\displaystyle \textbf{Question 1: } \text{Find the distance between the following pairs of points:}
\displaystyle \text{(i) }P(1,-1,0)\text{ and }Q(2,1,2)\qquad\text{(ii) }A(3,2,-1)\text{ and }B(-1,-1,-1)
\displaystyle \text{Answer:}

\displaystyle \text{(i) }P(1,-1,0)\text{ and }Q(2,1,2)
\displaystyle PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}
\displaystyle =\sqrt{(2-1)^2+(1-(-1))^2+(2-0)^2}
\displaystyle =\sqrt{1^2+2^2+2^2}
\displaystyle =\sqrt{9}=3\text{ units.}

\displaystyle \text{(ii) }A(3,2,-1)\text{ and }B(-1,-1,-1)
\displaystyle AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}
\displaystyle =\sqrt{(-1-3)^2+(-1-2)^2+(-1-(-1))^2}
\displaystyle =\sqrt{(-4)^2+(-3)^2+0^2}
\displaystyle =\sqrt{16+9+0}
\displaystyle =\sqrt{25}=5\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 2: } \text{Find the distance between the points }P\text{ and }Q\text{ having coordinates }(-2,3,1)
\displaystyle \text{and }(2,1,2).
\displaystyle \text{Answer:}
\displaystyle PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}
\displaystyle =\sqrt{(2-(-2))^2+(1-3)^2+(2-1)^2}
\displaystyle =\sqrt{4^2+(-2)^2+1^2}
\displaystyle =\sqrt{16+4+1}
\displaystyle =\sqrt{21}\text{ units.}
\displaystyle \\

\displaystyle \textbf{Question 3: } \text{Using the distance formula, prove that the following points are collinear:}
\displaystyle \text{(i) }A(4,-3,-1),\ B(5,-7,6)\text{ and }C(3,1,-8)
\displaystyle \text{(ii) }P(0,7,-7),\ Q(1,4,-5)\text{ and }R(-1,10,-9)
\displaystyle \text{(iii) }A(3,-5,1),\ B(-1,0,8)\text{ and }C(7,-10,-6)
\displaystyle \text{Answer:}

\displaystyle \text{(i) Given }A(4,-3,-1),\ B(5,-7,6)\text{ and }C(3,1,-8).
\displaystyle AB=\sqrt{(5-4)^2+(-7+3)^2+(6+1)^2}
\displaystyle =\sqrt{1^2+(-4)^2+7^2}
\displaystyle =\sqrt{1+16+49}=\sqrt{66}\text{ units.}
\displaystyle BC=\sqrt{(3-5)^2+(1+7)^2+(-8-6)^2}
\displaystyle =\sqrt{(-2)^2+8^2+(-14)^2}
\displaystyle =\sqrt{4+64+196}=\sqrt{264}=2\sqrt{66}\text{ units.}
\displaystyle AC=\sqrt{(3-4)^2+(1+3)^2+(-8+1)^2}
\displaystyle =\sqrt{(-1)^2+4^2+(-7)^2}
\displaystyle =\sqrt{1+16+49}=\sqrt{66}\text{ units.}
\displaystyle AB+AC=\sqrt{66}+\sqrt{66}=2\sqrt{66}=BC.
\displaystyle \therefore A\text{ lies between }B\text{ and }C,\text{ and the points }A,\ B\text{ and }C\text{ are collinear.}

\displaystyle \text{(ii) Given }P(0,7,-7),\ Q(1,4,-5)\text{ and }R(-1,10,-9).
\displaystyle PQ=\sqrt{(1-0)^2+(4-7)^2+(-5+7)^2}
\displaystyle =\sqrt{1^2+(-3)^2+2^2}
\displaystyle =\sqrt{1+9+4}=\sqrt{14}\text{ units.}
\displaystyle QR=\sqrt{(-1-1)^2+(10-4)^2+(-9+5)^2}
\displaystyle =\sqrt{(-2)^2+6^2+(-4)^2}
\displaystyle =\sqrt{4+36+16}=\sqrt{56}=2\sqrt{14}\text{ units.}
\displaystyle PR=\sqrt{(-1-0)^2+(10-7)^2+(-9+7)^2}
\displaystyle =\sqrt{(-1)^2+3^2+(-2)^2}
\displaystyle =\sqrt{1+9+4}=\sqrt{14}\text{ units.}
\displaystyle PQ+PR=\sqrt{14}+\sqrt{14}=2\sqrt{14}=QR.
\displaystyle \therefore P\text{ lies between }Q\text{ and }R,\text{ and the points }P,\ Q\text{ and }R\text{ are collinear.}

\displaystyle \text{(iii) Given }A(3,-5,1),\ B(-1,0,8)\text{ and }C(7,-10,-6).
\displaystyle AB=\sqrt{(-1-3)^2+(0+5)^2+(8-1)^2}
\displaystyle =\sqrt{(-4)^2+5^2+7^2}
\displaystyle =\sqrt{16+25+49}=\sqrt{90}=3\sqrt{10}\text{ units.}
\displaystyle BC=\sqrt{(7+1)^2+(-10-0)^2+(-6-8)^2}
\displaystyle =\sqrt{8^2+(-10)^2+(-14)^2}
\displaystyle =\sqrt{64+100+196}=\sqrt{360}=6\sqrt{10}\text{ units.}
\displaystyle AC=\sqrt{(7-3)^2+(-10+5)^2+(-6-1)^2}
\displaystyle =\sqrt{4^2+(-5)^2+(-7)^2}
\displaystyle =\sqrt{16+25+49}=\sqrt{90}=3\sqrt{10}\text{ units.}
\displaystyle AB+AC=3\sqrt{10}+3\sqrt{10}=6\sqrt{10}=BC.
\displaystyle \therefore A\text{ lies between }B\text{ and }C,\text{ and the points }A,\ B\text{ and }C\text{ are collinear.}
\displaystyle \\

