NCERT Solutions · Class 11 Mathematics Introduction to Three Dimensional Geometry

Distance formula in space. For \(\displaystyle \mathrm{P}(x_1,y_1,z_1)\) and \(\displaystyle \mathrm{Q}(x_2,y_2,z_2)\), \[\mathrm{PQ}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}. \](i) \(\displaystyle (2,3,5)\) and \(\displaystyle (4,3,1)\). The differences are \(\displaystyle 2,\,0,\,-4\): \[\sqrt{2^2+0^2+(-4)^2}=\sqrt{4+0+16}=\sqrt{20}=2\sqrt{5}. \](ii) \(\displaystyle (-3,7,2)\) and \(\displaystyle (2,4,-1)\). The differences are \(\displaystyle 5,\,-3,\,-3\): \[\sqrt{5^2+(-3)^2+(-3)^2}=\sqrt{25+9+9}=\sqrt{43}. \](iii) \(\displaystyle (-1,3,-4)\) and \(\displaystyle (1,-3,4)\). The differences are \(\displaystyle 2,\,-6,\,8\): \[\sqrt{2^2+(-6)^2+8^2}=\sqrt{4+36+64}=\sqrt{104}=2\sqrt{26}. \](iv) \(\displaystyle (2,-1,3)\) and \(\displaystyle (-2,1,3)\). The differences are \(\displaystyle -4,\,2,\,0\): \[\sqrt{(-4)^2+2^2+0^2}=\sqrt{16+4+0}=\sqrt{20}=2\sqrt{5}. \](i) \(\displaystyle 2\sqrt{5}\) (ii) \(\displaystyle \sqrt{43}\) (iii) \(\displaystyle 2\sqrt{26}\) (iv) \(\displaystyle 2\sqrt{5}\).

Collinearity by the distance test. Three points are collinear precisely when one of the three distances equals the sum of the other two; if instead the largest is strictly less than the sum of the other two, they form a genuine triangle.Let \(\displaystyle \mathrm{A}(-2,3,5)\), \(\displaystyle \mathrm{B}(1,2,3)\), \(\displaystyle \mathrm{C}(7,0,-1)\).\[\mathrm{AB}=\sqrt{3^2+(-1)^2+(-2)^2}=\sqrt{9+1+4}=\sqrt{14} \] \[\mathrm{BC}=\sqrt{6^2+(-2)^2+(-4)^2}=\sqrt{36+4+16}=\sqrt{56}=2\sqrt{14} \] \[\mathrm{AC}=\sqrt{9^2+(-3)^2+(-6)^2}=\sqrt{81+9+36}=\sqrt{126}=3\sqrt{14} \]Now \[\mathrm{AB}+\mathrm{BC}=\sqrt{14}+2\sqrt{14}=3\sqrt{14}=\mathrm{AC}. \]Since the distance from A to C is exactly the distance from A to B plus that from B to C, no detour is possible: B lies on the segment AC.Hence the three points are collinear, with B between A and C.

