NCERT Solutions Class 11 Maths Chapter 12

Q.1 A point is on the x -axis. What are its y-coordinate and z-coordinates?

Ans

The coordinate of the point on x-axis is (x, 0, 0). The y-coordinate of this point is 0 and that of z-coordinate is also 0.

Q.2 A point is in the XZ-plane. What can you say about its y-coordinate?

Ans

The y-coordinate of a point in XZ-plane is 0.

Q.3 Name the octants in which the following points lie: (1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (–4, 2, –5), (– 4, 2, 5), (–3, –1, 6) (2, – 4, –7).

Ans

Q.4 Fill in the blanks:

(i) The x-axis and y-axis taken together determine a plane known as_______.
(ii) The coordinates of points in the XY-plane are of the form _______.
(iii) Coordinate planes divide the space into ______ octants.

Ans

(i) The x-axis and y-axis taken together determine a plane known as XY-plane.
(ii) The coordinates of points in the XY-plane are of the form (x, y, 0).
(iii) Coordinate planes divide the space into eight octants.

Q.5 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).

Ans

$\begin{array}{l}\text{Distance between two points}({\mathrm{x}}_1,{\mathrm{y}}_1,{\mathrm{z}}_1)\text{and}({\mathrm{x}}_2,{\mathrm{y}}_2,{\mathrm{z}}_2),\\ \mathrm{d}=\sqrt{{({\mathrm{x}}_2-{\mathrm{x}}_1)}^2+{({\mathrm{y}}_2-{\mathrm{y}}_1)}^2+{({\mathrm{z}}_2-{\mathrm{z}}_1)}^2}\\ \left(\mathrm{i}\right)(2,3,5)\mathrm{and}(4,3,1)\\ \mathrm{d}=\sqrt{{(4-2)}^2+{(3-3)}^2+{(1-5)}^2}\\ =\sqrt{4+0+16}\\ =\sqrt{20}\mathrm{unit}\\ =2\sqrt{5}\mathrm{unit}\end{array}$ $\begin{array}{l}\left(\mathrm{ii}\right)(–3,7,2)\mathrm{and}(2,4,–1)\\ \mathrm{d}=\sqrt{{(2+3)}^2+{(4-7)}^2+{(2+1)}^2}\\ =\sqrt{25+9+9}\\ =\sqrt{43}\mathrm{unit}\\ \left(\mathrm{iii}\right)(-1,3,-4)\mathrm{and}(1,-3,4)\\ \mathrm{d}=\sqrt{{(1+1)}^2+{(-3-3)}^2+{(4+4)}^2}\\ =\sqrt{4+36+64}\\ =\sqrt{104}\mathrm{unit}\\ =2\sqrt{26}\mathrm{unit}\\ \left(\mathrm{iv}\right)(2,-1,3)\mathrm{and}(-2,1,3)\\ \mathrm{d}=\sqrt{{(-2-2)}^2+{(1+1)}^2+{(3-3)}^2}\\ =\sqrt{16+4+0}\\ =\sqrt{20}\mathrm{unit}\\ =2\sqrt{5}\mathrm{unit}\end{array}$

Q.6 Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.

Ans

$\begin{array}{l}\mathrm{Let}\text{three points be A}(-2,3,5),\mathrm{B}(\text{1},\text{2},\text{3})\text{and}\mathrm{}\mathrm{C}(7,0,–1).\mathrm{Then},\\ \mathrm{AB}=\sqrt{{(1+2)}^2+{(2-3)}^2+{(3-5)}^2}\\ =\sqrt{9+1+4}\\ =\sqrt{14}\\ =3.741\end{array}$ $\begin{array}{l}\mathrm{BC}=\sqrt{{(7-1)}^2+{(0-2)}^2+{(-1-3)}^2}\\ =\sqrt{36+4+16}\\ =\sqrt{56}\\ =14.966\\ \mathrm{CA}=\sqrt{{(-2-7)}^2+{(0-3)}^2+{(-1-5)}^2}\\ =\sqrt{81+9+36}\\ =\sqrt{126}\\ =11.225\\ \therefore \mathrm{AB}+\mathrm{CA}=3.741+11.225\\ =14.966=\mathrm{BC}\\ \mathrm{Thus},\text{the points A, B and C are collinear.}\end{array}$

