NCERT Solutions for Class 11 Maths Chapter 12

Exercise 12.1

1. A point is on the $\mathbf{x}$- axis. What are its $\mathbf{y}$-coordinate and $\mathbf{z}$ -coordinates?

Ans: When a point is on the $x$-axis, then the $y$-coordinate and $z$-coordinate of that point are both zero.

2. A point is in the $\mathbf{xz}$-plane. What can you say about its $\mathbf{y}$-coordinate?

Ans: When a point is on the $xz$-plane, then $y$-coordinate of that point is zero.

3. Name the octant in which the following points lie:

$\left( \mathbf{1,2,3} \right)\mathbf{,}\left( \mathbf{4,-2,3} \right)\mathbf{,}\left( \mathbf{4,-2,-5} \right)\mathbf{,}\left( \mathbf{4,2,-5} \right)$, $\left( -\mathbf{4},\mathbf{2},-\mathbf{5} \right),\left( -\mathbf{4},\mathbf{2},\mathbf{5} \right),$ $\left( -\mathbf{3},-\mathbf{1},\mathbf{6} \right),$$\left( -\mathbf{2},-\mathbf{4},-\mathbf{7} \right)$.

Ans: Consider the following table.

By following rules given in the above table, we can conclude the following results.

Since, all the three coordinates in the point $\left( 1,2,3 \right)$ are positive, so this point is in the octant $I$.

Since in the point $\left( 4,-2,3 \right)$, the $x$ and $z$-coordinate are positive and the $y$-coordinate is negative, so this point is in the octant $IV$.

Since in the point $\left( 4,-2,-5 \right)$, the $y$ and $z$-coordinate are negative and the $x$-coordinate is positive, so this point is in the octant $VIII$.

Since in the point $\left( 4,2,-5 \right)$, the $x$ and $y$-coordinate are positive and the $z$-coordinate is negative, so this point is in the octant $V$.

Since in the point $\left( -4,2,-5 \right)$, the $x$ and $z$-coordinate are negative and the $y$-coordinate is positive, so this point is in the octant $VI$.

Since in the point $\left( -3,-1,6 \right)$, the $x$ and $y$-coordinate are negative and the $z$-coordinate is positive, so this point is in the octant $II$.

Since in the point $\left( -2,-4,-7 \right)$, all the three coordinates in the point are negative, so this point is in the octant $VII$.

4. Fill in the following blanks:

(i) The $\mathbf{x}$-axis and $\mathbf{y}$-axis taken together determine a plane known as ________.

Ans: The $x$-axis and $y$-axis taken together determine a plane known as $XY$ plane.

(ii) The coordinates of points in the $\mathbf{XY}$-plane are of the form __________.

Ans: The coordinates of points in the  $XY$-plane are of the form $\left( x,y,0 \right)$.

(iii) Coordinate planes divided the space into ________ octants.

Ans: Coordinate planes divided the space into eight octants.

Exercise 12.2

1. Find the distance between the following pairs of points:

(i) $\left( \mathbf{2,3,5} \right)$ and $\left( \mathbf{4,3,1} \right)$.

Ans: Recall that, distance between any two points $P\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and \[Q\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)\] is $PQ=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$

Therefore, distance between the points $\left( 2,3,5 \right)$ and $\left( 4,3,1 \right)$ is 

$  =\sqrt{{{\left( 4-2 \right)}^{2}}+{{\left( 3-3 \right)}^{2}}+{{\left( 1-5 \right)}^{2}}}$

$ =\sqrt{{{2}^{2}}+{{0}^{2}}+{{\left( -4 \right)}^{2}}} $

$ =\sqrt{4+16} $

$ =\sqrt{20} $

$=2\sqrt{5}$ units.

(ii) $\left( \mathbf{-3,7,2} \right)$ and $\left( \mathbf{2,4,-1} \right)$.