Verification by side lengths. Throughout, \(\displaystyle \mathrm{PQ}^2=(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2\); squares are easier to compare than surds, so we work with them.(i) \(\displaystyle \mathrm{A}(0,7,-10)\), \(\displaystyle \mathrm{B}(1,6,-6)\), \(\displaystyle \mathrm{C}(4,9,-6)\) — isosceles. \[\mathrm{AB}^2=1^2+(-1)^2+4^2=18,\qquad \mathrm{BC}^2=3^2+3^2+0^2=18,\qquad \mathrm{CA}^2=(-4)^2+(-2)^2+(-4)^2=36 \] So \(\displaystyle \mathrm{AB}=\mathrm{BC}=3\sqrt{2}\) while \(\displaystyle \mathrm{CA}=6\). Two sides are equal and \(\displaystyle \mathrm{AB}+\mathrm{BC}=6\sqrt{2}>6\), so the points are not collinear and do form a triangle. Hence ABC is an isosceles triangle.(ii) \(\displaystyle \mathrm{A}(0,7,10)\), \(\displaystyle \mathrm{B}(-1,6,6)\), \(\displaystyle \mathrm{C}(-4,9,6)\) — right angled. \[\mathrm{AB}^2=(-1)^2+(-1)^2+(-4)^2=18,\qquad \mathrm{BC}^2=(-3)^2+3^2+0^2=18,\qquad \mathrm{CA}^2=4^2+(-2)^2+4^2=36 \] Then \[\mathrm{AB}^2+\mathrm{BC}^2=18+18=36=\mathrm{CA}^2, \] so the converse of Pythagoras' theorem applies and the angle opposite CA is a right angle. Hence ABC is a right angled triangle, right angled at B.(iii) \(\displaystyle \mathrm{A}(-1,2,1)\), \(\displaystyle \mathrm{B}(1,-2,5)\), \(\displaystyle \mathrm{C}(4,-7,8)\), \(\displaystyle \mathrm{D}(2,-3,4)\) — parallelogram. \[\mathrm{AB}^2=2^2+(-4)^2+4^2=36,\qquad \mathrm{CD}^2=(-2)^2+4^2+(-4)^2=36 \] \[\mathrm{BC}^2=3^2+(-5)^2+3^2=43,\qquad \mathrm{DA}^2=(-3)^2+5^2+(-3)^2=43 \] so \(\displaystyle \mathrm{AB}=\mathrm{CD}=6\) and \(\displaystyle \mathrm{BC}=\mathrm{DA}=\sqrt{43}\): opposite sides are equal.In space, equal opposite sides alone would still allow a skew (non-planar) quadrilateral, so we clinch it with the diagonals — a quadrilateral is a parallelogram exactly when its diagonals bisect each other, i.e. when AC and BD share a midpoint: \[\text{mid}(\mathrm{AC})=\left(\tfrac{-1+4}{2},\tfrac{2-7}{2},\tfrac{1+8}{2}\right)=\left(\tfrac32,-\tfrac52,\tfrac92\right) \] \[\text{mid}(\mathrm{BD})=\left(\tfrac{1+2}{2},\tfrac{-2-3}{2},\tfrac{5+4}{2}\right)=\left(\tfrac32,-\tfrac52,\tfrac92\right) \] The midpoints coincide. Hence ABCD is a parallelogram (and since \(\displaystyle \mathrm{AC}^2=155\neq 3=\mathrm{BD}^2\), the diagonals are unequal, so it is not a rectangle).

Equidistance condition, squared. Let \(\displaystyle \mathrm{P}(x,y,z)\) be a point of the set, \(\displaystyle \mathrm{A}(1,2,3)\) and \(\displaystyle \mathrm{B}(3,2,-1)\). Distances are non-negative, so \(\displaystyle \mathrm{PA}=\mathrm{PB}\) is equivalent to \(\displaystyle \mathrm{PA}^2=\mathrm{PB}^2\), which avoids the surds:\[(x-1)^2+(y-2)^2+(z-3)^2=(x-3)^2+(y-2)^2+(z+1)^2 \]The \(\displaystyle (y-2)^2\) terms cancel. Expanding the rest, \[x^2-2x+1+z^2-6z+9 = x^2-6x+9+z^2+2z+1 \] \[-2x-6z+10 = -6x+2z+10 \] \[4x-8z=0 \] \[x-2z=0 \]Check: \(\displaystyle (2,0,1)\) satisfies \(\displaystyle x-2z=0\), and indeed \(\displaystyle \mathrm{PA}^2=1+4+4=9=1+4+4=\mathrm{PB}^2\).Note that \(\displaystyle y\) has vanished from the equation, so the set is a plane perpendicular to AB through its midpoint \(\displaystyle (2,2,1)\) — the perpendicular bisector plane of AB.The required equation is \(\displaystyle x-2z=0\).

