Applying the right-triangle rule twice to get out of the plane

What to assume they know

• The coordinates of a point in space as three perpendicular distances, one per coordinate plane (Reading a triple as three perpendicular distances, one per plane)
• The three axes, three coordinate planes and the sign conventions of §11.2 (Three perpendicular planes, and the eight regions they cut space into)
• Pythagoras' theorem in the plane, in the form relating the squares of the three sides
• That a line perpendicular to a plane makes a right angle with every line of that plane through its foot
• How far apart two plotted points lie in the plane, from the earlier coordinate geometry of this book
• Expanding squared brackets and collecting like terms in three variables
• That the length of a segment is non-negative, and that squaring a difference erases its sign

What they should be able to do

• Describe the box Fig 11.4 builds on the segment PQ and identify the two extra corners the derivation needs
• Write the path from P to Q as three moves, each parallel to one axis, and state the coordinates of the two intermediate points
• State why the angle at the first intermediate corner is a right angle, using the segment-to-plane fact the chapter omits
• State why the angle at the second intermediate corner is a right angle, and say which plane that one lives in
• Combine the two Pythagoras statements and identify the quantity that cancels
• Write the distance between two general points in space, and specialise it to a distance from the origin without treating that as a separate result
• Compute a distance from two given triples, including cases where one or two coordinates agree
• Decide whether three points lie on a line by comparing three computed lengths, and explain why the comparison is conclusive
• Test a triangle for a right angle by checking all three of the possible pairings, and say why checking one is not enough
• Test four points for being a parallelogram, and say why equal opposite sides alone does not settle it in space
• Turn a stated condition on distances into an equation in three variables, by naming the moving point and expanding

Where it usually goes wrong

• "The formula is Pythagoras with an extra term stuck on." It is Pythagoras applied twice, in two planes that meet along a line. The extra term is not decoration; it is the second triangle's contribution, and skipping the second triangle leaves the result unproved.
• "You can see that the angle at A is a right angle, so it is." You cannot see it — the figure is a flat drawing of a solid, and drawn angles there are unreliable. The reason is that PA runs perpendicular to a whole plane containing AQ. Section 4 is the section that makes this a proof rather than an appeal to the picture.
• "Both right angles are in the same plane." They are not, and if they were the argument would collapse into two dimensions. The two triangles share only the segment AQ.
• "The coordinate differences are lengths, so they cannot be negative." As the chapter writes them, they can be. They are squared immediately, which is why nothing goes wrong, and saying so is cheaper than letting a student silently doubt the working.
• "Swapping which point is first changes the distance." It negates all three differences at once, and squaring undoes that. The formula is symmetric in the two points.
• "The derivation needs a real box, so it fails when two coordinates agree." The drawing fails; the formula does not. Exercise 11.2 q1(i) and q1(iv) are exactly these flattened cases, and both work out — worth checking rather than asserting.
• "Adding two lengths and getting the third is a coincidence." It is a characterisation. Two distances can only add to the third when the middle point sits on the segment joining the other two; anything else leaves a genuine triangle and a strict inequality.
• "One failed right-angle test settles it." Example 5 tests one of the three possible vertices and stops. A triangle has three angles, and any of them could have been the right one, so a complete answer needs all three checks. Do them — they are three additions.
• "Equal opposite sides make a parallelogram." In the plane that is safe; in space it is not, because four points with matching opposite sides need not lie in one plane at all. This is why the chapter's own Note offers the diagonal argument.
• "Every question in the chapter can be answered from the chapter." Three of the four Miscellaneous items need a rule for the midpoint or the centroid that this edition does not print anywhere in the chapter.

Questions to check understanding

• Compute the distance between two given points in space, including pairs sharing one or two coordinates
• Compute the distance of a point from the origin
• Show that three given points lie on a line by comparing three distances, and name which point is between the other two
• Decide whether three given points form a right-angled triangle, testing all three vertices
• Decide whether three given points form an isosceles triangle
• Show that four given points form a parallelogram, and decide whether it is a rectangle
• Given a condition on the distances from a moving point to one or two fixed points, form an equation satisfied by every such point and by no other
• Find a point of a triangle's or a parallelogram's data from the rest — noting that these items need the midpoint or centroid rule, which the chapter does not supply

Examples worth working on the board

Inputs only. Anything marked verified is an added derivation from data printed inside pp. 208–216; the chapter's answers live in a separate file that was not opened.

