Exercise 11.2 — Introduction to Three Dimensional Geometry

1. 11.2.14 marksNCERT Class 11 Mathematics, Introduction to Three Dimensional Geometry, Reprint 2026-27

   Find the distance between the following pairs of points: (i) (2,3,5) and (4,3,1) (ii) (-3,7,2) and (2,4,-1) (iii) (-1,3,-4) and (1,-3,4) (iv) (2,-1,3) and (-2,1,3).

   Hint. Apply the 3D distance formula to each pair in turn, substituting carefully and simplifying the surd at the end.

   (i) d=sqrt[(4-2)^2+(3-3)^2+(1-5)^2]=sqrt[4+0+16]=sqrt(20)=2sqrt(5). (ii) d=sqrt[(2+3)^2+(4-7)^2+(-1-2)^2]=sqrt[25+9+9]=sqrt(43). (iii) d=sqrt[(1+1)^2+(-3-3)^2+(4+4)^2]=sqrt[4+36+64]=sqrt(104)=2sqrt(26). (iv) d=sqrt[(-2-2)^2+(1+1)^2+(3-3)^2]=sqrt[16+4+0]=sqrt(20)=2sqrt(5).

   ✦ Working through each part gives: (i) 2sqrt(5); (ii) sqrt(43); (iii) 2sqrt(26); (iv) 2sqrt(5).

2. 11.2.24 marksNCERT Class 11 Mathematics, Introduction to Three Dimensional Geometry, Reprint 2026-27

   Show that the points (-2,3,5), (1,2,3), and (7,0,-1) are collinear.

   Hint. Compute all three pairwise distances and check whether the two shorter ones sum to the longest.

   Let P(-2,3,5), Q(1,2,3), R(7,0,-1). PQ=sqrt[(1+2)^2+(2-3)^2+(3-5)^2]=sqrt[9+1+4]=sqrt(14). QR=sqrt[(7-1)^2+(0-2)^2+(-1-3)^2]=sqrt[36+4+16]=sqrt(56)=2sqrt(14). PR=sqrt[(7+2)^2+(0-3)^2+(-1-5)^2]=sqrt[81+9+36]=sqrt(126)=3sqrt(14). Since PQ+QR=sqrt(14)+2sqrt(14)=3sqrt(14)=PR, point Q lies between P and R.

   ✦ Working through each part gives: The points are collinear, since PQ+QR=PR (all equal to a multiple of sqrt(14)).

3. 11.2.36 marksNCERT Class 11 Mathematics, Introduction to Three Dimensional Geometry, Reprint 2026-27

   Verify the following: (i) (0,7,-10), (1,6,-6), (4,9,-6) are the vertices of an isosceles triangle. (ii) (0,7,10), (-1,6,6), (-4,9,6) are the vertices of a right angled triangle. (iii) (-1,2,1), (1,-2,5), (4,-7,8), (2,-3,4) are the vertices of a parallelogram.

   Hint. For (i) and (ii), compute all three side lengths using the distance formula and check for equal sides or the Pythagorean relation. For (iii), compute the midpoints of both diagonals and check they coincide.

   (i) With A(0,7,-10), B(1,6,-6), C(4,9,-6): AB=sqrt[1+1+16]=sqrt(18); BC=sqrt[9+9+0]=sqrt(18); CA=sqrt[16+4+16]=sqrt(36)=6. Since AB=BC, the triangle is isosceles. (ii) With A(0,7,10), B(-1,6,6), C(-4,9,6): AB=sqrt[1+1+16]=sqrt(18); BC=sqrt[9+9+0]=sqrt(18); CA=sqrt[16+4+16]=sqrt(36)=6. Since AB^2+BC^2=18+18=36=CA^2, the triangle is right-angled at B. (iii) With A(-1,2,1), B(1,-2,5), C(4,-7,8), D(2,-3,4): midpoint of AC=((-1+4)/2,(2-7)/2,(1+8)/2)=(3/2,-5/2,9/2). Midpoint of BD=((1+2)/2,(-2-3)/2,(5+4)/2)=(3/2,-5/2,9/2). Since both diagonals share the same midpoint, they bisect each other, so ABCD is a parallelogram.