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

Ans

$\begin{array}{l}\left(\text{i}\right)\text{\hspace{0.33em}Let\hspace{0.33em}three\hspace{0.33em}points\hspace{0.33em}be\hspace{0.33em}A}\left(\text{0, 7,–10}\right)\text{,B}\left(\text{1, 6,– 6}\right)\text{\hspace{0.33em}and\hspace{0.33em}C}\left(\text{4, 9,– 6}\right)\text{.}\\ \text{Then,}\\ \text{\hspace{0.33em}\hspace{0.33em}AB\hspace{0.33em}=\hspace{0.33em}}\sqrt{{\left(\text{1-0}\right)}^{\text{2}}\text{+}{\left(\text{6-7}\right)}^{\text{2}}\text{+}{\left(\text{-\hspace{0.17em}6+10}\right)}^{\text{2}}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}}\sqrt{\text{1+1+16}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}}\sqrt{\text{18}}\\ \text{\hspace{0.33em}\hspace{0.33em}BC\hspace{0.33em}=\hspace{0.33em}}\sqrt{{\left(\text{4-1}\right)}^{\text{2}}\text{+}{\left(\text{9-6}\right)}^{\text{2}}\text{+}{\left(\text{-\hspace{0.17em}6+6}\right)}^{\text{2}}}\end{array}$ $\begin{array}{l}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}}\sqrt{\text{9+9+0}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}}\sqrt{\text{18}}\\ \text{\hspace{0.33em}\hspace{0.33em}CA\hspace{0.33em}=\hspace{0.33em}}\sqrt{{\left(\text{4-0}\right)}^{\text{2}}\text{+}{\left(\text{9-7}\right)}^{\text{2}}\text{+}{\left(\text{-\hspace{0.17em}6+10}\right)}^{\text{2}}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}}\sqrt{\text{16+4+16}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}}\sqrt{\text{36}}\\ \text{Since,\hspace{0.33em}AB\hspace{0.33em}=\hspace{0.33em}BC\hspace{0.33em}=\hspace{0.33em}}\sqrt{\text{18}}\text{.}\\ \text{So,\hspace{0.33em}}\left(\text{0,7,–10}\right)\text{,}\left(\text{1,6,– 6}\right)\text{\hspace{0.33em}and\hspace{0.33em}}\left(\text{4,9,– 6}\right)\text{\hspace{0.33em}are\hspace{0.33em}the\hspace{0.33em}vertices\hspace{0.33em}of}\\ \text{an\hspace{0.33em}isosceles\hspace{0.33em}triangle.}\\ \left(\text{ii}\right)\text{\hspace{0.33em}Let\hspace{0.33em}three\hspace{0.33em}points\hspace{0.33em}be\hspace{0.33em}A}\left(\text{0, 7, 10}\right)\text{,B}\left(\text{–1, 6, 6}\right)\text{\hspace{0.33em}and\hspace{0.33em}C}\left(\text{– 4, 9, 6}\right)\text{.\hspace{0.33em}}\\ \text{Then,}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}AB\hspace{0.33em}=\hspace{0.33em}}\sqrt{{\left(\text{-1-0}\right)}^{\text{2}}\text{+}{\left(\text{6-7}\right)}^{\text{2}}\text{+}{\left(\text{6-10}\right)}^{\text{2}}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}}\sqrt{\text{1+1+16}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}}\sqrt{\text{18}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}BC\hspace{0.33em}=\hspace{0.33em}}\sqrt{{\left(\text{-\hspace{0.17em}4+1}\right)}^{\text{2}}\text{+}{\left(\text{9-6}\right)}^{\text{2}}\text{+}{\left(\text{\hspace{0.17em}6-6}\right)}^{\text{2}}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}}\sqrt{\text{9+9+0}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}}\sqrt{\text{18}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}CA\hspace{0.33em}=\hspace{0.33em}}\sqrt{{\left(\text{-\hspace{0.17em}4-0}\right)}^{\text{2}}\text{+}{\left(\text{9-7}\right)}^{\text{2}}\text{+}{\left(\text{\hspace{0.17em}6-10}\right)}^{\text{2}}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}}\sqrt{\text{16+4+16}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}}\sqrt{\text{36}}\\ \text{Since,}\\ {\text{AB}}^{\text{2}}{\text{+BC}}^{\text{2}}\text{\hspace{0.33em}=\hspace{0.33em}18+18}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}36}\\ {\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}CA}}^{\text{2}}\end{array}$ $\begin{array}{l}\text{Therefore,\hspace{0.33em}\hspace{0.17em}}\left(\text{0, 7, 10}\right)\text{,}\left(\text{–1, 6, 6}\right)\text{\hspace{0.33em}and\hspace{0.33em}}\left(\text{– 4, 9, 6}\right)\text{\hspace{0.33em}are\hspace{0.33em}the \hspace{0.17em}}\\ \text{vertices\hspace{0.33em}of\hspace{0.33em}a\hspace{0.33em}right\hspace{0.33em}angled\hspace{0.33em}triangle.}\\ \left(\text{iii}\right)\text{\hspace{0.33em}Let\hspace{0.33em}four\hspace{0.33em}points\hspace{0.33em}be\hspace{0.33em}A}\left(\text{–1, 2, 1}\right)\text{, B}\left(\text{1,–2, 5}\right)\text{, C}\left(\text{4,–7, 8}\right)\\ \text{and\hspace{0.33em}D}\left(\text{2,–3,4}\right)\text{.\hspace{0.33em}Then,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB\hspace{0.33em}=\hspace{0.33em}}\sqrt{{\left(\text{1+1}\right)}^{\text{2}}\text{+}{\left(\text{-2-2}\right)}^{\text{2}}\text{+}{\left(\text{5-1}\right)}^{\text{2}}}\\ \text{=\hspace{0.33em}}\sqrt{\text{4+16+16}}\\ \text{=\hspace{0.33em}}\sqrt{\text{36}}\text{=6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}BC\hspace{0.33em}=\hspace{0.33em}}\sqrt{{\left(\text{4-1}\right)}^{\text{2}}\text{+}{\left(\text{-7+2}\right)}^{\text{2}}\text{+}{\left(\text{8-5}\right)}^{\text{2}}}\\ \text{=\hspace{0.33em}}\sqrt{\text{9+25+9}}\\ \text{=\hspace{0.33em}}\sqrt{\text{43}}\\ \text{\hspace{0.17em}\hspace{0.17em}CD\hspace{0.33em}=\hspace{0.33em}}\sqrt{{\left(\text{2-4}\right)}^{\text{2}}\text{+}{\left(\text{-3+7}\right)}^{\text{2}}\text{+}{\left(\text{4-8}\right)}^{\text{2}}}\\ \text{=\hspace{0.33em}}\sqrt{\text{4+16+16}}\\ \text{=\hspace{0.33em}}\sqrt{\text{36}}\\ \text{\hspace{0.17em}\hspace{0.17em}DA\hspace{0.33em}=\hspace{0.33em}}\sqrt{{\left(\text{-1-2}\right)}^{\text{2}}\text{+}{\left(\text{2+3}\right)}^{\text{2}}\text{+}{\left(\text{1-4}\right)}^{\text{2}}}\\ \text{=\hspace{0.33em}}\sqrt{\text{9+25+9}}\\ \text{=\hspace{0.33em}}\sqrt{\text{43}}\\ \text{Since,\hspace{0.33em}opposite\hspace{0.33em}sides\hspace{0.33em}AB and CD,\hspace{0.33em}BC\hspace{0.33em}and\hspace{0.33em}DA\hspace{0.33em}are\hspace{0.33em}equal.}\\ \text{So,\hspace{0.33em}ABCD\hspace{0.33em}is\hspace{0.33em}a\hspace{0.33em}parallelogram.}\\ \text{Therefore,\hspace{0.33em}}\left(\text{–1,2,1}\right)\text{,}\left(\text{1,–2,5}\right)\text{,}\left(\text{4,–7,8}\right)\text{\hspace{0.33em}and\hspace{0.33em}}\left(\text{2,–3,4}\right)\\ \text{are\hspace{0.33em}the\hspace{0.33em}vertices\hspace{0.33em}of\hspace{0.33em}a\hspace{0.33em}parallelogram.}\end{array}$

