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== 1.2#h(0.6em)Three Planes: Point, Line, or Nothing


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Objectives]


      - Classify the solution set of three equations in three unknowns as a point, a line, a plane, or empty.
- Recognize inconsistency geometrically, as three planes with no common point.
- Connect the special role of dependent equations to the collapse from a point to a line of solutions.
]


    Add a third equation and the story sharpens. Three planes in space #emph[usually] pin down a single point: the first two intersect in a line, and the third plane crosses that line in one spot. But "usually" is doing real work in that sentence. This section is about the exceptional cases — and about seeing exactly how special they are.


    The figure below shows the planes #math.equation(block: false, alt: "z equals x plus 1")[$z = x + 1$] and #math.equation(block: false, alt: "z equals y minus 1")[$z = y − 1$], whose intersection is the dashed amber line, together with a third plane #math.equation(block: false, alt: "z equals 0.5 x plus 0.5 y plus d")[$z = 0.5 x + 0.5 y + d$] controlled by the slider #math.equation(block: false, alt: "d")[$d$]. A fourth plane, a tilted replacement for the third, is included but hidden.


    
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      Two fixed planes meet along the dashed amber line. The slider #math.equation(block: false, alt: "d")[$d$] raises and lowers the third plane; only at #math.equation(block: false, alt: "d equals 0")[$d = 0$] does it capture the line. Toggle the hidden tilted variant to see the generic one-point case.
    
    
=== Explore the figure


    + Start at #math.equation(block: false, alt: "d equals 0.9")[$d = 0.9$]. Orbit and check each #emph[pair] of planes: every pair crosses in a line. Now look for a single point on all three planes. Can you find one?
+ Drag #math.equation(block: false, alt: "d")[$d$] toward #math.equation(block: false, alt: "0")[$0$] and watch the third plane sweep down onto the amber line. At exactly #math.equation(block: false, alt: "d equals 0")[$d = 0$], what is the solution set of the system — a point, a line, or nothing?
+ Push #math.equation(block: false, alt: "d")[$d$] below zero. The moment #math.equation(block: false, alt: "d")[$d$] leaves #math.equation(block: false, alt: "0")[$0$], where does the third plane sit relative to the amber line?
+ Hide the third plane and reveal the hidden tilted variant #math.equation(block: false, alt: "z equals minus x minus y plus d")[$z = − x − y + d$]. Now drag #math.equation(block: false, alt: "d")[$d$] again. Does the system ever lose its solution? What shape is the solution set at each #math.equation(block: false, alt: "d")[$d$]?


    
=== Why the third plane is special


    Write the three equations with the unknowns on one side: #math.equation(block: false, alt: "x minus z equals minus 1")[$x − z = − 1$], then #math.equation(block: false, alt: "y minus z equals 1")[$y − z = 1$], then #math.equation(block: false, alt: "0.5 x plus 0.5 y minus z equals minus d")[$0.5 x + 0.5 y − z = − d$]. Take half of the first equation plus half of the second: the left side is #math.equation(block: false, alt: "0.5 x plus 0.5 y minus z")[$0.5 x + 0.5 y − z$] — #emph[exactly] the left side of the third equation — while the right side is #math.equation(block: false, alt: "0.5 open parenthesis minus 1 close parenthesis plus 0.5 open parenthesis 1 close parenthesis equals 0")[$0.5 ( − 1 ) + 0.5 ( 1 ) = 0$]. So the first two equations already force #math.equation(block: false, alt: "0.5 x plus 0.5 y minus z equals 0")[$0.5 x + 0.5 y − z = 0$] at every one of their common points. The third equation demands that this same quantity equal #math.equation(block: false, alt: "minus d")[$− d$]. If #math.equation(block: false, alt: "d not equal to 0")[$d ≠ 0$] the demands clash and no point can satisfy all three: the system is #strong[inconsistent], even though every pair of planes meets happily. If #math.equation(block: false, alt: "d equals 0")[$d = 0$] the third equation is redundant — it adds no new information — and the solution set stays the entire amber line #math.equation(block: false, alt: "open parenthesis t minus 1 , t plus 1 , t close parenthesis")[$( t − 1 , #h(0.278em) t + 1 , #h(0.278em) t )$].


    The tilted variant behaves differently because its left side, #math.equation(block: false, alt: "minus x minus y minus z")[$− x − y − z$] shifted to standard form as #math.equation(block: false, alt: "x plus y plus z equals d")[$x + y + z = d$], is #emph[not] a combination of the first two left sides. Three independent conditions in three unknowns cut the solution down to a single point, and sliding #math.equation(block: false, alt: "d")[$d$] merely moves that point.


    #examplebox("Example 1")[The unique solution, when the third plane is independent][
With the tilted plane #math.equation(block: false, alt: "z equals minus x minus y plus d")[$z = − x − y + d$] replacing the third equation, find the single solution of the system in terms of #math.equation(block: false, alt: "d")[$d$].


      From the first two planes, #math.equation(block: false, alt: "x equals z minus 1")[$x = z − 1$] and #math.equation(block: false, alt: "y equals z plus 1")[$y = z + 1$]. Substitute into #math.equation(block: false, alt: "z equals minus x minus y plus d")[$z = − x − y + d$]: we get #math.equation(block: false, alt: "z equals minus open parenthesis z minus 1 close parenthesis minus open parenthesis z plus 1 close parenthesis plus d equals minus 2 z plus d")[$z = − ( z − 1 ) − ( z + 1 ) + d = − 2 z + d$], so #math.equation(block: false, alt: "3 z equals d")[$3 z = d$] and #math.equation(block: false, alt: "z equals d / 3")[$z = d / 3$]. Then #math.equation(block: false, alt: "x equals d / 3 minus 1")[$x = d / 3 − 1$] and #math.equation(block: false, alt: "y equals d / 3 plus 1")[$y = d / 3 + 1$]. The solution is the single point #math.equation(block: false, alt: "open parenthesis d / 3 minus 1 , d / 3 plus 1 , d / 3 close parenthesis")[$( d / 3 − 1 , #h(0.278em) d / 3 + 1 , #h(0.278em) d / 3 )$] — check that at #math.equation(block: false, alt: "d equals 0.9")[$d = 0.9$] it is #math.equation(block: false, alt: "open parenthesis minus 0.7 , 1.3 , 0.3 close parenthesis")[$( − 0.7 , #h(0.278em) 1.3 , #h(0.278em) 0.3 )$], which you can spot in the figure sitting on the amber line.
]


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The dictionary you should carry forward: #emph[unique solution] means the planes are independent and meet in a point; #emph[infinitely many solutions] means at least one equation was redundant; #emph[no solution] means the equations contradict — the left sides combine but the right sides refuse to.
]