\displaystyle \textbf{Question 4: } \text{Determine the points in (i) the }XY\text{-plane, (ii) the }YZ\text{-plane and (iii) the}
\displaystyle ZX\text{-plane which are equidistant from the points }A(1,-1,0),\ B(2,1,2)\text{ and }C(3,2,-1).
\displaystyle \text{Answer:}

\displaystyle \text{(i) Let }P(x,y,0)\text{ be the required point in the }XY\text{-plane such that }PA=PB=PC.
\displaystyle PA^2=PB^2
\displaystyle (x-1)^2+(y+1)^2=(x-2)^2+(y-1)^2+(0-2)^2
\displaystyle x^2-2x+1+y^2+2y+1=x^2-4x+4+y^2-2y+1+4
\displaystyle 2x+4y=7\qquad\text{... ... ... ... ... (i)}
\displaystyle PB^2=PC^2
\displaystyle (x-2)^2+(y-1)^2+(0-2)^2=(x-3)^2+(y-2)^2+(0+1)^2
\displaystyle x^2-4x+4+y^2-2y+1+4=x^2-6x+9+y^2-4y+4+1
\displaystyle 2x+2y=5\qquad\text{... ... ... ... ... (ii)}
\displaystyle \text{Subtracting (ii) from (i), we get }2y=2\Rightarrow y=1.
\displaystyle \text{Substituting }y=1\text{ in (ii), we get }2x+2=5\Rightarrow x=\frac{3}{2}.
\displaystyle \therefore \text{The required point in the }XY\text{-plane is }\left(\frac{3}{2},1,0\right).

\displaystyle \text{(ii) Let }P(0,y,z)\text{ be the required point in the }YZ\text{-plane such that }PA=PB=PC.
\displaystyle PA^2=PB^2
\displaystyle (0-1)^2+(y+1)^2+z^2=(0-2)^2+(y-1)^2+(z-2)^2
\displaystyle 1+y^2+2y+1+z^2=4+y^2-2y+1+z^2-4z+4
\displaystyle 4y+4z=7\qquad\text{... ... ... ... ... (i)}
\displaystyle PB^2=PC^2
\displaystyle (0-2)^2+(y-1)^2+(z-2)^2=(0-3)^2+(y-2)^2+(z+1)^2
\displaystyle 4+y^2-2y+1+z^2-4z+4=9+y^2-4y+4+z^2+2z+1
\displaystyle 2y-6z=5\qquad\text{... ... ... ... ... (ii)}
\displaystyle \text{From (i), }y+z=\frac{7}{4},\text{ and from (ii), }y-3z=\frac{5}{2}.
\displaystyle \text{Subtracting, }-4z=\frac{3}{4}\Rightarrow z=-\frac{3}{16}.
\displaystyle y=\frac{7}{4}+\frac{3}{16}=\frac{31}{16}.
\displaystyle \therefore \text{The required point in the }YZ\text{-plane is }\left(0,\frac{31}{16},-\frac{3}{16}\right).

\displaystyle \text{(iii) Let }P(x,0,z)\text{ be the required point in the }ZX\text{-plane such that }PA=PB=PC.
\displaystyle PA^2=PB^2
\displaystyle (x-1)^2+(0+1)^2+z^2=(x-2)^2+(0-1)^2+(z-2)^2
\displaystyle x^2-2x+1+1+z^2=x^2-4x+4+1+z^2-4z+4
\displaystyle 2x+4z=7\qquad\text{... ... ... ... ... (i)}
\displaystyle PB^2=PC^2
\displaystyle (x-2)^2+(0-1)^2+(z-2)^2=(x-3)^2+(0-2)^2+(z+1)^2
\displaystyle x^2-4x+4+1+z^2-4z+4=x^2-6x+9+4+z^2+2z+1
\displaystyle 2x-6z=5\qquad\text{... ... ... ... ... (ii)}
\displaystyle \text{Subtracting (ii) from (i), we get }10z=2\Rightarrow z=\frac{1}{5}.
\displaystyle \text{Substituting }z=\frac{1}{5}\text{ in (i), we get }2x+\frac{4}{5}=7.
\displaystyle 2x=\frac{31}{5}\Rightarrow x=\frac{31}{10}.
\displaystyle \therefore \text{The required point in the }ZX\text{-plane is }\left(\frac{31}{10},0,\frac{1}{5}\right).
\displaystyle \\

\displaystyle \textbf{Question 5: } \text{Determine the point on the }z\text{-axis which is equidistant from the points }(1,5,7)
\displaystyle \text{and }(5,1,-4).
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(1,5,7)\text{ and }B(5,1,-4).
\displaystyle \text{Let }M(0,0,z)\text{ be the required point on the }z\text{-axis.}
\displaystyle AM=\sqrt{(0-1)^2+(0-5)^2+(z-7)^2}
\displaystyle =\sqrt{(-1)^2+(-5)^2+z^2-14z+49}
\displaystyle =\sqrt{z^2-14z+75}.
\displaystyle BM=\sqrt{(0-5)^2+(0-1)^2+(z+4)^2}
\displaystyle =\sqrt{(-5)^2+(-1)^2+z^2+8z+16}
\displaystyle =\sqrt{z^2+8z+42}.
\displaystyle \text{Since }M\text{ is equidistant from }A\text{ and }B,\ AM=BM.
\displaystyle \sqrt{z^2-14z+75}=\sqrt{z^2+8z+42}
\displaystyle \text{Squaring both sides,}
\displaystyle z^2-14z+75=z^2+8z+42
\displaystyle -14z-8z=42-75
\displaystyle -22z=-33
\displaystyle z=\frac{33}{22}=\frac{3}{2}.
\displaystyle \therefore \text{The required point is }\left(0,0,\frac{3}{2}\right).
\displaystyle \\

