature is: 1/R = 1/1 = 1

and the length is:

2pi.R = 2pi

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