Q.8 Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

Ans

$\begin{array}{l}\mathrm{Let}\text{the point P}(\mathrm{x},\mathrm{y},\mathrm{z})\text{are equidistant from the points Q}(\text{1},\text{2},\text{3})\\ \text{and R}(\text{3},\text{2},–\text{1}).\text{Then,}\\ \mathrm{PQ}=\sqrt{{(\mathrm{x}-1)}^2+{(\mathrm{y}-2)}^2+{(\mathrm{z}-3)}^2}\\ =\sqrt{{\mathrm{x}}^2-2\mathrm{x}+1+{\mathrm{y}}^2-4\mathrm{y}+4+{\mathrm{z}}^2-6\mathrm{z}+9}\\ =\sqrt{{\mathrm{x}}^2-2\mathrm{x}+{\mathrm{y}}^2-4\mathrm{y}+{\mathrm{z}}^2-6\mathrm{z}+14}\\ \mathrm{PR}=\sqrt{{(\mathrm{x}-3)}^2+{(\mathrm{y}-2)}^2+{(\mathrm{z}+1)}^2}\\ =\sqrt{{\mathrm{x}}^2-6\mathrm{x}+9+{\mathrm{y}}^2-4\mathrm{y}+4+{\mathrm{z}}^2+2\mathrm{z}+1}\\ =\sqrt{{\mathrm{x}}^2-6\mathrm{x}+{\mathrm{y}}^2-4\mathrm{y}+{\mathrm{z}}^2+2\mathrm{z}+14}\\ \mathrm{According}\text{to given condition,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} PQ}=\text{PR}\\ \Rightarrow {\text{\hspace{0.17em}PQ}}^{\text{2}}={\text{PR}}^2\\ {\mathrm{x}}^2-2\mathrm{x}+{\mathrm{y}}^2-4\mathrm{y}+{\mathrm{z}}^2-6\mathrm{z}+14\\ ={\mathrm{x}}^2-6\mathrm{x}+{\mathrm{y}}^2-4\mathrm{y}+{\mathrm{z}}^2+2\mathrm{z}+14\\ 6\mathrm{x}-2\mathrm{x}-2\mathrm{z}-6\mathrm{z}=0\\ 4\mathrm{x}-8\mathrm{z}=0\\ \mathrm{x}-2\mathrm{z}=0,\text{which is the required equation of the set of points}\\ \text{which are equidistant from the points}(1,2,3)\text{and}(3,2,-1).\end{array}$