\displaystyle \textbf{Question 6: } \text{Find the point on the }y\text{-axis which is equidistant from the points }(3,1,2)
\displaystyle \text{and }(5,5,2).
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(3,1,2),\ Q(5,5,2)\text{ and }M(0,y,0)\text{ be the required point on the }y\text{-axis.}
\displaystyle \text{Since }M\text{ is equidistant from }P\text{ and }Q,\ PM=QM.
\displaystyle \sqrt{(0-3)^2+(y-1)^2+(0-2)^2}=\sqrt{(0-5)^2+(y-5)^2+(0-2)^2}
\displaystyle \sqrt{9+y^2-2y+1+4}=\sqrt{25+y^2-10y+25+4}
\displaystyle \sqrt{y^2-2y+14}=\sqrt{y^2-10y+54}
\displaystyle \text{Squaring both sides,}
\displaystyle y^2-2y+14=y^2-10y+54
\displaystyle -2y+10y=54-14
\displaystyle 8y=40
\displaystyle y=5.
\displaystyle \therefore \text{The required point on the }y\text{-axis is }(0,5,0).
\displaystyle \\

\displaystyle \textbf{Question 7: } \text{Find the points on the }z\text{-axis which are at a distance }\sqrt{21}\text{ from the}
\displaystyle \text{point }(1,2,3).
\displaystyle \text{Answer:}
\displaystyle \text{Let the required point on the }z\text{-axis be }A(0,0,z).
\displaystyle AP=\sqrt{21}
\displaystyle \Rightarrow \sqrt{(0-1)^2+(0-2)^2+(z-3)^2}=\sqrt{21}
\displaystyle \Rightarrow (-1)^2+(-2)^2+(z-3)^2=21
\displaystyle \Rightarrow 1+4+(z-3)^2=21
\displaystyle \Rightarrow (z-3)^2=16
\displaystyle \Rightarrow z-3=\pm4
\displaystyle \Rightarrow z=7\text{ or }z=-1.
\displaystyle \therefore \text{The required points are }(0,0,7)\text{ and }(0,0,-1).
\displaystyle \\

\displaystyle \textbf{Question 8: } \text{Prove that the triangle formed by joining the three points }(1,2,3),\ (2,3,1)
\displaystyle \text{and }(3,1,2)\text{ is an equilateral triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(1,2,3),\ B(2,3,1)\text{ and }C(3,1,2)\text{ be the vertices of }\triangle ABC.
\displaystyle AB=\sqrt{(2-1)^2+(3-2)^2+(1-3)^2}
\displaystyle =\sqrt{1^2+1^2+(-2)^2}
\displaystyle =\sqrt{1+1+4}=\sqrt{6}.
\displaystyle BC=\sqrt{(3-2)^2+(1-3)^2+(2-1)^2}
\displaystyle =\sqrt{1^2+(-2)^2+1^2}
\displaystyle =\sqrt{1+4+1}=\sqrt{6}.
\displaystyle CA=\sqrt{(1-3)^2+(2-1)^2+(3-2)^2}
\displaystyle =\sqrt{(-2)^2+1^2+1^2}
\displaystyle =\sqrt{4+1+1}=\sqrt{6}.
\displaystyle \therefore AB=BC=CA.
\displaystyle \therefore \triangle ABC\text{ is an equilateral triangle.}
\displaystyle \\

\displaystyle \textbf{Question 9: } \text{Show that the points }(0,7,10),\ (-1,6,6)\text{ and }(-4,9,6)\text{ are the}
\displaystyle \text{vertices of an isosceles right-angled triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(0,7,10),\ B(-1,6,6)\text{ and }C(-4,9,6)\text{ be the vertices of }\triangle ABC.
\displaystyle AB=\sqrt{(-1-0)^2+(6-7)^2+(6-10)^2}
\displaystyle =\sqrt{(-1)^2+(-1)^2+(-4)^2}
\displaystyle =\sqrt{1+1+16}=\sqrt{18}=3\sqrt{2}.
\displaystyle BC=\sqrt{(-4+1)^2+(9-6)^2+(6-6)^2}
\displaystyle =\sqrt{(-3)^2+3^2+0^2}
\displaystyle =\sqrt{9+9}=\sqrt{18}=3\sqrt{2}.
\displaystyle CA=\sqrt{(0+4)^2+(7-9)^2+(10-6)^2}
\displaystyle =\sqrt{4^2+(-2)^2+4^2}
\displaystyle =\sqrt{16+4+16}=\sqrt{36}=6.
\displaystyle AB^2+BC^2=(3\sqrt{2})^2+(3\sqrt{2})^2=18+18=36=6^2=CA^2.
\displaystyle \therefore \angle ABC=90^\circ\text{ and }AB=BC.
\displaystyle \therefore \triangle ABC\text{ is an isosceles right-angled triangle.}
\displaystyle \\