   ✦ Working through each part gives: (i) Isosceles, since AB=BC=sqrt(18); (ii) Right-angled at B, since AB^2+BC^2=CA^2; (iii) A parallelogram, since both diagonals share midpoint (3/2,-5/2,9/2).

4. 11.2.44 marksNCERT Class 11 Mathematics, Introduction to Three Dimensional Geometry, Reprint 2026-27

   Find the equation of the set of points which are equidistant from the points (1,2,3) and (3,2,-1).

   Hint. Set PA=PB, square both sides, expand, and simplify — the squared terms in each variable will cancel, leaving a linear equation.

   Let P(x,y,z), A(1,2,3), B(3,2,-1). PA^2=(x-1)^2+(y-2)^2+(z-3)^2, PB^2=(x-3)^2+(y-2)^2+(z+1)^2. Setting PA^2=PB^2, the (y-2)^2 term cancels from both sides, leaving (x-1)^2+(z-3)^2=(x-3)^2+(z+1)^2. Expanding: x^2-2x+1+z^2-6z+9 = x^2-6x+9+z^2+2z+1. Simplifying: -2x-6z+10=-6x+2z+10, so 4x-8z=0, i.e. x-2z=0.

   ✦ Working through each part gives: x - 2z = 0.

5. 11.2.54 marksNCERT Class 11 Mathematics, Introduction to Three Dimensional Geometry, Reprint 2026-27

   Find the equation of the set of points P, the sum of whose distances from A(4,0,0) and B(-4,0,0) is equal to 10.

   Hint. Set PA+PB=10, isolate one square root, square, simplify, then square again — the same two-squaring technique used to derive the 2D ellipse equation, extended to three dimensions.

   Let P(x,y,z). PA=sqrt[(x-4)^2+y^2+z^2], PB=sqrt[(x+4)^2+y^2+z^2]. From PA+PB=10: PA^2-PB^2=(PA-PB)(PA+PB)=10(PA-PB). Directly, PA^2-PB^2=(x-4)^2-(x+4)^2=-16x. So PA-PB=-16x/10=-8x/5. Combined with PA+PB=10: 2PA=10-8x/5, so PA=5-4x/5. Squaring: (x-4)^2+y^2+z^2=25-8x+16x^2/25. Expanding the left side: x^2-8x+16+y^2+z^2=25-8x+16x^2/25. The -8x terms cancel, leaving x^2-16x^2/25+y^2+z^2=9, i.e. 9x^2/25+y^2+z^2=9. Dividing by 9 gives the standard form.

   ✦ Working through each part gives: 9x^2 + 25y^2 + 25z^2 = 225, i.e. x^2/25 + y^2/9 + z^2/9 = 1.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh111.pdf) — Exercise 11.1 (4 questions), Exercise 11.2 (5 questions), plus the chapter's Miscellaneous Exercise (4 questions), 13 questions total, the shortest solutions file in this project to date, matching the book's own shortest chapter (9 pages, only two numbered exercises). Confirmed against the CBSE curriculum PDF that the section formula is formative-only for this chapter and confirmed via full-text search that it is completely absent from the book (zero hits for 'section formula', 'internal division', or 'ratio m'). The old stub had fabricated a third numbered exercise ('Exercise 11.3', 5 questions on the section formula) that does not exist in the book at all — removed. Midpoint and centroid, while never given a named boxed formula in the book, are confirmed as genuinely required techniques via the book's own Example 7 (parallelogram diagonals bisecting each other), Example 9 (centroid used to find a missing vertex), and the Miscellaneous Exercise's own Q1 (parallelogram's fourth vertex) and Q2 (median lengths).. Questions are referenced from the NCERT textbook for identification.