• Fig 11.4 (§11.4, p. 211), read off the printed page and then off the printed page because the lettering sits inside the artwork. The axes are drawn with Z up, Y to the right and X toward the lower left; O is lettered at the origin. A box is drawn away from the origin. Measured on the printed page: the box's two faces of constant first coordinate are drawn offset up and to the right of one another, and since the positive x direction runs toward the lower left — toward the reader — the higher, righter face is the far one. P and A sit on that far face; Q and N sit on the near face. A redraw that reverses this mirrors the printed figure, and it also breaks the chapter's own leg identification, in which the step from A to N runs the way the first coordinate increases.

P and Q are diagonally opposite corners of the box, A is lettered to the right of P at the same level and N below A. Two angles are marked with arcs and both are labelled 90° — one at A, one at N.

Two diagonals are drawn, not one. Five lines leave Q: three box edges, the space diagonal down to P, and a fifth running down and to the right from Q to A, which is the diagonal of the box face on which the second coordinate is constant. Both are solid and printed. QA is the segment the derivation squares and then eliminates, and it is what closes triangle PAQ and triangle ANQ — without it the right angle marked at A has only one arm and neither triangle exists.

• The two intermediate corners, in coordinates. With P at (x₁, y₁, z₁) and Q at (x₂, y₂, z₂), the chapter's own leg identifications give PA the y-difference, AN the x-difference and NQ the z-difference. Verified: that forces A = (x₁, y₂, z₁) and N = (x₂, y₂, z₁).
• Why the angle at A is right — the step the chapter states and does not support. Verified: PA changes only the y-entry, so PA runs along the y-direction. A and Q share the same y-entry, so the whole segment AQ lies in the plane through A on which the second coordinate is constant. That plane is perpendicular to the y-direction, so PA is perpendicular to every line drawn in it through A, and AQ is one such line. The argument never mentions where Q is, which is the point: the right angle is structural, not a feature of the drawing.
• Why the angle at N is right. Verified: AN changes only the x-entry and NQ only the z-entry, so the two run along two perpendicular axis directions and meet at N at a right angle. Note: this second right angle sits in the plane on which the second coordinate is constant, and the first one does not — the two triangles are not coplanar, and that is exactly why one application of Pythagoras is not enough.
• The combination (§11.4, pp. 211–212). Triangle PAQ gives the square on PQ as the sum of the squares on PA and AQ; triangle ANQ gives the square on AQ as the sum of the squares on AN and NQ. Verified: substituting the second into the first removes AQ entirely and leaves the square on PQ as the sum of three squares — one for each coordinate difference. The quantity that vanishes, AQ, is the face diagonal; it is the scaffolding, and section 6 should show it disappearing.
• The legs, and the sign question. The chapter writes the three legs as plain coordinate differences rather than as their sizes. Verified: every one of them is immediately squared, so writing y₂ − y₁ where the length |y₂ − y₁| is meant changes nothing in the result — but it is worth naming, because a student who has just been told these are lengths will notice that one of them can come out negative.
• The origin case (§11.4, p. 212). Setting the first point at the origin. Verified: the three differences become the three coordinates of the second point, so the distance from the origin is the square root of the sum of the three squared coordinates. It is the same formula with one point pinned, not a new one.
• Example 3 (p. 212): P(1, −3, 4) and Q(−4, 1, 2). Verified: the three differences are −5, 4 and −2; their squares are 25, 16 and 4, totalling 45, so the distance is 3√5.
• Example 4 (p. 212): P(−2, 3, 5), Q(1, 2, 3), R(7, 0, −1), asked to be shown in a line. Verified: PQ² = 9 + 1 + 4 = 14; QR² = 36 + 4 + 16 = 56, so QR = 2√14; PR² = 81 + 9 + 36 = 126, so PR = 3√14. Since √14 + 2√14 = 3√14 exactly, the shorter two lengths add to the longest. Verified justification, which the chapter omits: for any three points the two shorter distances add to at least the third, with equality only when the middle point lies on the segment joining the other two — so the exact equality is what rules out a genuine triangle. Note also that the equality identifies Q as the middle point, which is more than "in a line".
• Example 5 (pp. 212–213): A(3, 6, 9), B(10, 20, 30), C(25, −41, 5), asked whether the triangle has a right angle. Verified: AB² = 49 + 196 + 441 = 686; BC² = 225 + 3721 + 625 = 4571; CA² = 484 + 2209 + 16 = 2709. The printed solution compares CA² + AB² against BC² — that is, tests only the angle at A — finds 3395 ≠ 4571, and concludes. Verified completion, which the chapter does not print: AB² + BC² = 5257 ≠ 2709 and BC² + CA² = 7280 ≠ 686, so the angles at B and at C are not right either, and only now is the conclusion earned.