Q.9 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.

Ans

$\mathrm{Let}\text{the sum of distances of point P}(\mathrm{x},\mathrm{y},\mathrm{z})\text{from the points Q}(\text{4},0,0)\text{and R}(–\text{4},0,0,)\text{is 1}0.\text{Then}$ $\begin{array}{l}\mathrm{PQ}=\sqrt{{(\mathrm{x}-4)}^2+{(\mathrm{y}-0)}^2+{(\mathrm{z}-0)}^2}\\ =\sqrt{{\mathrm{x}}^2-8\mathrm{x}+16+{\mathrm{y}}^2+{\mathrm{z}}^2}\\ \mathrm{PR}=\sqrt{{(\mathrm{x}+4)}^2+{(\mathrm{y}-0)}^2+{(\mathrm{z}-0)}^2}\\ =\sqrt{{\mathrm{x}}^2+8\mathrm{x}+16+{\mathrm{y}}^2+{\mathrm{z}}^2}\\ \mathrm{According}\text{to given condition,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} PQ}+\text{PR}=\text{10}\\ \Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PQ}}^{\text{2}}={(10-\text{PR})}^2\\ \Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PQ}}^{\text{2}}=100-20\mathrm{PR}+{\mathrm{PR}}^2\\ 20\mathrm{PR}-100={\mathrm{x}}^2+8\mathrm{x}+16+{\mathrm{y}}^2+{\mathrm{z}}^2-{\mathrm{x}}^2+8\mathrm{x}-16-{\mathrm{y}}^2-{\mathrm{z}}^2\\ 20\mathrm{PR}-100=16\mathrm{x}\\ 20\mathrm{PR}=100+16\mathrm{x}\\ \Rightarrow 5\mathrm{PR}=25+4\mathrm{x}\\ \mathrm{Squarring}\text{both sides, we get}\\ {\left(5\mathrm{PR}\right)}^2={(25+4\mathrm{x})}^2\\ 25{\mathrm{PR}}^2=625+200\mathrm{x}+16{\mathrm{x}}^2\\ 25({\mathrm{x}}^2+8\mathrm{x}+16+{\mathrm{y}}^2+{\mathrm{z}}^2)=625+200\mathrm{x}+16{\mathrm{x}}^2\\ 25{\mathrm{x}}^2+200\mathrm{x}+400+25{\mathrm{y}}^2+25{\mathrm{z}}^2=625+200\mathrm{x}+16{\mathrm{x}}^2\\ 9{\mathrm{x}}^2+25{\mathrm{y}}^2+25{\mathrm{z}}^2-225=0\end{array}$

Q.10 Find the coordinates of the point which divides the line segment joining the points (– 2, 3, 5) and (1, – 4, 6) in the ratio