Ans: The distance between the points $\left( -3,7,2 \right)$ and $\left( 2,4,-1 \right)$ is

$ =\sqrt{{{\left( 2+3 \right)}^{2}}+{{\left( 4-7 \right)}^{2}}+{{\left( -1-2 \right)}^{2}}} $

$ =\sqrt{{{5}^{2}}+{{\left( -3 \right)}^{2}}+{{\left( -3 \right)}^{2}}} $

$ =\sqrt{25+9+9} $ 

$=\sqrt{43}$ units

(iii) $\left( -\mathbf{1},\mathbf{3},-\mathbf{4} \right)$ and $\left( \mathbf{1},-\mathbf{3},\mathbf{4} \right)$.

Ans:  The distance between the points $\left( -1,3,-4 \right)$ and $\left( 1,-3,4 \right)$ is

$=\sqrt{{{\left( 1+1 \right)}^{2}}+{{\left( -3-3 \right)}^{2}}+{{\left( 4+4 \right)}^{2}}} $

$ =\sqrt{{{2}^{2}}+{{\left( -6 \right)}^{2}}+{{8}^{2}}} $ 

$ =\sqrt{4+36+64} $

$ =\sqrt{104} $ 

$=2\sqrt{26}$ units

(iv) $\left( \mathbf{2},-\mathbf{1},\mathbf{3} \right)$ and $\left( \mathbf{-2,1,3} \right)$.

Ans: The distance between the points $\left( 2,-1,3 \right)$ and $\left( -2,1,3 \right)$ is

$=\sqrt{{{\left( -2-2 \right)}^{2}}+{{\left( 1+1 \right)}^{2}}+{{\left( 3-3 \right)}^{2}}} $

$ =\sqrt{{{\left( -4 \right)}^{2}}+{{2}^{2}}+{{0}^{2}}} $

$ =\sqrt{16+4}$

$ =\sqrt{20}$

$=2\sqrt{5}$ units

2. Show that the points $\left( \mathbf{-2,3,5} \right),\left( \mathbf{1},\mathbf{2},\mathbf{3} \right)$ and $\left( \mathbf{7},\mathbf{0},-\mathbf{1} \right)$ are collinear.

Ans: Recall that, any points $P$, $Q$, $R$ are said to be collinear if they lie on a line.

Now, suppose that the given points are $P\left( -2,3,5 \right)$, $Q\left( 1,2,3 \right)$, and $R\left( 7,0,-1 \right)$.

Then, $PQ=\sqrt{{{\left( 1+2 \right)}^{2}}+{{\left( 2-3 \right)}^{2}}+{{\left( 3-5 \right)}^{2}}}$

$ =\sqrt{{{3}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( -2 \right)}^{2}}} $

$ =\sqrt{9+1+4} $

$=\sqrt{14}$ units

$QR=\sqrt{{{\left( 7-1 \right)}^{2}}+{{\left( 0-2 \right)}^{2}}+{{\left( -1-3 \right)}^{2}}}$

$=\sqrt{{{6}^{2}}+{{\left( -2 \right)}^{2}}+{{\left( -4 \right)}^{2}}} $ 

$ =\sqrt{36+4+16} $

$ =\sqrt{56} $

$=2\sqrt{14}$ units

Also, $PR=\sqrt{{{\left( 7+2 \right)}^{2}}+{{\left( 0-3 \right)}^{2}}+{{\left( -1-5 \right)}^{2}}}$

$=\sqrt{{{9}^{2}}+{{\left( -3 \right)}^{2}}+{{\left( -6 \right)}^{2}}} $ 

$ =\sqrt{81+9+36} $ 

$=\sqrt{126}$

$=3\sqrt{14}$ units.

Notice that, $PQ+QR=\sqrt{14}+2\sqrt{14}=3\sqrt{14}=PR$.

Thus, the points lie in the same line.

Hence, the given points are collinear.