\displaystyle \textbf{Question 10: } \text{Show that the points }A(3,3,3),\ B(0,6,3),\ C(1,7,7)\text{ and }D(4,4,7)
\displaystyle \text{are the vertices of a square.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(3,3,3),\ B(0,6,3),\ C(1,7,7)\text{ and }D(4,4,7)\text{ be the vertices of quadrilateral }ABCD.
\displaystyle AB=\sqrt{(0-3)^2+(6-3)^2+(3-3)^2}
\displaystyle =\sqrt{(-3)^2+3^2+0^2}=\sqrt{18}=3\sqrt{2}.
\displaystyle BC=\sqrt{(1-0)^2+(7-6)^2+(7-3)^2}
\displaystyle =\sqrt{1^2+1^2+4^2}=\sqrt{18}=3\sqrt{2}.
\displaystyle CD=\sqrt{(4-1)^2+(4-7)^2+(7-7)^2}
\displaystyle =\sqrt{3^2+(-3)^2+0^2}=\sqrt{18}=3\sqrt{2}.
\displaystyle DA=\sqrt{(4-3)^2+(4-3)^2+(7-3)^2}
\displaystyle =\sqrt{1^2+1^2+4^2}=\sqrt{18}=3\sqrt{2}.
\displaystyle \therefore AB=BC=CD=DA.
\displaystyle AC=\sqrt{(1-3)^2+(7-3)^2+(7-3)^2}
\displaystyle =\sqrt{(-2)^2+4^2+4^2}=\sqrt{36}=6.
\displaystyle BD=\sqrt{(4-0)^2+(4-6)^2+(7-3)^2}
\displaystyle =\sqrt{4^2+(-2)^2+4^2}=\sqrt{36}=6.
\displaystyle \therefore AC=BD.
\displaystyle \text{Since all four sides are equal and the diagonals are equal, }ABCD\text{ is a square.}
\displaystyle \\

\displaystyle \textbf{Question 11: } \text{Prove that the points }A(1,3,0),\ B(-5,5,2),\ C(-9,-1,2)\text{ and}
\displaystyle D(-3,-3,0)\text{, taken in order, are the vertices of a parallelogram. Also, show that }ABCD
\displaystyle \text{is not a rectangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(1,3,0),\ B(-5,5,2),\ C(-9,-1,2)\text{ and }D(-3,-3,0)\text{ be the vertices of quadrilateral }ABCD.
\displaystyle AB=\sqrt{(-5-1)^2+(5-3)^2+(2-0)^2}
\displaystyle =\sqrt{(-6)^2+2^2+2^2}=\sqrt{44}=2\sqrt{11}.
\displaystyle BC=\sqrt{(-9+5)^2+(-1-5)^2+(2-2)^2}
\displaystyle =\sqrt{(-4)^2+(-6)^2+0^2}=\sqrt{52}=2\sqrt{13}.
\displaystyle CD=\sqrt{(-3+9)^2+(-3+1)^2+(0-2)^2}
\displaystyle =\sqrt{6^2+(-2)^2+(-2)^2}=\sqrt{44}=2\sqrt{11}.
\displaystyle DA=\sqrt{(1+3)^2+(3+3)^2+(0-0)^2}
\displaystyle =\sqrt{4^2+6^2+0^2}=\sqrt{52}=2\sqrt{13}.
\displaystyle \therefore AB=CD\text{ and }BC=DA.
\displaystyle \text{Since both pairs of opposite sides are equal, }ABCD\text{ is a parallelogram.}
\displaystyle AC=\sqrt{(-9-1)^2+(-1-3)^2+(2-0)^2}
\displaystyle =\sqrt{(-10)^2+(-4)^2+2^2}=\sqrt{120}=2\sqrt{30}.
\displaystyle BD=\sqrt{(-3+5)^2+(-3-5)^2+(0-2)^2}
\displaystyle =\sqrt{2^2+(-8)^2+(-2)^2}=\sqrt{72}=6\sqrt{2}.
\displaystyle \therefore AC\ne BD.
\displaystyle \text{Since the diagonals of the parallelogram are not equal, }ABCD\text{ is not a rectangle.}
\displaystyle \\

\displaystyle \textbf{Question 12: } \text{Show that the points }A(1,3,4),\ B(-1,6,10),\ C(-7,4,7)\text{ and}
\displaystyle D(-5,1,1)\text{ are the vertices of a rhombus.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(1,3,4),\ B(-1,6,10),\ C(-7,4,7)\text{ and }D(-5,1,1)\text{ be the vertices of quadrilateral }ABCD.
\displaystyle AB=\sqrt{(-1-1)^2+(6-3)^2+(10-4)^2}
\displaystyle =\sqrt{4+9+36}=\sqrt{49}=7.
\displaystyle BC=\sqrt{(-7+1)^2+(4-6)^2+(7-10)^2}
\displaystyle =\sqrt{36+4+9}=\sqrt{49}=7.
\displaystyle CD=\sqrt{(-5+7)^2+(1-4)^2+(1-7)^2}
\displaystyle =\sqrt{4+9+36}=\sqrt{49}=7.
\displaystyle DA=\sqrt{(1+5)^2+(3-1)^2+(4-1)^2}
\displaystyle =\sqrt{36+4+9}=\sqrt{49}=7.
\displaystyle \therefore AB=CD\text{ and }BC=DA.
\displaystyle \text{Hence both pairs of opposite sides are equal. Therefore, }ABCD\text{ is a parallelogram.}
\displaystyle \text{Also, }AB=BC=CD=DA.
\displaystyle \text{Therefore, the parallelogram }ABCD\text{ has all its sides equal. Hence }ABCD\text{ is a rhombus.}
\displaystyle \\

