Find the equation for the plane through the points Upper P 0 (4 comma 5 comma negative 3 )P0(4,5,−3), Upper Q 0 (1 comma 4 comma 5 )Q0(1,4,5), and Upper R 0 (0 comma 2 comma 1 )R0(0,2,1).

  4x -4y +z = -7

Step-by-step explanation:

The direction vector will be perpendicular to any/all vectors between the given points. It can be found, for example, using the cross product:

  d = (P0Q0) × (P0R0) = (-3, -1, 8) × (-4, -3, 4)

The cross product can be computed by hand or by any of a number of calculators or web tools. Here it will be ...

  d = (20, -20, 5)

These can be the coefficients of x, y, and z in the plane equation. However, the equation is better written using coefficients that are mutually prime, so we choose the direction vector to be ...

  d = (4, -4, 1)

The constant C in the plane equation

  d·(x, y, z) = C

will be the dot product of this direction vector (d) any any of the given points. Using R0, we find the constant to be ...

 C = (4, -4, 1)·(0, 2, 1) = 4·0 +(-4)·2 +1·1 = -8+1 = -7

so, the plane equation is ...

  4x -4y +z = -7

no association

step-by-step explanation:

if you have eyes and you can use them efficiently you can see that it just looks like random dots like do a line of best fit and like they're so spread out it's such a weak association you might as well consider it nonexistent