3. Verify the following statements:

(i) $\left( \mathbf{0,7,-10} \right)\mathbf{,}\left( \mathbf{1,6,-6} \right)$ and $\left( \mathbf{4},\mathbf{9},-\mathbf{6} \right)$ are the vertices of an isosceles triangle.

Ans: Recall that, in an isosceles triangle any two sides are of equal length.

Now, let the given points are $P\left( 0,7,-10 \right)$, $Q\left( 1,6,-6 \right)$ and $R\left( 4,9,-6 \right)$.

Then, $PQ=\sqrt{{{\left( 1-0 \right)}^{2}}+{{\left( 6-7 \right)}^{2}}+{{\left( -6+10 \right)}^{2}}}$

$=\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}+{{4}^{2}}}$

$=\sqrt{1+1+16}$ 

$ =\sqrt{18} $

$=3\sqrt{2}$ units.

$QR=\sqrt{{{\left( 4-1 \right)}^{2}}+{{\left( 9-6 \right)}^{2}}+{{\left( -6+6 \right)}^{2}}}$

$ =\sqrt{{{3}^{2}}+{{3}^{2}}+{{0}^{2}}}$

$ =\sqrt{9+9}$ 

$ =\sqrt{18}$ 

$=3\sqrt{2}$ units

Also, $RP=\sqrt{{{\left( 0-4 \right)}^{2}}+{{\left( 7-9 \right)}^{2}}+{{\left( -10+6 \right)}^{2}}}$

$ =\sqrt{{{\left( -4 \right)}^{2}}+{{\left( -2 \right)}^{2}}+{{\left( -4 \right)}^{2}}} $

$=\sqrt{16+4+16} $

$=\sqrt{36}$

$=6$ units.

Note that, $PQ=QR\ne RP$

Hence, the provided points are the vertices of an isosceles triangle.

(ii) $\left( \mathbf{0},\mathbf{7},\mathbf{10} \right),\left( -\mathbf{1},\mathbf{6},\mathbf{6} \right)$ and $\left( -\mathbf{4},\mathbf{9},\mathbf{6} \right)$ are the vertices of a right-angled triangle.

Ans: Recall that, according to the Pythagorean theorem, a triangle is said to be a right-angled if the sum of the squares of two sides equal to the square of the third side.

Now, let the given points are $P\left( 0,7,-10 \right)$, $Q\left( 1,6,-6 \right)$ and $R\left( 4,9,-6 \right)$.

Then, $PQ=\sqrt{{{\left( 1-0 \right)}^{2}}+{{\left( 6-7 \right)}^{2}}+{{\left( -6+10 \right)}^{2}}}$

$=\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}+{{4}^{2}}}$

$=\sqrt{1+1+16}$ 

$ =\sqrt{18}$ 

$=3\sqrt{2}$ units.

$ QR=\sqrt{{{\left( 4-1 \right)}^{2}}+{{\left( 9-6 \right)}^{2}}+{{\left( -6+6 \right)}^{2}}}$ 

$ =\sqrt{{{3}^{2}}+{{3}^{2}}+{{0}^{2}}} $ 

$ =\sqrt{9+9} $ 

$ =\sqrt{18} $ 

$=3\sqrt{2}$ units

Also, $RP=\sqrt{{{\left( 0-4 \right)}^{2}}+{{\left( 7-9 \right)}^{2}}+{{\left( -10+6 \right)}^{2}}}$

$=\sqrt{{{\left( -4 \right)}^{2}}+{{\left( -2 \right)}^{2}}+{{\left( -4 \right)}^{2}}}$

$ =\sqrt{16+4+16}$ 

$=\sqrt{36}$

$=6$ units.

Now, note that,

$ P{{Q}^{2}}+Q{{R}^{2}}={{\left( 3\sqrt{2} \right)}^{2}}+{{\left( 3\sqrt{2} \right)}^{2}}$

$ =18+18$ 

$ =36 $ 

$ ={{\left( RP \right)}^{2}}$

That is, $P{{Q}^{2}}+Q{{R}^{2}}=R{{P}^{2}}$.