• Example 6 (p. 213): A(3, 4, 5) and B(−1, 3, −7), and a moving point whose two squared distances to them add to 2k². Verified: naming the moving point (x, y, z) and expanding gives 2x² + 2y² + 2z² − 4x − 14y + 4z + 109 = 2k², so the constant that comes out of the six squared brackets is 109.
• Exercise 11.2 q1 (p. 213), four pairs. (i) (2, 3, 5) with (4, 3, 1); (ii) (−3, 7, 2) with (2, 4, −1); (iii) (−1, 3, −4) with (1, −3, 4); (iv) (2, −1, 3) with (−2, 1, 3). Verified: 2√5; √43; 2√26; 2√5. Verified structural point: pair (i) shares its y-value and pair (iv) shares its z-value, so in both the box of Fig 11.4 collapses to a flat rectangle and the derivation's picture no longer exists — even though the formula still gives the right answer. The chapter's very first exercise item is a case its own figure cannot draw.
• Exercise 11.2 q2 (p. 213) prints the same three points as Example 4 and asks the same thing. Worth flagging to the teacher so the explanation does not present it as fresh practice.
• Exercise 11.2 q3 (p. 213), three verifications. (i) (0, 7, −10), (1, 6, −6), (4, 9, −6) offered as an isosceles triangle. Verified: the three squared lengths are 18, 18 and 36, so two sides are equal at 3√2 and the third is 6. (ii) (0, 7, 10), (−1, 6, 6), (−4, 9, 6) offered as a right-angled triangle. Verified: the squared lengths are again 18, 18 and 36, and 18 + 18 = 36, so the right angle sits at the middle vertex. Verified observation: items (i) and (ii) are congruent copies of one triangle — negating the first and third coordinates of every point in (ii) produces (i) — so the chapter asks about one shape twice and each answer is true of both. Negating exactly two of the three coordinates is a half-turn about the second axis, which preserves orientation, so the map between them is a rotation, not a reflection; calling it a reflection invites a student to flip the figure over. (iii) (−1, 2, 1), (1, −2, 5), (4, −7, 8), (2, −3, 4) offered as a parallelogram. Verified: opposite sides come out 6 and 6, √43 and √43, and the two diagonals share the midpoint (1.5, −2.5, 4.5), so it really is one.
• Exercise 11.2 q4 (p. 213): points at equal distance from (1, 2, 3) and (3, 2, −1). Verified: squaring both distances and cancelling leaves 4x − 8z = 0, that is x = 2z — a plane, and it contains the midpoint (2, 2, 1) as it must.
• Exercise 11.2 q5 (p. 213): A(4, 0, 0), B(−4, 0, 0), and points whose two distances to them add to 10. Verified: the difference of the two squared distances is −16x, so the difference of the distances is −8x/5; adding that to the sum gives one distance as 5 − 4x/5, and squaring and simplifying yields 9x² + 25y² + 25z² = 225. Worth saying what that surface is: a stretched sphere, longest along the axis through A and B.
• Example 7 (pp. 213–214): A(1, 2, 3), B(−1, −2, −1), C(2, 3, 2), D(4, 7, 6). Verified: AB = 6, BC = √43, CD = 6, DA = √43, so opposite sides match; the diagonals are AC = √3 and BD = √155, which differ, so it is not a rectangle. The chapter adds a Note saying the parallelogram could instead be established from the diagonals cutting each other in half. Verified, and this matters: the midpoint of AC and the midpoint of BD are both (1.5, 2.5, 2.5). In space the length argument alone is genuinely insufficient — four points with equal opposite sides need not even lie in one plane — whereas the shared midpoint settles both the flatness and the parallelogram at once. The Note is not an alternative; it is the argument that works.
• Example 8 (p. 214): points equally far from A(3, 4, −5) and B(−2, 1, 4). Verified: expanding and cancelling the squared terms gives 10x + 6y − 18z − 29 = 0.
• Example 9 (pp. 214–215): a triangle whose centroid is (1, 1, 1), with two vertices at (3, −5, 7) and (−1, 7, −6), asked for the third. Verified: averaging each coordinate and solving gives (1, 1, 2). The rule being used — that the centroid's coordinates are the averages of the three vertices' — is nowhere in pp. 208–216.
• Miscellaneous Exercise on Chapter 11 (p. 215), all four items with data intact. q1: a parallelogram ABCD with A(3, −1, 2), B(1, 2, −4), C(−1, 1, 2), fourth vertex wanted. Verified: the diagonals share a midpoint, so D = (1, −2, 8). q2: the medians of the triangle on A(0, 0, 6), B(0, 4, 0) and (6, 0, 0) — the third vertex is printed without a letter. Verified: the medians measure 7, √34 and 7. q3: the origin as centroid of P(2a, 2, 6), Q(−4, 3b, −10), R(8, 14, 2c). Verified: a = −2, b = −16/3, c = 2. q4: A(3, 4, 5) and B(−1, 3, −7) again, with the two squared distances now adding to k² rather than 2k². Verified: the same expansion as Example 6 gives 2x² + 2y² + 2z² − 4x − 14y + 4z = k² − 109. q1, q2 and q3 all need the midpoint or centroid rule, which this chapter does not print.
• The Summary (p. 215) closes with the distance rule in the same form §11.4 derived it, and lists nothing else from this section.