\displaystyle \textbf{Question 13: } \text{Prove that the tetrahedron with vertices at the points }O(0,0,0),
\displaystyle A(0,1,1),  \ B(1,0,1)\text{ and }C(1,1,0)\text{ is a regular one.}
\displaystyle \text{Answer:}
\displaystyle \text{A tetrahedron is regular if all its six edges are equal.}
\displaystyle OA=\sqrt{(0-0)^2+(1-0)^2+(1-0)^2}=\sqrt{0+1+1}=\sqrt{2}.
\displaystyle OB=\sqrt{(1-0)^2+(0-0)^2+(1-0)^2}=\sqrt{1+0+1}=\sqrt{2}.
\displaystyle OC=\sqrt{(1-0)^2+(1-0)^2+(0-0)^2}=\sqrt{1+1+0}=\sqrt{2}.
\displaystyle AB=\sqrt{(1-0)^2+(0-1)^2+(1-1)^2}=\sqrt{1+1+0}=\sqrt{2}.
\displaystyle BC=\sqrt{(1-1)^2+(1-0)^2+(0-1)^2}=\sqrt{0+1+1}=\sqrt{2}.
\displaystyle CA=\sqrt{(1-0)^2+(1-1)^2+(0-1)^2}=\sqrt{1+0+1}=\sqrt{2}.
\displaystyle \therefore OA=OB=OC=AB=BC=CA=\sqrt{2}.
\displaystyle \text{Hence all the six edges are equal. Therefore, the tetrahedron }OABC\text{ is a regular tetrahedron.}
\displaystyle \\

\displaystyle \textbf{Question 14: } \text{Show that the points }(3,2,2),\ (-1,4,2),\ (0,5,6)\text{ and }(2,1,2)
\displaystyle \text{lie on a sphere whose centre is }(1,3,4).\text{ Find also its radius.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(3,2,2),\ B(-1,4,2),\ C(0,5,6),\ D(2,1,2)\text{ and let the centre be }P(1,3,4).
\displaystyle AP=\sqrt{(1-3)^2+(3-2)^2+(4-2)^2}
\displaystyle =\sqrt{(-2)^2+1^2+2^2}=\sqrt{4+1+4}=3.
\displaystyle BP=\sqrt{(1+1)^2+(3-4)^2+(4-2)^2}
\displaystyle =\sqrt{2^2+(-1)^2+2^2}=\sqrt{4+1+4}=3.
\displaystyle CP=\sqrt{(1-0)^2+(3-5)^2+(4-6)^2}
\displaystyle =\sqrt{1^2+(-2)^2+(-2)^2}=\sqrt{1+4+4}=3.
\displaystyle DP=\sqrt{(1-2)^2+(3-1)^2+(4-2)^2}
\displaystyle =\sqrt{(-1)^2+2^2+2^2}=\sqrt{1+4+4}=3.
\displaystyle \therefore AP=BP=CP=DP=3.
\displaystyle \text{Hence all the four points lie on the sphere whose centre is }(1,3,4)\text{ and whose radius is }3.
\displaystyle \\

\displaystyle \textbf{Question 15: } \text{Find the coordinates of the point which is equidistant from the} 
\displaystyle \text{four points }O(0,0,0), \ A(2,0,0),\ B(0,3,0)\text{ and }C(0,0,8).
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,y,z)\text{ be the point which is equidistant from }O(0,0,0),\ A(2,0,0),\ B(0,3,0)
\displaystyle \text{and }C(0,0,8).
\displaystyle \text{Since }OP=AP,
\displaystyle OP^2=AP^2
\displaystyle \Rightarrow x^2+y^2+z^2=(x-2)^2+y^2+z^2
\displaystyle \Rightarrow x^2=x^2-4x+4
\displaystyle \Rightarrow 4x=4
\displaystyle \Rightarrow x=1.
\displaystyle \text{Also, since }OP=BP,
\displaystyle OP^2=BP^2
\displaystyle \Rightarrow x^2+y^2+z^2=x^2+(y-3)^2+z^2
\displaystyle \Rightarrow y^2=y^2-6y+9
\displaystyle \Rightarrow 6y=9
\displaystyle \Rightarrow y=\frac{3}{2}.
\displaystyle \text{Further, since }OP=CP,
\displaystyle OP^2=CP^2
\displaystyle \Rightarrow x^2+y^2+z^2=x^2+y^2+(z-8)^2
\displaystyle \Rightarrow z^2=z^2-16z+64
\displaystyle \Rightarrow 16z=64
\displaystyle \Rightarrow z=4.
\displaystyle \therefore \text{The required point is }P\left(1,\frac{3}{2},4\right).
\displaystyle \\

\displaystyle \textbf{Question 16: } \text{If }A(-2,2,3)\text{ and }B(13,-3,13)\text{ are two points, find the locus of a point }P
\displaystyle \text{which moves in such a way that }3PA=2PB.
\displaystyle \text{Answer:}
\displaystyle \text{Let the coordinates of the moving point }P\text{ be }(x,y,z).
\displaystyle \text{Given }3PA=2PB.
\displaystyle 3\sqrt{(x+2)^2+(y-2)^2+(z-3)^2}
\displaystyle =2\sqrt{(x-13)^2+(y+3)^2+(z-13)^2}.
\displaystyle \text{Squaring both sides,}
\displaystyle 9\left[(x+2)^2+(y-2)^2+(z-3)^2\right]
\displaystyle =4\left[(x-13)^2+(y+3)^2+(z-13)^2\right].
\displaystyle 9\left(x^2+y^2+z^2+4x-4y-6z+17\right)
\displaystyle =4\left(x^2+y^2+z^2-26x+6y-26z+347\right).
\displaystyle 9x^2+9y^2+9z^2+36x-36y-54z+153
\displaystyle =4x^2+4y^2+4z^2-104x+24y-104z+1388.
\displaystyle 5x^2+5y^2+5z^2+140x-60y+50z-1235=0.
\displaystyle \Rightarrow x^2+y^2+z^2+28x-12y+10z-247=0.
\displaystyle \Rightarrow (x+14)^2+(y-6)^2+(z+5)^2=504.
\displaystyle \therefore \text{The locus of }P\text{ is the sphere }(x+14)^2+(y-6)^2+(z+5)^2=504.
\displaystyle \\