Hence, according to the Pythagorean theorem, the given points form a right-angled triangle.

(iii) $\left( -\mathbf{1},\mathbf{2},\mathbf{1} \right),\left( \mathbf{1},-\mathbf{2},\mathbf{5} \right),\left( \mathbf{4},-\mathbf{7},\mathbf{8} \right)$ and $\left( \mathbf{2},-\mathbf{3},\mathbf{4} \right)$ are the vertices of a parallelogram.

Ans: Recall that, a quadrilateral is said to be a parallelogram if the opposite sides are equal.

Now, suppose that, the given points are $P\left( -1,2,1 \right)$, $Q\left( 1,-2,5 \right)$, $R\left( 4,-7,8 \right)$, and $S\left( 2,-3,4 \right)$.

Then, $PQ=\sqrt{{{\left( 1+1 \right)}^{2}}+{{\left( -2-2 \right)}^{2}}+{{\left( 5-1 \right)}^{2}}}$

$=\sqrt{{{2}^{2}}+{{\left( -4 \right)}^{2}}+{{4}^{2}}}$ 

$ =\sqrt{4+16+16}$

$=\sqrt{36}$

$=6$ units.

$QR=\sqrt{{{\left( 4-1 \right)}^{2}}+{{\left( -7+2 \right)}^{2}}+{{\left( 8-5 \right)}^{2}}}$

$=\sqrt{{{3}^{2}}+{{\left( -5 \right)}^{2}}+{{3}^{2}}}$ 

$ =\sqrt{9+25+9}$ 

$=\sqrt{43}$ units

$RS=\sqrt{{{\left( 2-4 \right)}^{2}}+{{\left( -3+7 \right)}^{2}}+{{\left( 4-8 \right)}^{2}}}$

$ =\sqrt{{{\left( -2 \right)}^{2}}+{{4}^{2}}+{{\left( -4 \right)}^{2}}} $ 

$ =\sqrt{4+16+16} $ 

$=\sqrt{36}$

$=6$ units

Also, $SP=\sqrt{{{\left( -1-2 \right)}^{2}}+{{\left( 2+3 \right)}^{2}}+{{\left( 1-4 \right)}^{2}}}$

$ =\sqrt{{{\left( -3 \right)}^{2}}+{{5}^{2}}+{{\left( -3 \right)}^{2}}}$

$ =\sqrt{9+25+9}$ 

$=\sqrt{43}$ units.

Therefore, we have

$PQ=RS=6$ units and $QR=SP=\sqrt{43}$ units.

Thus, in the quadrilateral $PQRS$, the opposite sides are equal.

Hence, $PQRS$ is a parallelogram, that is, the provided points are the vertices of a parallelogram.

4. Find the equation of the set of points which are equidistant from the points $\left( \mathbf{1},\mathbf{2},\mathbf{3} \right)$ and $\left( \mathbf{3},\mathbf{2},-\mathbf{1} \right)$.

Ans: Suppose that, the points $A\left( 1,2,3 \right)$ and $B\left( 3,2,-1 \right)$ are equidistant from the point $P\left( x,y,z \right)$.

Then, we have, $AP=BP$

$\Rightarrow A{{P}^{2}}=B{{P}^{2}}$

$\Rightarrow {{\left( x-1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}+{{\left( z-3 \right)}^{2}}={{\left( x-3 \right)}^{2}}+{{\left( y-2 \right)}^{2}}+{{\left( z+1 \right)}^{2}}$

$\Rightarrow{{x}^{2}}-2x+1+{{y}^{2}}-4y+4+{{z}^{2}}-6z+9={{x}^{2}}-6x+9+{{y}^{2}}-4y+4+{{z}^{2}}+2z+1 $ 

$ \Rightarrow -2x-4y-6z+14=-6x-4y+2z+14$ 

$ \Rightarrow -2x-6z+6x-2z=0$

$ \Rightarrow 4x-8z=0$

$\Rightarrow x-2z=0$

Hence, the equation of the set of points that are equidistant from the given points is given by

$x-2z=0$.