Figures to have open

• Fig 11.4 redrawn with the box, both printed diagonals — PQ across the box and QA across the face — both marked right angles, and the corners A and N carrying their coordinate triples. The chapter's own figure; sections 2 to 6 cannot proceed without the two intermediate corners being explicit.
• A two-plane exploded view showing that triangle PAQ and triangle ANQ lie in different planes and share only AQ. Standard schematic; it is the answer to the commonest misreading of the proof.
• A segment standing perpendicular to a tinted plane while a second segment sweeps around inside that plane, the right angle mark staying put. Standard schematic; it carries section 4.
• The collapsing box for Exercise 11.2 q1(i), where the box becomes a flat rectangle. Standard schematic built from the chapter's data.
• For section 12, a surface plot of 9x² + 25y² + 25z² = 225 with A(4, 0, 0) and B(−4, 0, 0) marked on the long axis. Standard schematic; the chapter asks for the equation and draws nothing.

Where this sits in the book

• NCERT Mathematics, Textbook for Class XI, Chapter 11 "Introduction to Three Dimensional Geometry", §11.4 Distance between Two Points, pp. 211–212 — Fig 11.4, the two right triangles, the combination, the naming of the three legs, and the origin case
• Examples 3, 4, 5 and 6, pp. 212–213
• Exercise 11.2, p. 213, all five questions
• Miscellaneous Examples 7, 8 and 9, pp. 213–215, including the Note on the diagonals of a parallelogram at p. 214
• Miscellaneous Exercise on Chapter 11, p. 215, all four questions
• Summary, p. 215 — the distance rule as the chapter's closing statement of this section
• Deliberate cross-references outside this chapter: the plane distance formula §11.4 builds on belongs to the earlier coordinate geometry of this book, and the perpendicular-to-a-plane fact used in section 4 is assumed school solid geometry. Both are used here without being re-derived.