\displaystyle \textbf{Question 17: } \text{Find the locus of }P\text{ if }PA^2+PB^2=2k^2,\text{ where }A\text{ and }B\text{ are the}
\displaystyle \text{points }(3,4,5)\text{ and }(-1,3,-7)\text{ respectively.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the coordinates of the moving point }P\text{ be }(x,y,z).
\displaystyle \text{Given }PA^2+PB^2=2k^2.
\displaystyle (x-3)^2+(y-4)^2+(z-5)^2+(x+1)^2+(y-3)^2+(z+7)^2=2k^2.
\displaystyle x^2-6x+9+y^2-8y+16+z^2-10z+25
\displaystyle +x^2+2x+1+y^2-6y+9+z^2+14z+49=2k^2.
\displaystyle 2x^2+2y^2+2z^2-4x-14y+4z+109=2k^2.
\displaystyle \Rightarrow x^2+y^2+z^2-2x-7y+2z+\frac{109}{2}-k^2=0.
\displaystyle \Rightarrow (x^2-2x+1)+\left(y^2-7y+\frac{49}{4}\right)+(z^2+2z+1)
\displaystyle =k^2-\frac{109}{2}+1+\frac{49}{4}+1.
\displaystyle \Rightarrow (x-1)^2+\left(y-\frac{7}{2}\right)^2+(z+1)^2=k^2-\frac{161}{4}.
\displaystyle \therefore \text{The locus of }P\text{ is the sphere whose centre is }\left(1,\frac{7}{2},-1\right)
\displaystyle \text{and whose radius is }\sqrt{k^2-\frac{161}{4}}.
\displaystyle \\

\displaystyle \textbf{Question 18: } \text{Show that the points }(a,b,c),\ (b,c,a)\text{ and }(c,a,b)\text{ are the}
\displaystyle \text{vertices of an equilateral triangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(a,b,c),\ B(b,c,a)\text{ and }C(c,a,b)\text{ be the vertices of }\triangle ABC.
\displaystyle AB=\sqrt{(b-a)^2+(c-b)^2+(a-c)^2}
\displaystyle =\sqrt{b^2-2ab+a^2+c^2-2bc+b^2+a^2-2ac+c^2}
\displaystyle =\sqrt{2a^2+2b^2+2c^2-2ab-2bc-2ac}
\displaystyle =\sqrt{2(a^2+b^2+c^2-ab-bc-ac)}.
\displaystyle BC=\sqrt{(c-b)^2+(a-c)^2+(b-a)^2}
\displaystyle =\sqrt{c^2-2bc+b^2+a^2-2ac+c^2+b^2-2ab+a^2}
\displaystyle =\sqrt{2a^2+2b^2+2c^2-2ab-2bc-2ac}
\displaystyle =\sqrt{2(a^2+b^2+c^2-ab-bc-ac)}.
\displaystyle CA=\sqrt{(a-c)^2+(b-a)^2+(c-b)^2}
\displaystyle =\sqrt{a^2-2ac+c^2+b^2-2ab+a^2+c^2-2bc+b^2}
\displaystyle =\sqrt{2a^2+2b^2+2c^2-2ab-2bc-2ac}
\displaystyle =\sqrt{2(a^2+b^2+c^2-ab-bc-ac)}.
\displaystyle \therefore AB=BC=CA.
\displaystyle \therefore \triangle ABC\text{ is an equilateral triangle.}
\displaystyle \\

\displaystyle \textbf{Question 19: } \text{Are the points }A(3,6,9),\ B(10,20,30)\text{ and }C(25,-41,5)\text{ the}
\displaystyle \text{vertices of a right-angled triangle?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(3,6,9),\ B(10,20,30)\text{ and }C(25,-41,5)\text{ be the vertices of }\triangle ABC.
\displaystyle AB=\sqrt{(10-3)^2+(20-6)^2+(30-9)^2}
\displaystyle =\sqrt{7^2+14^2+21^2}
\displaystyle =\sqrt{49+196+441}=\sqrt{686}=7\sqrt{14}.
\displaystyle BC=\sqrt{(25-10)^2+(-41-20)^2+(5-30)^2}
\displaystyle =\sqrt{15^2+(-61)^2+(-25)^2}
\displaystyle =\sqrt{225+3721+625}=\sqrt{4571}.
\displaystyle AC=\sqrt{(25-3)^2+(-41-6)^2+(5-9)^2}
\displaystyle =\sqrt{22^2+(-47)^2+(-4)^2}
\displaystyle =\sqrt{484+2209+16}=\sqrt{2709}=3\sqrt{301}.
\displaystyle BC^2=4571,\quad AB^2=686,\quad AC^2=2709.
\displaystyle AB^2+AC^2=686+2709=3395\ne4571=BC^2.
\displaystyle \therefore \triangle ABC\text{ is not a right-angled triangle.}
\displaystyle \\

\displaystyle \textbf{Question 20: } \text{Verify the following:}
\displaystyle \text{(i) }(0,7,-10),\ (1,6,-6)\text{ and }(4,9,-6)\text{ are the vertices of an isosceles triangle.}
\displaystyle \text{(ii) }(0,7,10),\ (-1,6,6)\text{ and }(-4,9,6)\text{ are the vertices of a right-angled triangle.}
\displaystyle \text{(iii) }(-1,2,1),\ (1,-2,5),\ (4,-7,8)\text{ and }(2,-3,4)\text{ are the vertices of a parallelogram.}
\displaystyle \text{(iv) }(5,-1,1),\ (7,-4,7),\ (1,-6,10)\text{ and }(-1,-3,4)\text{ are the vertices of a rhombus.}
\displaystyle \text{Answer:}

