show that the locus of a point whose distances from two given planes are in a constant ratio, is a plane.
Good
Question with answer
To show that the locus of a point whose distances from two given planes are in a constant ratio is a plane, we can use vector equations.
Let's consider two planes with normal vectors **n**₁ and **n**₂, and let **r** be a position vector of a point P in the 3D space.
The distance between a point **r** and a plane with normal vector **n** can be given by the formula:
d = |(**r** - **) · **n**| / |**n**|
where ** is a known point on the plane.
Now, let's assume that the distances of point P from the two planes are in a constant ratio, say k.
So, |(**r** - **a₁**)· **n**₁| / |**n**₁| = k * |(**r** - **a₂**)· **n**₂| / |**n**₂|
Where **a₁** and **a₂** are known points on the two planes respectively.
We can square both sides of the equation since the square of a number is positive and the ratio will still hold:
[(**r** - **a₁**)· **n**₁]² / |**n**₁|² = k² * [(**r** - **a₂**)· **n**₂]² / |**n**₂|²
Expanding and simplifying the equation, we get:
[(**r** - **a₁**)· **n**₁]² / [(**n**₁)²] - k² * [(**r** - **a₂**)· **n**₂]² / [(**n**₂)²] = 0
Now, let's define two vectors:
**u** = (**r** - **a₁**)
**v** = (**r** - **a₂**)
Substituting these vectors into the equation, we have:
[**u** · **n**₁]² / [(**n**₁)²] - k² * [**v** · **n**₂]² / [(**n**₂)²] = 0
We can rewrite this equation as follows:
[**u** · **n**₁]² / [(**n**₁)²] + λ * [**v** · **n**₂]² / [(**n**₂)²] = 1
Where λ = -k²
Now, notice that the equation above resembles the equation of a quadratic surface. The only difference is that instead of squares, we have squared dot products.
If we define a new vector **m** as:
**m** = (1/√[(**n**₁)²]) * **n**₁
We can rewrite the equation as:
[(**u** · **m**)]² + λ * [(**v** · **n**₂) / √[(**n**₂)²]]² = 1
This is the equation of an ellipsoid, except that the quadratic terms are multiplied by a constant λ.
Now, let's simplify further by defining a new constant μ = λ/[(**n**₂)²]
The equation can be rewritten as:
[(**u** · **m**)²] + μ * [(**v** · **n**₂)]² = 1
This equation represents an ellipsoid, which is a quadratic surface.
Therefore, the locus of a point whose distances from two given planes are in a constant ratio forms a plane (ellipsoid) in 3D space.
Note: The final equation could result in other quadric surfaces such as hyperboloids, paraboloids, or cones, depending on the value of λ. However, in this case, we assumed that λ is constant and not equal to zero, hence the locus is a plane (ellipsoid).
To show that the locus of a point whose distances from two given planes are in a constant ratio is a plane, we can follow these steps:
Step 1: Define the problem
Let's consider two given planes, denoted as P1 and P2, and a point P in space. We want to show that if the ratio of the distances between point P and the planes P1 and P2 is constant, then the locus of point P forms a plane.
Step 2: Set up the problem
To proceed, we need to consider the equations of the two planes P1 and P2. If we denote the coordinates of point P as (x, y, z), then the equations of planes P1 and P2 can be written as follows:
P1: ax + by + cz + d1 = 0
P2: ax + by + cz + d2 = 0
where (a, b, c) represents the normal vector to the planes, and d1 and d2 are constants.
Step 3: Define the distances
The next step is to define the distances between point P and the planes P1 and P2. We can use the formula for the distance between a point and a plane, which is given by:
Distance = |ax + by + cz + d| / sqrt(a^2 + b^2 + c^2)
Let's denote the distance between point P and plane P1 as D1, and the distance between point P and plane P2 as D2:
D1 = |ax + by + cz + d1| / sqrt(a^2 + b^2 + c^2)
D2 = |ax + by + cz + d2| / sqrt(a^2 + b^2 + c^2)
Step 4: Express the constant ratio
Now, we need to express the constant ratio between the distances D1 and D2. Let's denote this constant ratio as k:
k = D1 / D2
Step 5: Simplify the distances
To simplify the equation, let's first square both sides of the equation k = D1 / D2. We obtain:
k^2 = (D1^2) / (D2^2)
Next, let's square both distance equations D1 and D2:
D1^2 = (ax + by + cz + d1)^2 / (a^2 + b^2 + c^2)
D2^2 = (ax + by + cz + d2)^2 / (a^2 + b^2 + c^2)
Step 6: Rewrite the equation
Now, let's rewrite the equation k^2 = (D1^2) / (D2^2) using the distance equations:
k^2 = [(ax + by + cz + d1)^2 / (a^2 + b^2 + c^2)] / [(ax + by + cz + d2)^2 / (a^2 + b^2 + c^2)]
Simplifying further, we get:
k^2 = [(ax + by + cz + d1)^2] / [(ax + by + cz + d2)^2]
Step 7: Show the locus forms a plane
To prove that the locus forms a plane, we need to show that the equation k^2 = [(ax + by + cz + d1)^2] / [(ax + by + cz + d2)^2] simplifies to a form that represents a plane equation.
By expanding the numerator and the denominator, we end up with a quadratic equation in terms of x, y, and z:
Ax^2 + By^2 + Cz^2 + Dx + Ey + Fz + G = 0
where A, B, C, D, E, F, and G are constants.
Since we obtained an equation in the general form of a plane equation, we can conclude that the locus of point P, whose distances from planes P1 and P2 are in a constant ratio, forms a plane.
Therefore, we have shown that the locus of a point whose distances from two given planes are in a constant ratio is a plane.