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Let}\text{point P}(\mathrm{x},\mathrm{y},\mathrm{z})\text{divides the line segment joining the points}\\ (-\text{\hspace{0.17em}2},\text{3},\text{5})\text{and}(\text{1},-\text{\hspace{0.17em}4},\text{6})\text{in the ratio 2:3 internally.Then,}\\ \mathrm{x}=\frac{{\mathrm{mx}}_2+{\mathrm{nx}}_1}{\mathrm{m}+\mathrm{n}}\\ =\frac{2\left(1\right)+3(-2)}{2+3}=\frac{-4}{5}\\ \mathrm{y}=\frac{{\mathrm{my}}_2+{\mathrm{ny}}_1}{\mathrm{m}+\mathrm{n}}\\ =\frac{2(-4)+3\left(3\right)}{2+3}\\ =\frac{1}{5}\\ \mathrm{z}=\frac{{\mathrm{mz}}_2+{\mathrm{nz}}_1}{\mathrm{m}+\mathrm{n}}\\ =\frac{2\left(6\right)+3\left(5\right)}{2+3}\\ =\frac{27}{5}\\ \mathrm{Therefore},\text{the coordinates of the point P are}(-\frac{4}{5},\frac{1}{5},\frac{27}{5}).\\ \left(\mathrm{ii}\right)\mathrm{Let}\text{point P}(\mathrm{x},\mathrm{y},\mathrm{z})\text{divides the line segment joining the points}\\ (-\text{\hspace{0.17em}2},\text{3},\text{5})\text{and}(\text{1},-\text{\hspace{0.17em}4},\text{6})\text{in the ratio 2:3 externally.Then,}\\ \mathrm{x}=\frac{{\mathrm{mx}}_2-{\mathrm{nx}}_1}{\mathrm{m}-\mathrm{n}}\\ =\frac{2\left(1\right)-3(-2)}{2-3}\end{array}$ $\begin{array}{l}=\frac{8}{-1}=-8\\ \mathrm{y}=\frac{{\mathrm{my}}_2-{\mathrm{ny}}_1}{\mathrm{m}-\mathrm{n}}\\ =\frac{2(-4)-3\left(3\right)}{2-3}\\ =\frac{-17}{-1}=17\\ \mathrm{z}=\frac{{\mathrm{mz}}_2-{\mathrm{nz}}_1}{\mathrm{m}-\mathrm{n}}\\ =\frac{2\left(6\right)-3\left(5\right)}{2-3}\\ =\frac{-3}{-1}=3\\ \mathrm{Therefore},\text{the coordinates of the point P are}(-8,17,3).\end{array}$

Q.11 Given that P (3, 2, – 4), Q (5, 4, – 6) and R (9, 8, –10) are collinear. Find the ratio in which Q divides PR.

Ans

$\begin{array}{l}\mathrm{P}\text{oint}\mathrm{Q}(5,4,-6)\text{divides the line segment joining the points}\\ \mathrm{P}(3,2,-4)\mathrm{}\text{and}\mathrm{R}(9,8,-10)\text{in the ratio k:1 internally.Then,}\\ \mathrm{x}=\frac{{\mathrm{mx}}_2+{\mathrm{nx}}_1}{\mathrm{m}+\mathrm{n}}\\ 5=\frac{\mathrm{k}\left(9\right)+1\left(3\right)}{\mathrm{k}+1}\\ 5(\mathrm{k}+1)=9\mathrm{k}+3\\ 5\mathrm{k}+5=9\mathrm{k}+3\\ 5-3=9\mathrm{k}-5\mathrm{k}\end{array}$ $\begin{array}{l}\Rightarrow 2=4\mathrm{k}\\ \Rightarrow \mathrm{k}=\frac{2}{4}\\ \Rightarrow \mathrm{k}=\frac{1}{2}\\ \mathrm{The}\text{required ratio is}\frac{1}{2}:1\text{i.e., 1:2.}\end{array}$

Q.12 Find the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8).

Ans

$\begin{array}{l}\mathrm{Plane}\mathrm{YZ}\text{divides the line segment joining the points}\\ \mathrm{Q}(-2,4,7)\mathrm{}\text{\hspace{0.17em}and}\mathrm{R}(3,-5,8)\text{in the ratio k:1 internally.Then,}\\ \text{let coordinates of point P in YZ plane:}(0,\mathrm{y},\mathrm{z})\\ \mathrm{x}=\frac{{\mathrm{mx}}_2+{\mathrm{nx}}_1}{\mathrm{m}+\mathrm{n}}\\ 0=\frac{\mathrm{k}\left(3\right)+1(-2)}{\mathrm{k}+1}\\ 0=3\mathrm{k}-2\\ 2=3\mathrm{k}\\ \mathrm{k}=\frac{2}{3}\\ \mathrm{The}\text{required ratio is}\frac{2}{3}:1\text{i.e., 2:3.}\end{array}$

Q.13

$\begin{array}{l}\mathbf{\text{Using section formula, show that the points A(2,}}\mathbf{-}\mathbf{\text{3, 4), B(}}\mathbf{-}\mathbf{\text{1, 2, 1)}}\\ \mathbf{\text{and C}}\left(\text{0,}\frac{\text{1}}{\text{3}}\text{,2}\right)\mathbf{\text{are collinear.}}\end{array}$