\displaystyle \text{(i) Let }A(0,7,-10),\ B(1,6,-6)\text{ and }C(4,9,-6)\text{ be the vertices of }\triangle ABC.
\displaystyle AB=\sqrt{(1-0)^2+(6-7)^2+(-6+10)^2}
\displaystyle =\sqrt{1^2+(-1)^2+4^2}=\sqrt{18}=3\sqrt{2}.
\displaystyle BC=\sqrt{(4-1)^2+(9-6)^2+(-6+6)^2}
\displaystyle =\sqrt{3^2+3^2+0^2}=\sqrt{18}=3\sqrt{2}.
\displaystyle CA=\sqrt{(0-4)^2+(7-9)^2+(-10+6)^2}
\displaystyle =\sqrt{(-4)^2+(-2)^2+(-4)^2}=\sqrt{36}=6.
\displaystyle \therefore AB=BC.
\displaystyle \therefore \text{The given points are the vertices of an isosceles triangle.}

\displaystyle \text{(ii) Let }A(0,7,10),\ B(-1,6,6)\text{ and }C(-4,9,6)\text{ be the vertices of }\triangle ABC.
\displaystyle AB=\sqrt{(-1-0)^2+(6-7)^2+(6-10)^2}
\displaystyle =\sqrt{(-1)^2+(-1)^2+(-4)^2}=\sqrt{18}=3\sqrt{2}.
\displaystyle BC=\sqrt{(-4+1)^2+(9-6)^2+(6-6)^2}
\displaystyle =\sqrt{(-3)^2+3^2+0^2}=\sqrt{18}=3\sqrt{2}.
\displaystyle AC=\sqrt{(-4-0)^2+(9-7)^2+(6-10)^2}
\displaystyle =\sqrt{(-4)^2+2^2+(-4)^2}=\sqrt{36}=6.
\displaystyle AB^2+BC^2=18+18=36=AC^2.
\displaystyle \therefore \angle ABC=90^\circ.
\displaystyle \therefore \text{The given points are the vertices of a right-angled triangle.}

\displaystyle \text{(iii) Let }A(-1,2,1),\ B(1,-2,5),\ C(4,-7,8)\text{ and }D(2,-3,4)
\displaystyle \text{be the vertices of quadrilateral }ABCD.
\displaystyle AB=\sqrt{(1+1)^2+(-2-2)^2+(5-1)^2}
\displaystyle =\sqrt{2^2+(-4)^2+4^2}=\sqrt{36}=6.
\displaystyle BC=\sqrt{(4-1)^2+(-7+2)^2+(8-5)^2}
\displaystyle =\sqrt{3^2+(-5)^2+3^2}=\sqrt{43}.
\displaystyle CD=\sqrt{(2-4)^2+(-3+7)^2+(4-8)^2}
\displaystyle =\sqrt{(-2)^2+4^2+(-4)^2}=\sqrt{36}=6.
\displaystyle DA=\sqrt{(-1-2)^2+(2+3)^2+(1-4)^2}
\displaystyle =\sqrt{(-3)^2+5^2+(-3)^2}=\sqrt{43}.
\displaystyle \therefore AB=CD\text{ and }BC=DA.
\displaystyle \text{Since both pairs of opposite sides are equal, }ABCD\text{ is a parallelogram.}

\displaystyle \text{(iv) Let }A(5,-1,1),\ B(7,-4,7),\ C(1,-6,10)\text{ and }D(-1,-3,4)
\displaystyle \text{be the vertices of quadrilateral }ABCD.
\displaystyle AB=\sqrt{(7-5)^2+(-4+1)^2+(7-1)^2}
\displaystyle =\sqrt{2^2+(-3)^2+6^2}=\sqrt{49}=7.
\displaystyle BC=\sqrt{(1-7)^2+(-6+4)^2+(10-7)^2}
\displaystyle =\sqrt{(-6)^2+(-2)^2+3^2}=\sqrt{49}=7.
\displaystyle CD=\sqrt{(-1-1)^2+(-3+6)^2+(4-10)^2}
\displaystyle =\sqrt{(-2)^2+3^2+(-6)^2}=\sqrt{49}=7.
\displaystyle DA=\sqrt{(5+1)^2+(-1+3)^2+(1-4)^2}
\displaystyle =\sqrt{6^2+2^2+(-3)^2}=\sqrt{49}=7.
\displaystyle \therefore AB=CD\text{ and }BC=DA.
\displaystyle \text{Hence both pairs of opposite sides are equal, so }ABCD\text{ is a parallelogram.}
\displaystyle \text{Also, }AB=BC=CD=DA=7.
\displaystyle \therefore \text{The parallelogram }ABCD\text{ has all its sides equal. Hence }ABCD\text{ is a rhombus.}
\displaystyle \\

\displaystyle \textbf{Question 21: } \text{Find the locus of the points which are equidistant from the} \\ \text{points}(1,2,3)\text{ and }(3,2,-1).
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,y,z)\text{ be any point which is equidistant from }A(1,2,3)\text{ and }B(3,2,-1).
\displaystyle \therefore PA=PB.
\displaystyle \sqrt{(x-1)^2+(y-2)^2+(z-3)^2}
\displaystyle =\sqrt{(x-3)^2+(y-2)^2+(z+1)^2}.
\displaystyle \text{Squaring both sides,}
\displaystyle (x-1)^2+(y-2)^2+(z-3)^2=(x-3)^2+(y-2)^2+(z+1)^2.
\displaystyle x^2-2x+1+y^2-4y+4+z^2-6z+9
\displaystyle =x^2-6x+9+y^2-4y+4+z^2+2z+1.
\displaystyle -2x-6z+14=-6x+2z+14.
\displaystyle \Rightarrow 4x-8z=0.
\displaystyle \Rightarrow x-2z=0.
\displaystyle \therefore \text{The required locus is the plane }x-2z=0.
\displaystyle \\