Focal-distance condition (an ellipsoid of revolution). Let \(\displaystyle \mathrm{P}(x,y,z)\), \(\displaystyle \mathrm{A}(4,0,0)\), \(\displaystyle \mathrm{B}(-4,0,0)\), and write \[r_1=\mathrm{PA}=\sqrt{(x-4)^2+y^2+z^2},\qquad r_2=\mathrm{PB}=\sqrt{(x+4)^2+y^2+z^2}. \] We are given \(\displaystyle r_1+r_2=10\).Squaring both surds separately is messy; instead use the difference of squares, where the surds disappear at once: \[r_1^2-r_2^2=(x-4)^2-(x+4)^2=-16x. \] Since \(\displaystyle r_1^2-r_2^2=(r_1+r_2)(r_1-r_2)=10(r_1-r_2)\), \[r_1-r_2=-\frac{16x}{10}=-\frac{8x}{5}. \]Adding this to \(\displaystyle r_1+r_2=10\) and halving, \[r_1=5-\frac{4x}{5}. \]Now square this single equation: \[(x-4)^2+y^2+z^2=\left(5-\frac{4x}{5}\right)^2 \] \[x^2-8x+16+y^2+z^2=25-8x+\frac{16x^2}{25} \] \[x^2-\frac{16x^2}{25}+y^2+z^2=9 \] \[\frac{9x^2}{25}+y^2+z^2=9 \]Multiplying through by \(\displaystyle 25\): \[9x^2+25y^2+25z^2=225. \]Check: the point \(\displaystyle (1,\sqrt{216}/5,0)\) satisfies this, and its distances to A and B do sum to \(\displaystyle 10\).The required equation is \(\displaystyle 9x^2+25y^2+25z^2=225\).

Diagonals of a parallelogram bisect each other. In parallelogram ABCD the diagonals are AC and BD, and they cross at their common midpoint. So\[\text{mid}(\mathrm{AC})=\text{mid}(\mathrm{BD}). \]With \(\displaystyle \mathrm{A}(3,-1,2)\) and \(\displaystyle \mathrm{C}(-1,1,2)\), \[\text{mid}(\mathrm{AC})=\left(\frac{3+(-1)}{2},\ \frac{-1+1}{2},\ \frac{2+2}{2}\right)=(1,\,0,\,2). \]Let \(\displaystyle \mathrm{D}(x,y,z)\). With \(\displaystyle \mathrm{B}(1,2,-4)\), \[\left(\frac{1+x}{2},\ \frac{2+y}{2},\ \frac{-4+z}{2}\right)=(1,0,2) \]Comparing coordinates: \[1+x=2\Rightarrow x=1,\qquad 2+y=0\Rightarrow y=-2,\qquad -4+z=4\Rightarrow z=8. \]Check: \(\displaystyle \mathrm{AB}^2=(1-3)^2+(2+1)^2+(-4-2)^2=4+9+36=49\) and \(\displaystyle \mathrm{DC}^2=(-1-1)^2+(1+2)^2+(2-8)^2=4+9+36=49\), so \(\displaystyle \mathrm{AB}=\mathrm{DC}=7\), as a parallelogram requires.The fourth vertex is \(\displaystyle \mathrm{D}(1,-2,8)\).

Median = vertex to the midpoint of the opposite side. Let \(\displaystyle \mathrm{A}(0,0,6)\), \(\displaystyle \mathrm{B}(0,4,0)\), \(\displaystyle \mathrm{C}(6,0,0)\), and let D, E, F be the midpoints of BC, CA, AB respectively.Median AD. \(\displaystyle \mathrm{D}=\left(\frac{0+6}{2},\frac{4+0}{2},\frac{0+0}{2}\right)=(3,2,0)\), so \[\mathrm{AD}=\sqrt{(3-0)^2+(2-0)^2+(0-6)^2}=\sqrt{9+4+36}=\sqrt{49}=7. \]Median BE. \(\displaystyle \mathrm{E}=\left(\frac{6+0}{2},\frac{0+0}{2},\frac{0+6}{2}\right)=(3,0,3)\), so \[\mathrm{BE}=\sqrt{(3-0)^2+(0-4)^2+(3-0)^2}=\sqrt{9+16+9}=\sqrt{34}. \]Median CF. \(\displaystyle \mathrm{F}=\left(\frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2}\right)=(0,2,3)\), so \[\mathrm{CF}=\sqrt{(0-6)^2+(2-0)^2+(3-0)^2}=\sqrt{36+4+9}=\sqrt{49}=7. \]The medians have lengths \(\displaystyle 7\), \(\displaystyle \sqrt{34}\) and \(\displaystyle 7\).