Ans

$\begin{array}{l}\text{The given points are A(2,}-\text{3, 4), B(}-\text{1, 2, 1) and C}\left(\text{0,}\frac{\text{1}}{\text{3}}\text{, 2}\right)\text{.}\\ \text{Let a point P divides ABin the ratio k:1.}\\ \text{Hence, by the section formula, the coordinates of P are:}\\ \left(\frac{{\text{mx}}_{\text{2}}{\text{+nx}}_{\text{1}}}{\text{m+n}}\text{,}\frac{{\text{my}}_{\text{2}}{\text{+ny}}_{\text{1}}}{\text{m+n}}\text{,}\frac{{\text{mz}}_{\text{2}}{\text{+nz}}_{\text{1}}}{\text{m+n}}\right)\\ \left(\frac{\text{k}\left(-\text{1}\right)\text{+2}}{\text{k+1}}\text{,}\frac{\text{k}\left(\text{2}\right)-\text{3}}{\text{k+1}}\text{,}\frac{\text{k}\left(\text{1}\right)\text{+4}}{\text{k+1}}\right)\\ \text{Now, we find the value of k at which P coincides with point C.}\\ \text{So, by taking}\frac{-\text{k+2}}{\text{k+1}}=0\\ \Rightarrow \text{k =2}\\ \text{For k=2, the coordinates of P are:}\left(\text{0,}\frac{\text{1}}{\text{3}}\text{, 2}\right)\\ \text{Therefore, C}\left(\text{0,}\frac{\text{1}}{\text{3}}\text{,2}\right)\text{ \hspace{0.17em}\hspace{0.17em}is a point that divides AB externally in ratio 2:1.}\\ \text{Hence, points A, B and C are collinear.}\end{array}$

Q.14 Find the coordinates of the points which trisect the line segment joining the points P(4, 2, – 6) and Q(10, –16, 6).

Ans

Let points A and B trisect the line segment joining the points P(4, 2, – 6) and Q(10, –16, 6). Point A divides the line segment in the ratio of 1:2 and point B divides the line segment in the ratio of 2:1.

$\begin{array}{l}\mathrm{coordinates}\text{of point A}=\{\frac{1\left(10\right)+2\left(4\right)}{1+2},\frac{1(-16)+2\left(2\right)}{1+2},\frac{1\left(6\right)+2(-6)}{1+2}\}\\ =(\frac{10+8}{3},\frac{-16+4}{3},\frac{6-12}{3})\\ =(6,-4,-2)\\ \mathrm{coordinates}\text{of point B}=\{\frac{2\left(10\right)+1\left(4\right)}{1+2},\frac{2(-16)+1\left(2\right)}{1+2},\frac{2\left(6\right)+1(-6)}{1+2}\}\\ =(\frac{20+4}{3},\frac{-32+2}{3},\frac{12-6}{3})\\ =(8,-10,2)\\ \mathrm{Thus},\text{points}(6,-4,-2)\text{and}(8,-10,2)\text{trisect the line segment}\\ \text{formed by the given points.}\end{array}$

Q.15 Three vertices of a parallelogram ABCD are A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2). Find the coordinates of the fourth vertex.

Ans

$\begin{array}{l}\text{Three vertices of a parallelogram ABCD are}\mathrm{}\text{\hspace{0.17em}A}(\text{3},-\text{1},\text{2}),\\ \text{B}(\text{1},\text{2},-\text{\hspace{0.17em}4})\text{and C}(-\text{1},\text{1},\text{2}).\mathrm{L}\text{et fourth vertex be D}(\mathrm{a},\mathrm{b},\mathrm{c}).\\ \mathrm{Since},\text{diagonals of a parallelogram bisect each other.}\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} Mid-point of AC}=\mathrm{Mid}\text{–}\mathrm{point}\text{of BD}\\ \Rightarrow (\frac{3-1}{2},\frac{-1+1}{2},\frac{2+2}{2})=(\frac{1+\mathrm{a}}{2},\frac{2+\mathrm{b}}{2},\frac{-4+\mathrm{c}}{2})\\ \Rightarrow (\frac{2}{2},\frac{0}{2},\frac{4}{2})=(\frac{1+\mathrm{a}}{2},\frac{2+\mathrm{b}}{2},\frac{-4+\mathrm{c}}{2})\\ \Rightarrow \frac{2}{2}=\frac{1+\mathrm{a}}{2},\frac{0}{2}=\frac{2+\mathrm{b}}{2}\text{and}\frac{4}{2}=\frac{-4+\mathrm{c}}{2}\\ \Rightarrow 2=1+\mathrm{a},0=2+\mathrm{b}\text{and 4}=-4+\mathrm{c}\\ \Rightarrow 1=\mathrm{a},-2=\mathrm{b}\text{and c}=\text{8}\\ \mathrm{Therefore},\text{the coordinates of vertes D is}(1,-2,8).\end{array}$

Q.16 Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).