\displaystyle \textbf{Question 22: } \text{Find the locus of the point, the sum of whose distances from}
\displaystyle \text{the points }A(4,0,0) \ \text{and }B(-4,0,0)\text{ is equal to }10.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,y,z)\text{ be any point such that }PA+PB=10.
\displaystyle \sqrt{(x-4)^2+y^2+z^2}+\sqrt{(x+4)^2+y^2+z^2}=10.
\displaystyle \Rightarrow \sqrt{(x+4)^2+y^2+z^2}=10-\sqrt{(x-4)^2+y^2+z^2}.
\displaystyle \Rightarrow \sqrt{x^2+8x+16+y^2+z^2}
\displaystyle =10-\sqrt{x^2-8x+16+y^2+z^2}.
\displaystyle \text{Squaring both sides,}
\displaystyle x^2+8x+16+y^2+z^2
\displaystyle =100+x^2-8x+16+y^2+z^2-20\sqrt{x^2-8x+16+y^2+z^2}.
\displaystyle \Rightarrow 16x-100=-20\sqrt{x^2-8x+16+y^2+z^2}.
\displaystyle \Rightarrow 4x-25=-5\sqrt{x^2-8x+16+y^2+z^2}.
\displaystyle \text{Squaring both sides again,}
\displaystyle (4x-25)^2=25\left(x^2-8x+16+y^2+z^2\right).
\displaystyle 16x^2-200x+625=25x^2-200x+400+25y^2+25z^2.
\displaystyle \Rightarrow 9x^2+25y^2+25z^2-225=0.
\displaystyle \Rightarrow \frac{x^2}{25}+\frac{y^2}{9}+\frac{z^2}{9}=1.
\displaystyle \therefore \text{The required locus is the ellipsoid }\frac{x^2}{25}+\frac{y^2}{9}+\frac{z^2}{9}=1.
\displaystyle \\

\displaystyle \textbf{Question 23: } \text{Show that the points }A(1,2,3),\ B(-1,-2,-1),\ C(2,3,2)\text{ and }D(4,7,6)
\displaystyle \text{are the vertices of a parallelogram }ABCD,\text{ but not a rectangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(1,2,3),\ B(-1,-2,-1),\ C(2,3,2)\text{ and }D(4,7,6)
\displaystyle \text{be the vertices of quadrilateral }ABCD.
\displaystyle AB=\sqrt{(-1-1)^2+(-2-2)^2+(-1-3)^2}
\displaystyle =\sqrt{(-2)^2+(-4)^2+(-4)^2}=\sqrt{36}=6.
\displaystyle BC=\sqrt{(2+1)^2+(3+2)^2+(2+1)^2}
\displaystyle =\sqrt{3^2+5^2+3^2}=\sqrt{43}.
\displaystyle CD=\sqrt{(4-2)^2+(7-3)^2+(6-2)^2}
\displaystyle =\sqrt{2^2+4^2+4^2}=\sqrt{36}=6.
\displaystyle DA=\sqrt{(1-4)^2+(2-7)^2+(3-6)^2}
\displaystyle =\sqrt{(-3)^2+(-5)^2+(-3)^2}=\sqrt{43}.
\displaystyle \therefore AB=CD\text{ and }BC=DA.
\displaystyle \text{Since both pairs of opposite sides are equal, }ABCD\text{ is a parallelogram.}
\displaystyle AC=\sqrt{(2-1)^2+(3-2)^2+(2-3)^2}
\displaystyle =\sqrt{1^2+1^2+(-1)^2}=\sqrt{3}.
\displaystyle BD=\sqrt{(4+1)^2+(7+2)^2+(6+1)^2}
\displaystyle =\sqrt{5^2+9^2+7^2}=\sqrt{155}.
\displaystyle \therefore AC\ne BD.
\displaystyle \text{Since the diagonals of the parallelogram are not equal, }ABCD\text{ is not a rectangle.}
\displaystyle \\

\displaystyle \textbf{Question 24: } \text{Find the equation of the set of points }P\text{ such that its distances}
\displaystyle \text{from the points} \ A(3,4,-5)\text{ and }B(-2,1,4)\text{ are equal.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,y,z)\text{ be any point which is equidistant from }A(3,4,-5)\text{ and }B(-2,1,4).
\displaystyle \therefore PA=PB.
\displaystyle \sqrt{(x-3)^2+(y-4)^2+(z+5)^2}
\displaystyle =\sqrt{(x+2)^2+(y-1)^2+(z-4)^2}.
\displaystyle \text{Squaring both sides,}
\displaystyle (x-3)^2+(y-4)^2+(z+5)^2=(x+2)^2+(y-1)^2+(z-4)^2.
\displaystyle x^2-6x+9+y^2-8y+16+z^2+10z+25
\displaystyle =x^2+4x+4+y^2-2y+1+z^2-8z+16.
\displaystyle \Rightarrow -10x-6y+18z+29=0.
\displaystyle \Rightarrow 10x+6y-18z-29=0.
\displaystyle \therefore \text{The required equation of the locus is }10x+6y-18z-29=0.
\displaystyle \\

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