Yefim S. answered • 05/28/20

Math Tutor with Experience

For point (-1 + 2t, 3 + 4t,1 + 5t) we know that it belong to plane for any t. For t = 0 we have point ( -1, 3, ,1) and t = 1 give us point (1, 7, 6). We have now 3 points on the plane: (-1,3,1), (1,7,6) and (2,0,-3), Then we create 2 vectors in plane: r_{1} = (- 3, 3, 4) and r_{2} = (- 1, 7, 9) _{ } This 2 vectors in plane and becose their cross product perpendicular to plane:r_{1}×r_{2} = det([i j k], [-3 3 4], [- 1 7 9]) = - i + 23j - 18k = (- 1, 23, - 18).We now can get equation of plane: - (x - 2) + 23(y - 0) - 18(z + 3) = 0 or - x + 23y - 18z - 52 = 0.

This is scalar equation of this plane: x - 23y + 18z + 52 = 0;

Let P(x,y,z) is arbitrary point on the plane; then we can take Then vector OP = r = (x, y, z), r_{0 }= (2, 0,- 3)

vector u = r_{1} = (- 3, 3, 4) and v = r_{2} = (- 1, 7, 9) and r = r_{0} + su + tv, su + tv = (- 3s - t, 3s + 7t, 4s + 9t); s and t any real numbers. This is vector equation of Plane: r = r_{0} + su + tv, where r_{0 }= (2, 0,- 3) and

su + tv = (- 3s - t, 3s + 7t, 4s + 9t).

Now parametric equation: x = 2 - 3s - t, y = 3s + 7t, z = - 3 + 4s + 9t, where s,t ∈ R