Recover every valid parallel-plane distance parameter

Join plane and projection foundations

Plane classification checks proportional normals and constants. Perpendicular distance uses projection onto a unit normal. The inverse problem combines them while the normal itself depends on a parameter

An original changing-plane pair

Let n=(λ,3,4). The planes are Π1: n·r=2 and Π2: (λ−15/4)n·r=(λ−15/4)(2+5λ). Find every λ for which they are both planes separated by distance 3

What happens at λ=15/4?

Normalise the whole equation

For λ≠15/4, divide both sides of Π2 by λ−15/4. Now Π1 and Π2 have the same nonzero normal n and constants 2 and 2+5λ. A negative multiplier changes orientation, not the underlying plane

At λ=2, what is the normalised second constant minus the first?

Use the normal length

The common normal has magnitude √(λ²+25), not 5: its first component also changes. Thus the perpendicular separation is |5λ|/√(λ²+25). The absolute value retains either side of the first plane

Solve both algebraic branches

Set |5λ|/√(λ²+25)=3. Squaring gives 25λ²=9λ²+225, so λ²=225/16 and λ=±15/4. The denominator is positive and both sides are nonnegative, so squaring preserves both distance branches

Give λ² from the distance equation

In the candidates ±C√R with R=1, give C.

Which candidate survives the ORIGINAL plane equations?

How many valid parameter values remain if the candidates are λ=±15/4 and λ=15/4 makes the second equation 0=0?

Check the surviving value without losing signs

At λ=−15/4, |n|=25/4 and the normalised constant difference is −75/4. The separation is (75/4)/(25/4)=3. The second normal multiplier is −15/2, which is nonzero, so both original equations are planes

An irrational two-branch example

If n=(λ,3,4), the normalised difference is 4λ, the target distance is 2 and the excluded value is 0, then 12λ²=100. Thus λ=±(5/3)√3. Neither is zero, so both survive. Rationalise the coefficient while leaving a positive square-free radicand

For λ²=25/3, give R in λ=±C√R.

Establish and solve the full geometric problem, ungraded

For the original pair, state the excluded parameter, prove parallelism elsewhere, derive the distance using a point on Π1, solve both branches and test both in the original equations. Explain why a negative scale factor is allowed and why λ=15/4 is not a coincident-plane solution

Model: exclusions survive normalisation

n=(λ,3,4) is always nonzero; (λ−15/4)n is nonzero exactly when λ≠15/4. Elsewhere the normals are proportional and the normalised constants differ by 5λ. For any point P on Π1, n·P=2, so its distance to normalised Π2 is |n·P−(2+5λ)|/|n|=|5λ|/√(λ²+25). Solving the target 3 gives ±15/4. The positive value erases the second plane; only −15/4 remains. Multiplying an equation by a negative nonzero scalar preserves all its solutions. A zero multiplier erases the constraint, so it cannot establish coincident planes

Assessment boundary

The family grades exact candidates and the validity of each branch. Complete normalisation and geometric proofs remain ungraded