5. Find the equation of the set of points $\mathbf{P}$, the sum of whose distances from $\mathbf{A}\left( \mathbf{4},\mathbf{0},\mathbf{0} \right)$ and $\mathbf{B}\left( -\mathbf{4},\mathbf{0},\mathbf{0} \right)$ is equal to $\mathbf{10}$.

Ans: Suppose that, the points $A\left( 4,0,0 \right)$ and $B\left( -4,0,0 \right)$ are equidistant from the point $P\left( x,y,z \right)$.

Then, by the given condition, we have

$AP+BP=10$

$\Rightarrow \sqrt{{{\left( x-4 \right)}^{2}}+{{\left( y-0 \right)}^{2}}+{{\left( z-0 \right)}^{2}}}+\sqrt{{{\left( x+4 \right)}^{2}}+{{\left( y-0 \right)}^{2}}+{{\left( z-0 \right)}^{2}}}=10$

$\Rightarrow \sqrt{{{\left( x-4 \right)}^{2}}+{{y}^{2}}+{{z}^{2}}}=10-\sqrt{{{\left( x+4 \right)}^{2}}+{{y}^{2}}+{{z}^{2}}}$

Squaring both sides of the equation, yields

${{\left( x-4 \right)}^{2}}+{{y}^{2}}+{{z}^{2}}=100-20\sqrt{{{\left( x+4 \right)}^{2}}+{{y}^{2}}+{{z}^{2}}}+{{\left( x+4 \right)}^{2}}+{{y}^{2}}+{{z}^{2}}$

$\Rightarrow {{\left( x-4 \right)}^{2}}+{{y}^{2}}+{{z}^{2}}=100-20\sqrt{{{\left( x+4 \right)}^{2}}+{{y}^{2}}+{{z}^{2}}}+{{x}^{2}}+8x+16+{{y}^{2}}+{{z}^{2}}$

$\Rightarrow 20\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+8x+16}=100+16x$

$ \Rightarrow 5\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+8x+16}=25+4x$

Again, squaring both sides of the equation, gives

$25\left( {{x}^{2}}+8x+16+{{y}^{2}}+{{z}^{2}} \right)=625+16{{x}^{2}}+200x$

$\Rightarrow 25{{x}^{2}}+200x+400+25{{y}^{2}}+25{{z}^{2}}=625+16{{x}^{2}}+200x$

$\Rightarrow 9{{x}^{2}}+25{{y}^{2}}+25{{z}^{2}}-225=0$

Hence, the equation of the set of points, the sum of whose distances from the given points is equal to $10$, is given by

$9{{x}^{2}}+25{{y}^{2}}+25{{z}^{2}}-225=0$.

Exercise 12.3

1. Find the coordinates of the point which divides the line segment joining the points $\left( \text{-2,3,5} \right)$ and $\left( \text{1,-4,6} \right)$ in the ratio

(i) $\text{2:3}$ internally,

Ans: If $\text{R}$ is the point that divides the line segment joining the points $\text{P}\left( {{\text{x}}_{1}},{{\text{y}}_{1}},{{\text{z}}_{1}} \right)$ and $\text{Q}\left( {{\text{x}}_{2}},{{\text{y}}_{2}},{{\text{z}}_{2}} \right)$ internally, then the coordinates of the point $\text{R}$ will be,

$\left( \frac{\text{m}{{\text{x}}_{2}}\text{+n}{{\text{x}}_{1}}}{\text{m+n}},\text{ }\frac{\text{m}{{\text{y}}_{2}}\text{+n}{{\text{y}}_{1}}}{\text{m+n}},\text{ }\frac{\text{m}{{\text{z}}_{2}}\text{+n}{{\text{z}}_{1}}}{\text{m+n}} \right)$

