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Question:

Find the vector equation of the plane determined by the points A(3, -1, 2), B(5, 2, 4) and C(-1, -1, 6). Also find the distance of point P(6, 5, 9) from this plane.

Solution:

Let the equation of the plane be ax + by + cz = d.
The vectors AB and AC are given by:
AB = (5-3, 2-(-1), 4-2) = (2, 3, 2)
AC = (-1-3, -1-(-1), 6-2) = (-4, 0, 4)
The normal vector to the plane is given by the cross product of AB and AC:
N = AB x AC = (34 - 20, 2*(-4) - 24, 20 - 3*(-4)) = (12, -16, 12)
So the equation of the plane is of the form 12x - 16y + 12z = d.
Using point A(3, -1, 2):
12(3) - 16(-1) + 12(2) = d
36 + 16 + 24 = d
76 = d
Therefore, the equation of the plane is 12x - 16y + 12z = 76, which simplifies to 3x - 4y + 3z = 19.
The vector equation of the plane is given by:
→r . →n = →a . →n
where →r = xi + yj + zk, →n is the normal vector (3i - 4j + 3k), and →a is a point on the plane (e.g., A(3, -1, 2)).
→r . (3i - 4j + 3k) = (3i - j + 2k) . (3i - 4j + 3k)
→r . (3i - 4j + 3k) = 9 + 4 + 6 = 19
So the vector equation of the plane is (xi + yj + zk) . (3i - 4j + 3k) = 19, or 3x - 4y + 3z = 19.
Now, let's find the distance of point P(6, 5, 9) from the plane 3x - 4y + 3z - 19 = 0.
The distance formula is given by:
Distance = |Ax + By + Cz + D| / √(A² + B² + C²)
Distance = |3(6) - 4(5) + 3(9) - 19| / √(3² + (-4)² + 3²)
Distance = |18 - 20 + 27 - 19| / √(9 + 16 + 9)
Distance = |6| / √34
Distance = 6 / √34