#set document(title: "1.1 Points and Octants", author: "XYZ Homework")
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== 1.1#h(0.6em)Points and Octants


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


      - Locate points in space by their coordinate sign patterns and use orbiting to resolve depth ambiguity.
- Decide whether a point lies inside, on, or outside a sphere by prediction first and the distance formula second.
- Translate between the parametric description of a sphere and its standard equation.
]


    A point in space is an ordered triple #math.equation(block: false, alt: "open parenthesis x , y , z close parenthesis")[$( x , y , z )$]: walk #math.equation(block: false, alt: "x")[$x$] units along the first axis, #math.equation(block: false, alt: "y")[$y$] along the second, then climb #math.equation(block: false, alt: "z")[$z$]. The three coordinate planes slice space into eight rooms called #strong[octants], and the fastest way to name a point's room is its #emph[sign pattern] — the octant where all three coordinates are positive is the pattern #math.equation(block: false, alt: "open parenthesis plus , plus , plus close parenthesis")[$( + , + , + )$], and there are seven others.


    Here is the uncomfortable truth about pictures of space: a printed diagram shows you two of those three coordinates honestly and asks you to take the third on faith. Two different points can occupy the same spot on the page. The interactive figure below turns that defect into the lesson.


    
      {"camera":{"fov":50,"position":\[9,7,6\],"projection":"perspective","target":\[0,-1,2\]},"grid":{"axisColors":{"x":"\#ef4444","y":"\#22c55e","z":"\#3b82f6"},"divisions":10,"gridColor":"\#e5e7eb","showAxes":true,"showLabels":true,"size":10,"visible":true},"id":"3a4ae502-2505-46f8-aa51-22065217179d","metadata":{"alt\_text":"Four labeled colored points scattered among the octants of a 3D coordinate system: red A at (2,-3,4), blue B at (-2,-3,4), green C at (2,3,-4), and amber D at (2,-1,2). A translucent sphere of radius 3 is centered on point A; point D sits inside it, while B and C sit outside.","created\_at":"2026-07-07T00:00:00.000Z","description":"Workbook scene: four points A(2,-3,4), B(-2,-3,4), C(2,3,-4), D(2,-1,2) plus a translucent parametric sphere of radius 3 centered at A, for octant identification and distance-formula predictions.","tags":\["workbook","calc3-vectors-lines-quadrics","points","octants","sphere","distance"\],"updated\_at":"2026-07-07T00:00:00.000Z"},"objects":\[{"color":"\#ef4444","id":"2c81c760-9c86-4f0a-ae8c-e471aa2e377b","kind":"point","label":{"text":"A (2, -3, 4)","visible":true},"radius":0.12,"visible":true,"x":2,"y":-3,"z":4},{"color":"\#3b82f6","id":"6ef1fb2d-58cc-4250-80e5-29c279321c46","kind":"point","label":{"text":"B (-2, -3, 4)","visible":true},"radius":0.12,"visible":true,"x":-2,"y":-3,"z":4},{"color":"\#22c55e","id":"ae28d55f-8825-449b-a825-1b8eb617fa2a","kind":"point","label":{"text":"C (2, 3, -4)","visible":true},"radius":0.12,"visible":true,"x":2,"y":3,"z":-4},{"color":"\#f59e0b","id":"10ee098a-6bdf-48fc-8024-56de5a1068c5","kind":"point","label":{"text":"D (2, -1, 2)","visible":true},"radius":0.12,"visible":true,"x":2,"y":-1,"z":2},{"id":"ee433f0e-8ae9-40fd-acf6-f1904de2b7eb","kind":"parametric-surface","label":{"text":"sphere: center A, radius 3","visible":true},"resolution":64,"style":{"colormap":"viridis","doubleSided":true,"opacity":0.35,"wireframe":false,"wireframeColor":"\#000000"},"uDomain":\[0,6.283185307179586\],"vDomain":\[0,3.141592653589793\],"visible":true,"xExpr":"2 + 3\*cos(u)\*sin(v)","yExpr":"-3 + 3\*sin(u)\*sin(v)","zExpr":"4 + 3\*cos(v)"}\],"title":"Lost in the Octants","version":1}
      Four labeled points — A at #math.equation(block: false, alt: "open parenthesis 2 , minus 3 , 4 close parenthesis")[$( 2 , − 3 , 4 )$], B at #math.equation(block: false, alt: "open parenthesis minus 2 , minus 3 , 4 close parenthesis")[$( − 2 , − 3 , 4 )$], C at #math.equation(block: false, alt: "open parenthesis 2 , 3 , minus 4 close parenthesis")[$( 2 , 3 , − 4 )$], and D at #math.equation(block: false, alt: "open parenthesis 2 , minus 1 , 2 close parenthesis")[$( 2 , − 1 , 2 )$] — with a translucent sphere of radius 3 centered at A. Click the figure to activate it, then drag to orbit and scroll to zoom.
    
    
=== Explore


    Work through these with the figure above active. Commit to each prediction in writing before you test it.