We have, the point $\text{R}$ dividing the line segment joining the points $\left( \text{-2,3,5} \right)$ and $\left( \text{1,-4,6} \right)$ internally in the ratio $\text{2:3}$

$\text{x=}\frac{\text{2}\left( \text{1} \right)\text{+3}\left( \text{-2} \right)}{\text{2+3}}$

$\text{y=}\frac{\text{2}\left( \text{-4} \right)\text{+3}\left( \text{3} \right)}{\text{2+3}}$

$\text{z=}\frac{\text{2}\left( \text{6} \right)\text{+3}\left( \text{5} \right)}{\text{2+3}}$

On solving we get,

$\text{x = }\frac{\text{-4}}{\text{5}}$

$\text{y = }\frac{\text{1}}{\text{5}}$

$\text{z = }\frac{\text{27}}{\text{5}}$

Therefore, the coordinates we obtain are, $\left( \frac{\text{-4}}{\text{5}},\frac{\text{1}}{\text{5}},\frac{\text{27}}{\text{5}} \right)$

(ii) $\text{2:3}$ externally,

Ans:  If $\text{R}$ is the point that divides the line segment joining the points $\text{P}\left( {{\text{x}}_{1}},{{\text{y}}_{1}},{{\text{z}}_{1}} \right)$ and $\text{Q}\left( {{\text{x}}_{2}},{{\text{y}}_{2}},{{\text{z}}_{2}} \right)$ externally, then the coordinates of the point $\text{R}$ will be,

$\left( \frac{\text{m}{{\text{x}}_{2}}\text{-n}{{\text{x}}_{1}}}{\text{m-n}},\text{ }\frac{\text{m}{{\text{y}}_{2}}\text{-n}{{\text{y}}_{1}}}{\text{m-n}},\text{ }\frac{\text{m}{{\text{z}}_{2}}\text{-n}{{\text{z}}_{1}}}{\text{m-n}} \right)$

We have, the point $\text{R}$ dividing the line segment joining the points $\left( \text{-2,3,5} \right)$ and $\left( \text{1,-4,6} \right)$ externally in the ratio $\text{2:3}$

$\text{x=}\frac{\text{2}\left( \text{1} \right)\text{-3}\left( \text{-2} \right)}{\text{2-3}}$

$\text{y=}\frac{\text{2}\left( \text{-4} \right)\text{-3}\left( \text{3} \right)}{\text{2-3}}$

$\text{z=}\frac{\text{2}\left( \text{6} \right)\text{-3}\left( \text{5} \right)}{\text{2-3}}$

On solving we get,

$\text{x = -8}$

$\text{y = 17}$

$\text{z = 3}$

Therefore, the coordinates we obtain are, $\left( \text{-8},\text{17},\text{3} \right)$

2. Given that $\text{P}\left( \text{3,2,-4} \right)$, $\text{Q}\left( \text{5,4,-6} \right)$ and $\text{R}\left( \text{9,8,-10} \right)$ are collinear. Find the ratio in which $\text{Q}$ divides $\text{PR}$.

Ans: Let the ratio in which the point $\text{Q}$ divides the line segment joining the points $\text{P}\left( \text{3,2,-4} \right)$ and $\text{R}\left( \text{9,8,-10} \right)$ be $\text{k:1}$.

Using section formula,

$\left( \text{5,4,-6} \right)\text{ = }\left( \frac{\text{k}\left( \text{9} \right)\text{+3}}{\text{k+1}},\text{ }\frac{\text{k}\left( 8 \right)\text{+2}}{\text{k+1}},\text{ }\frac{\text{k}\left( \text{-10} \right)-4}{\text{k+1}} \right)$

$\frac{\text{9k+3}}{\text{k+1}}\text{ = 5}$

$\text{9k+3 = 5k+5}$

$\text{4k = 2}$

$\text{k = }\frac{\text{2}}{\text{4}}$

$\text{k = }\frac{\text{1}}{\text{2}}$

Therefore, the ratio in which the point $\text{Q}$ divides the line segment joining the points $\text{P}\left( \text{3,2,-4} \right)$ and $\text{R}\left( \text{9,8,-10} \right)$ is $\text{1:2}$.

