Find the equation of the plane for which each point is equidistant from P(2,1,1) and Q (-4,7,9).
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Distance between two points in space,
D = √((x1-x0)²+(y1-y0)²+(z1-z0)²)
Equate the distance between any point in the required plane, (x,y,z) and P(xp,yp,zp) to that between (x,y,z) and Q(xq,yq,zq) to get:
(x-xp)²+(y-yp)²+(z-zp)² = (x-xq)²+(y-yq)²+(z-zq)²
2x(xp-xq)+2y(yp-yq)+2z(zp-zq) - (xp²-xq²+yp²-yq²+zp²-zq²) = 0
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To find the equation of the plane equidistant from two points, P(2, 1, 1) and Q(-4, 7, 9), we can follow these steps:
Step 1: Find the midpoint of the line segment connecting P and Q.
The midpoint, M, can be found as:
M = ((x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2)
M = ((2 + (-4))/2, (1 + 7)/2, (1 + 9)/2)
M = (-2/2, 8/2, 10/2)
M = (-1, 4, 5)
Step 2: Find the direction vector of the line segment connecting P and Q.
The direction vector, d, can be found as:
d = (x₂ - x₁, y₂ - y₁, z₂ - z₁)
d = (-4 - 2, 7 - 1, 9 - 1)
d = (-6, 6, 8)
Step 3: Write the equation of the plane using the midpoint and direction vector.
Since the plane is equidistant from P and Q, any point (x, y, z) on the plane will have the same distance to P and Q as the midpoint M.
The equation of the plane can be written as:
(x - x₁)/d₁ = (y - y₁)/d₂ = (z - z₁)/d₃
Substituting the values:
(x + 1)/-6 = (y - 4)/6 = (z - 5)/8
Alternatively, we can multiply both sides of the equation by -6, 6, and 8, respectively, to remove the fractions:
(x + 1) = -6(y - 4) = 8(z - 5)
Simplifying, we get:
x + 1 = -6y + 24 = 8z - 40
Rearranging the equation to the general form, Ax + By + Cz + D = 0:
x + 6y - 8z + 15 = 0
Therefore, the equation of the plane for which each point is equidistant from P(2, 1, 1) and Q(-4, 7, 9) is x + 6y - 8z + 15 = 0.