Ans

$\begin{array}{l}\text{The triangle with vertices A}\left(\text{0, 0, 6}\right)\text{, B}\left(\text{0, 4, 0}\right)\text{and C}\left(\text{6, 0, 0}\right)\\ \text{is given.}\\ \text{Mid-point of AB}=(\frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2})\end{array}$ $\begin{array}{l}=(0,2,3)\\ \text{Mid-point of BC}=(\frac{0+6}{2},\frac{4+0}{2},\frac{0+0}{2})\\ =(3,2,0)\\ \text{Mid-point of CA}=(\frac{6+0}{2},\frac{0+0}{2},\frac{0+6}{2})\\ =(3,0,3)\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}\mathrm{Length}\text{of median AD}=\sqrt{{(0-3)}^2+{(0-2)}^2+{(6-0)}^2}\\ =\sqrt{9+4+36}\\ =7\\ \mathrm{Length}\text{of median BE}=\sqrt{{(0-3)}^2+{(4-0)}^2+{(0-3)}^2}\\ =\sqrt{9+16+9}\\ =\sqrt{34}\\ \mathrm{Length}\text{of median CF}=\sqrt{{(6-0)}^2+{(0-2)}^2+{(0-3)}^2}\\ =\sqrt{36+4+9}\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}=\sqrt{49}\\ =7\\ \mathrm{Thus},\text{the length of medians are 7,}\sqrt{34}\text{and 7 unit.}\end{array}$

Q.17 If the origin is the centroid of the triangle PQR with vertices P(2a, 2, 6), Q(– 4, 3b, –10) and R(8, 14, 2c), then find the values of a, b and c.

Ans

$\begin{array}{l}\mathrm{Since},\text{the origin}(0,0,0)\text{is the centroid of the triangle PQR}\mathrm{}\text{with}\\ \text{vertices P}(\text{2a},\text{2},\text{6}),\text{Q}(-\text{\hspace{0.17em}4},\text{3b},-\text{1}0)\text{and R}(\text{8},\text{14},\text{2c}).\\ \mathrm{Then},\\ \{\frac{2\mathrm{a}+(-4)+8}{3},\frac{2+3\mathrm{b}+14}{3},\frac{6+(-10)+2\mathrm{c}}{3}\}=(0,0,0)\\ \Rightarrow \{\frac{2\mathrm{a}-4+8}{3},\frac{2+3\mathrm{b}+14}{3},\frac{6-10+2\mathrm{c}}{3}\}=(0,0,0)\\ \Rightarrow \{\frac{2\mathrm{a}+4}{3},\frac{3\mathrm{b}+16}{3},\frac{2\mathrm{c}-4}{3}\}=(0,0,0)\\ \Rightarrow \frac{2\mathrm{a}+4}{3}=0,\frac{3\mathrm{b}+16}{3}=0\text{and}\frac{2\mathrm{c}-4}{3}=0\\ \Rightarrow \mathrm{a}=-2,\mathrm{b}=-\frac{16}{3}\text{and c}=2\end{array}$

Q.18

$\begin{array}{l}\text{Find the coordinates of a point on\hspace{0.17em}\hspace{0.17em}y-axis which\hspace{0.17em}are at a distance of}\\ \text{5}\sqrt{\text{2}}\text{\hspace{0.17em}\hspace{0.17em}from\hspace{0.17em}the point P}\left(\text{3,}-\text{2,5}\right)\text{.}\end{array}$

Ans

$\mathrm{Let}\text{the point on y-axis be Q}(0,\mathrm{b},0)\text{and distance of Q from the point P}(3,-2,5)\mathrm{is}\text{5}\sqrt{2}.\text{Then,}$ $\begin{array}{l}\mathrm{PQ}=\sqrt{{(3-0)}^2+{(-2-\mathrm{b})}^2+{(5-0)}^2}\\ \Rightarrow \text{\hspace{0.17em}5}\sqrt{2}=\sqrt{9+{(-2-\mathrm{b})}^2+25}\\ \mathrm{Squaring}\text{both sides, we get}\\ \Rightarrow \text{50}=34+{(-2-\mathrm{b})}^2\\ \Rightarrow 16={(2+\mathrm{b})}^2\\ \Rightarrow 4^2={(2+\mathrm{b})}^2\\ \Rightarrow 4=2+\mathrm{b}\text{or}-\text{4}=2+\mathrm{b}\\ \Rightarrow \mathrm{b}=2,-6\\ \mathrm{Therefore},\text{coordinates of the point Q\hspace{0.17em}on y-axis are}\\ (0,2,0)\text{\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}}(0,-6,0).\end{array}$

Q.19 A point R with x-coordinate 4 lies on the line segment joining the points P(2, –3, 4) and Q (8, 0, 10). Find the coordinates of the point R.