3. Find the ratio in which $\text{YZ}$-plane divides the line segment formed by joining the points, $\left( \text{2,4,7} \right)$ and $\left( \text{3,-5,8} \right)$.

Ans: Let the ratio in which the $\text{YZ}$-plane divides the line segment joining the points $\left( \text{-2,4,7} \right)$ and $\left( \text{3,-5,8} \right)$ be $\text{k:1}$.

Using section formula,

$\left( \text{0,y,z} \right)\text{ = }\left( \frac{\text{k}\left( \text{3} \right)\text{-2}}{\text{k+1}},\text{ }\frac{\text{k}\left( -5 \right)\text{+4}}{\text{k+1}},\text{ }\frac{\text{k}\left( \text{8} \right)+7}{\text{k+1}} \right)$

The $\text{x}$ coordinate is $\text{0}$ on $\text{YZ}$-plane,

$\frac{\text{3k-2}}{\text{k+1}}\text{ = 0}$

$\text{3k-2 = 0}$

$\text{3k = 2}$

$\text{k = }\frac{\text{2}}{\text{3}}$

Therefore, the ratio in which the $\text{YZ}$-plane divides the line segment joining the points $\left( \text{-2,4,7} \right)$ and $\left( \text{3,-5,8} \right)$  is $\text{2:3}$.

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

Ans: We are given three points $\text{A}\left( \text{2,-3,4} \right)$, $\text{B}\left( \text{-1,2,1} \right)$ and $\text{C}\left( \text{0,}\frac{\text{1}}{\text{3}}\text{,2} \right)$

Let $\text{P}$ is point which divides the line segment $\text{AB}$in the ratio $\text{k:1}$.

Using section formula,

$\text{ }\left( \frac{\text{k}\left( \text{-1} \right)\text{+2}}{\text{k+1}},\text{ }\frac{\text{k}\left( 2 \right)\text{-3}}{\text{k+1}},\text{ }\frac{\text{k}\left( \text{1} \right)+4}{\text{k+1}} \right)$

The value of $\text{k}$ such that the point $\text{P}$ coincides with point $\text{C}$ will be,

$\frac{\text{-k+2}}{\text{k+1}}\text{ = 0}$

$\text{2-k = 0}$

$\text{-k = -2}$

$\text{k = 2}$

Now checking for $\text{k = 2}$.

The coordinates of point $\text{P}$ are $\text{ }\left( \frac{2\left( \text{-1} \right)\text{+2}}{\text{2+1}},\text{ }\frac{2\left( 2 \right)\text{-3}}{\text{2+1}},\text{ }\frac{2\left( \text{1} \right)+4}{\text{2+1}} \right)$

On solving, the coordinates of point $\text{P}$ are, $\left( \text{0,}\frac{\text{1}}{\text{3}}\text{,2} \right)$

Therefore, the point $\left( \text{0,}\frac{\text{1}}{\text{3}}\text{,2} \right)$ divides $\text{AB}$in the ratio $\text{2:1}$. Also, the point is same as point $\text{C}$.

Hence, proved that the points $\text{A}$, $\text{B}$ and $\text{C}$ are collinear.

5. Find the coordinates of the points which trisect the line segment joining the points, $\text{P}\left( \text{4,2,-6} \right)$ and $\text{Q}\left( \text{10,-16,6} \right)$.

Ans: Let $\text{A}$ and $\text{B}$ are the points which trisect the line segment joining the points $\text{P}\left( \text{4,2,-6} \right)$ and $\text{Q}\left( \text{10,-16,6} \right)$.

(Image Will Be Updated Soon)

The point $\text{A}$ divides the line segment $\text{PQ}$ in the ratio of $\text{1:2}$.