    + Without orbiting, assign each of the four points a sign pattern such as #math.equation(block: false, alt: "open parenthesis plus , minus , plus close parenthesis")[$( + , − , + )$]. At least two of them are genuinely ambiguous from the opening camera. Which two are you least sure of, and what information is the flat view withholding?
+ Now orbit until every point is unambiguous. A good technique: look straight down the #math.equation(block: false, alt: "z")[$z$]-axis to settle the #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] signs, then swing low to settle #math.equation(block: false, alt: "z")[$z$]. Which of your calls were wrong? Two of the four points share an octant — which two?
+ The sphere is centered at A with radius 3. Before computing anything, predict whether B, C, and D are each inside, on, or outside the sphere. Orbit to check by eye — B is deceptive from some angles.
+ Settle it with algebra: compute the distances from A to each point using the distance formula. Which of your predictions survive?
+ Select the sphere and read its three component expressions. The center's coordinates appear as the added constants and the radius as the multiplier on the trigonometric parts. Where exactly do you see the 2, the #math.equation(block: false, alt: "minus 3")[$− 3$], the 4, and the 3?


    
=== The distance formula wears every costume


    Everything in the figure is one formula. The distance between #math.equation(block: false, alt: "P sub 1 equals open parenthesis x sub 1 , y sub 1 , z sub 1 close parenthesis")[$P_(1) = ( x_(1) , y_(1) , z_(1) )$] and #math.equation(block: false, alt: "P sub 2 equals open parenthesis x sub 2 , y sub 2 , z sub 2 close parenthesis")[$P_(2) = ( x_(2) , y_(2) , z_(2) )$] is


    #math.equation(block: true, alt: "d equals the square root of open parenthesis x sub 2 minus x sub 1 close parenthesis squared plus open parenthesis y sub 2 minus y sub 1 close parenthesis squared plus open parenthesis z sub 2 minus z sub 1 close parenthesis squared")[$d = sqrt(( x_(2) − x_(1) )^(2) + ( y_(2) − y_(1) )^(2) + ( z_(2) − z_(1) )^(2))$]


    which is the Pythagorean theorem applied twice. A #strong[sphere] is defined by distance alone: it is the set of points at distance exactly #math.equation(block: false, alt: "r")[$r$] from a center #math.equation(block: false, alt: "C equals open parenthesis a , b , c close parenthesis")[$C = ( a , b , c )$], so "inside, on, outside" translates directly to "distance less than, equal to, greater than #math.equation(block: false, alt: "r")[$r$]", and squaring the distance condition gives the standard equation


    #math.equation(block: true, alt: "open parenthesis x minus a close parenthesis squared plus open parenthesis y minus b close parenthesis squared plus open parenthesis z minus c close parenthesis squared equals r squared .")[$( x − a )^(2) + ( y − b )^(2) + ( z − c )^(2) = r^(2) .$]


    The parametric expressions you inspected in the figure above build the same sphere constructively — start at the center, move a distance #math.equation(block: false, alt: "r")[$r$] in every direction — and the identity #math.equation(block: false, alt: "cos squared u plus sin squared u equals 1")[$cos^(2) u + sin^(2) u = 1$], applied twice, collapses them back to the standard equation.


    #examplebox("Example 1")[Inside, on, or outside][
The sphere in the figure above is centered at #math.equation(block: false, alt: "A equals open parenthesis 2 , minus 3 , 4 close parenthesis")[$A = ( 2 , − 3 , 4 )$] with radius 3. Classify #math.equation(block: false, alt: "B equals open parenthesis minus 2 , minus 3 , 4 close parenthesis")[$B = ( − 2 , − 3 , 4 )$] and #math.equation(block: false, alt: "D equals open parenthesis 2 , minus 1 , 2 close parenthesis")[$D = ( 2 , − 1 , 2 )$] as inside, on, or outside, and find the coordinates of one point that lies exactly on the sphere.


      The distance from A to B is #math.equation(block: false, alt: "the square root of open parenthesis minus 2 minus 2 close parenthesis squared plus 0 squared plus 0 squared equals the square root of 16 equals 4")[$sqrt(( − 2 − 2 )^(2) + 0^(2) + 0^(2)) = sqrt(16) = 4$], which is greater than 3, so B is outside. The distance from A to D is #math.equation(block: false, alt: "the square root of 0 squared plus open parenthesis minus 1 plus 3 close parenthesis squared plus open parenthesis 2 minus 4 close parenthesis squared equals the square root of 8 approximately equals 2.83")[$sqrt(0^(2) + ( − 1 + 3 )^(2) + ( 2 − 4 )^(2)) = sqrt(8) ≈ 2.83$], which is less than 3, so D is inside — even though, from many camera angles, B #emph[looks] like the closer point. For a point on the sphere, move exactly 3 units from A parallel to an axis: #math.equation(block: false, alt: "open parenthesis 5 , minus 3 , 4 close parenthesis")[$( 5 , − 3 , 4 )$] works, as do #math.equation(block: false, alt: "open parenthesis 2 , 0 , 4 close parenthesis")[$( 2 , 0 , 4 )$] and #math.equation(block: false, alt: "open parenthesis 2 , minus 3 , 7 close parenthesis")[$( 2 , − 3 , 7 )$]. Each satisfies #math.equation(block: false, alt: "open parenthesis x minus 2 close parenthesis squared plus open parenthesis y plus 3 close parenthesis squared plus open parenthesis z minus 4 close parenthesis squared equals 9")[$( x − 2 )^(2) + ( y + 3 )^(2) + ( z − 4 )^(2) = 9$].
]


    #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Depth is not optional information. Whenever a space picture surprises you in this book, your first move should be the one you practiced here: orbit until the ambiguity dissolves.
]