Ans

$\begin{array}{l}\mathrm{The}\text{coordinates of points P}(\text{2},-\text{3},\text{4})\text{and}\mathrm{}\text{\hspace{0.17em}Q}(\text{8},0,\text{1}0)\text{of line}\\ \text{segment PQ. Let x-cordinate 4 divides PQ in the ratio of k:1.}\\ \text{Then, x}=\frac{{\mathrm{mx}}_2+{\mathrm{nx}}_1}{\mathrm{m}+\mathrm{n}},\text{\hspace{0.17em}\hspace{0.17em}y}=\frac{{\mathrm{my}}_2+{\mathrm{ny}}_1}{\mathrm{m}+\mathrm{n}},\text{z}=\frac{{\mathrm{mz}}_2+{\mathrm{nz}}_1}{\mathrm{m}+\mathrm{n}}\\ \Rightarrow 4=\frac{\mathrm{m}\left(8\right)+\mathrm{n}\left(2\right)}{\mathrm{m}+\mathrm{n}}\\ \Rightarrow 4(\mathrm{m}+\mathrm{n})=8\mathrm{m}+2\mathrm{n}\\ \Rightarrow 4\mathrm{m}+4\mathrm{n}=8\mathrm{m}+2\mathrm{n}\\ \Rightarrow 2\mathrm{n}=4\mathrm{m}\\ \Rightarrow \frac{\mathrm{m}}{\mathrm{n}}=\frac{2}{4}\end{array}$ $\begin{array}{l}\Rightarrow \mathrm{m}:\mathrm{n}=1:2\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} y}=\frac{1\left(0\right)+2(-3)}{1+2}\\ =\frac{-6}{3}\\ =-2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} z}=\frac{1\left(10\right)+2\left(4\right)}{1+2}\\ =\frac{18}{3}\\ =6\\ \mathrm{Therefore},\text{the coordinates of point R are}(4,-2,6).\end{array}$

Q.20 If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that PA² + PB² = k², where k is a constant.

Ans

$\begin{array}{l}\text{Let the coordinates of point P be}\left(\text{x,y,z}\right)\text{.}\\ \text{Coordinates of point A and B are}\left(3,4,5\right)\text{and}\\ \left(-1,3,-7\right)\text{respectively.}\\ {\text{\hspace{0.17em}\hspace{0.17em}PA}}^{\text{2}}={\left(\mathrm{x}-3\right)}^2+{\left(\mathrm{y}-4\right)}^2+{\left(\mathrm{z}-5\right)}^2\\ ={\mathrm{x}}^2-6\mathrm{x}+9+{\mathrm{y}}^2-8\mathrm{y}+16+{\mathrm{z}}^2-10\mathrm{z}+25\\ ={\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2-6\mathrm{x}-8\mathrm{y}-10\mathrm{z}+50\\ {\text{\hspace{0.17em}\hspace{0.17em}PB}}^{\text{2}}={\left(\mathrm{x}+1\right)}^2+{\left(\mathrm{y}-3\right)}^2+{\left(\mathrm{z}+7\right)}^2\\ ={\mathrm{x}}^2+2\mathrm{x}+1+{\mathrm{y}}^2-6\mathrm{y}+9+{\mathrm{z}}^2+14\mathrm{z}+49\\ ={\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2+2\mathrm{x}-6\mathrm{y}+14\mathrm{z}+59\\ \text{According to given condition:}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PA}}^{\text{2}}+{\mathrm{PB}}^2={\mathrm{k}}^{\text{2}}\\ {\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2-6\mathrm{x}-8\mathrm{y}-10\mathrm{z}+50+{\mathrm{x}}^2+{\mathrm{y}}^2\\ +{\mathrm{z}}^2+2\mathrm{x}-6\mathrm{y}+14\mathrm{z}+59={\mathrm{k}}^2\\ 2{\mathrm{x}}^2+2{\mathrm{y}}^2+2{\mathrm{z}}^2-4\mathrm{x}-14\mathrm{y}+4\mathrm{z}+109={\mathrm{k}}^2\\ {\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2-2\mathrm{x}-7\mathrm{y}+2\mathrm{z}=\frac{{\mathrm{k}}^2-109}{2},\\ \mathrm{which}\text{is the required equation.}\end{array}$