show that the locus of a point whose distances from two given planes are in a constant ratio, is a plane.
To show that the locus of a point whose distances from two given planes are in a constant ratio is a plane, we can proceed as follows:
Let the two given planes be P₁ and P₂. Let a point in the locus be (x, y, z).
The distance from the point to plane P₁ is given by the formula:
d₁ = |Ax + By + Cz + D₁| / sqrt(A² + B² + C²)
Similarly, the distance from the point to plane P₂ is given by the formula:
d₂ = |Ax + By + Cz + D₂| / sqrt(A² + B² + C²)
Now, let the constant ratio of the distances be k. We have:
d₁/d₂ = k
(|Ax + By + Cz + D₁| / sqrt(A² + B² + C²)) / (|Ax + By + Cz + D₂| / sqrt(A² + B² + C²)) = k
Simplifying the above equation, we get:
(|Ax + By + Cz + D₁| / |Ax + By + Cz + D₂|) = k
Now, let's consider the two cases where the ratio is positive and where the ratio is negative:
Case 1: Positive ratio (k > 0):
In this case, the equation simplifies to:
(Ax + By + Cz + D₁) = k(Ax + By + Cz + D₂)
Expanding and simplifying the equation, we get:
(A - kA)x + (B - kB)y + (C - kC)z + (D₁ - kD₂) = 0
The above equation represents a plane, which means that for a positive ratio, the locus is a plane.
Case 2: Negative ratio (k < 0):
In this case, the equation simplifies to:
(Ax + By + Cz + D₁) = -k(Ax + By + Cz + D₂)
Expanding and simplifying the equation, we get:
(A + kA)x + (B + kB)y + (C + kC)z + (D₁ + kD₂) = 0
The above equation also represents a plane, which means that for a negative ratio, the locus is a plane.
Since the locus is a plane for both positive and negative ratios, we can conclude that the locus of a point whose distances from two given planes are in a constant ratio is a plane.
To show that the locus of a point whose distances from two given planes are in a constant ratio is a plane, we need to understand what the locus represents and how to derive its equation.
Let's start by determining the equation of the given planes. Suppose we have two planes with equations:
Plane 1: Ax + By + Cz + D1 = 0
Plane 2: Ax + By + Cz + D2 = 0
Now, let's consider a point P(x, y, z) that lies on the locus we are interested in. The distance between point P and Plane 1 can be found by using the formula for the distance between a point and a plane:
Distance1 = |Ax + By + Cz + D1| / sqrt(A^2 + B^2 + C^2)
Similarly, the distance between point P and Plane 2 is:
Distance2 = |Ax + By + Cz + D2| / sqrt(A^2 + B^2 + C^2)
We are given that the ratio of Distance1 to Distance2 is constant. Let's say this ratio is k:
Distance1 / Distance2 = k
Substituting the expressions for Distance1 and Distance2, we have:
(|Ax + By + Cz + D1| / sqrt(A^2 + B^2 + C^2)) / (|Ax + By + Cz + D2| / sqrt(A^2 + B^2 + C^2)) = k
Simplifying, we get:
(|Ax + By + Cz + D1|) / (|Ax + By + Cz + D2|) = k
Now, take the cases when Ax + By + Cz + D1 is positive and negative:
Case 1: Ax + By + Cz + D1 >= 0
In this case, the equation becomes:
(Ax + By + Cz + D1) / (Ax + By + Cz + D2) = k
Case 2: Ax + By + Cz + D1 < 0
In this case, the equation becomes:
-(Ax + By + Cz + D1) / -(Ax + By + Cz + D2) = k
Note that the negatives cancel out, and we end up with the same equation as in Case 1. Therefore, we can simplify the equation to:
(Ax + By + Cz + D1) / (Ax + By + Cz + D2) = k
Now, let's further simplify the equation:
Ax + By + Cz + D1 = k(Ax + By + Cz + D2)
Distribute k:
Ax + By + Cz + D1 = kAx + kBy + kCz + kD2
Rearrange terms:
(A - kA)x + (B - kB)y + (C - kC)z + (D1 - kD2) = 0
This equation represents a plane. Therefore, we have shown that the locus of a point whose distances from two given planes are in a constant ratio is indeed a plane.