Centroid formula. The centroid of a triangle with vertices \(\displaystyle (x_1,y_1,z_1)\), \(\displaystyle (x_2,y_2,z_2)\), \(\displaystyle (x_3,y_3,z_3)\) is \[\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3},\ \frac{z_1+z_2+z_3}{3}\right). \]Here the centroid is the origin \(\displaystyle (0,0,0)\), with \(\displaystyle \mathrm{P}(2a,2,6)\), \(\displaystyle \mathrm{Q}(-4,3b,-10)\), \(\displaystyle \mathrm{R}(8,14,2c)\). Equating each coordinate to \(\displaystyle 0\) (equivalently, each coordinate sum to \(\displaystyle 0\)):\[\frac{2a-4+8}{3}=0\ \Rightarrow\ 2a+4=0\ \Rightarrow\ a=-2 \] \[\frac{2+3b+14}{3}=0\ \Rightarrow\ 3b+16=0\ \Rightarrow\ b=-\frac{16}{3} \] \[\frac{6-10+2c}{3}=0\ \Rightarrow\ 2c-4=0\ \Rightarrow\ c=2 \]Check: the vertices become \(\displaystyle (-4,2,6)\), \(\displaystyle (-4,-16,-10)\), \(\displaystyle (8,14,4)\); the coordinate sums are \(\displaystyle -4-4+8=0\), \(\displaystyle 2-16+14=0\), \(\displaystyle 6-10+4=0\). ✔\(\displaystyle a=-2,\quad b=-\dfrac{16}{3},\quad c=2.\)

Write the condition in coordinates and expand. Let \(\displaystyle \mathrm{P}(x,y,z)\) be any point of the set, with \(\displaystyle \mathrm{A}(3,4,5)\) and \(\displaystyle \mathrm{B}(-1,3,-7)\). Then\[\mathrm{PA}^2+\mathrm{PB}^2=k^2 \] \[\big[(x-3)^2+(y-4)^2+(z-5)^2\big]+\big[(x+1)^2+(y-3)^2+(z+7)^2\big]=k^2 \]Expand each bracket: \[(x^2-6x+9)+(y^2-8y+16)+(z^2-10z+25)+(x^2+2x+1)+(y^2-6y+9)+(z^2+14z+49)=k^2 \]Collect like terms — the constants are \(\displaystyle 9+16+25+1+9+49=109\): \[2x^2+2y^2+2z^2-4x-14y+4z+109=k^2 \]Dividing by \(\displaystyle 2\): \[x^2+y^2+z^2-2x-7y+2z=\frac{k^2-109}{2} \]Completing the squares shows what the set is. Adding \(\displaystyle 1+\tfrac{49}{4}+1=\tfrac{57}{4}\) to both sides, \[(x-1)^2+\left(y-\tfrac72\right)^2+(z+1)^2=\frac{k^2-109}{2}+\frac{57}{4}=\frac{2k^2-161}{4}. \] So the set is a sphere centred at the midpoint \(\displaystyle \left(1,\tfrac72,-1\right)\) of AB, of radius \(\displaystyle \tfrac12\sqrt{2k^2-161}\) — a genuine sphere provided \(\displaystyle 2k^2>161\). (This is the median identity \(\displaystyle \mathrm{PA}^2+\mathrm{PB}^2=2\,\mathrm{PM}^2+\tfrac12\mathrm{AB}^2\) with \(\displaystyle \mathrm{AB}^2=161\).)The required equation is \(\displaystyle 2x^2+2y^2+2z^2-4x-14y+4z=k^2-109\), i.e. \(\displaystyle x^2+y^2+z^2-2x-7y+2z=\dfrac{k^2-109}{2}\).