Using section formula,

$\text{A}\left( \text{x,y,z} \right)\text{=}\left( \frac{\text{1}\left( \text{10} \right)\text{+2}\left( \text{4} \right)}{\text{1+2}}\text{, }\frac{\text{1}\left( \text{-16} \right)\text{+2}\left( \text{2} \right)}{\text{1+2}}\text{, }\frac{\text{1}\left( \text{6} \right)\text{+2}\left( \text{-4} \right)}{\text{1+2}} \right)\text{ }$

$\text{A}\left( \text{x,y,z} \right)\text{=}\left( \text{6,-4,-2} \right)$

Similarly, the point $\text{B}$ divides the line segment $\text{PQ}$ in the ratio of $\text{2:1}$.

$\text{B}\left( \text{x,y,z} \right)=\left( \frac{\text{2}\left( \text{10} \right)\text{+1}\left( \text{4} \right)}{\text{2+1}},\text{ }\frac{\text{2}\left( \text{-16} \right)\text{+1}\left( \text{2} \right)}{\text{2+1}},\text{ }\frac{\text{2}\left( \text{6} \right)\text{+1}\left( \text{-4} \right)}{\text{2+1}} \right)\text{ }$

$\text{B}\left( \text{x,y,z} \right)\text{=}\left( \text{8,-10,2} \right)$

Therefore, the point $\left( \text{6,-4,-2} \right)$ and $\left( \text{8,-10,2} \right)$ are the points which trisect the line segment joining the points $\text{P}\left( \text{4,2,-6} \right)$ and $\text{Q}\left( \text{10,-16,6} \right)$.

Miscellaneous Exercise

1. Three vertices of a parallelogram $\text{ABCD}$ are $\text{A}\left( \text{3,-1,2} \right)$, $\text{B}\left( \text{1,2,-4} \right)$ and $\text{C}\left( \text{-1,1,2} \right)$. Find the coordinates of the fourth vertex.

Ans: We are given the three vertices of a parallelogram $\text{ABCD}$ are $\text{A}\left( \text{3,-1,2} \right)$, $\text{B}\left( \text{1,2,-4} \right)$ and $\text{C}\left( \text{-1,1,2} \right)$.

Let the coordinates of the fourth vertex of the parallelogram $\text{ABCD}$ be $\text{D}\left( \text{x,y,z} \right)$.

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According to the property of parallelogram, the diagonals of the parallelogram bisect each other.

In this parallelogram $\text{ABCD}$, $\text{AC}$ and $\text{BD}$ at point $\text{O}$.

So, 

$\text{Mid-point of AC = Mid-point of BD}$

$\left( \frac{\text{3-1}}{\text{2}},\text{ }\frac{\text{-1+1}}{\text{2}},\text{ }\frac{\text{2+2}}{\text{2}} \right)\text{ = }\left( \frac{\text{x+1}}{\text{2}},\text{ }\frac{\text{y+1}}{\text{2}},\text{ }\frac{\text{z-4}}{\text{2}} \right)$

$\left( \text{1,0,2} \right)\text{ = }\left( \frac{\text{x+1}}{\text{2}},\text{ }\frac{\text{y+1}}{\text{2}},\text{ }\frac{\text{z-4}}{\text{2}} \right)$

$\frac{\text{x+1}}{\text{2}}\text{ = 1}$

$\frac{\text{y+2}}{\text{2}}\text{ = 0}$

$\frac{\text{z-4}}{\text{2}}\text{ = 2}$

We get, $\text{x = 1}$, $\text{y = 2}$ and $\text{z = 8}$

Therefore, the coordinates of the fourth vertex of the parallelogram $\text{ABCD}$ are $\text{D}\left( \text{1,-2,8} \right)$.

2. Find the lengths of the medians of the triangle with $\text{A}\left( \text{0,0,6} \right)$, $\text{B}\left( \text{0,4,0} \right)$ and $\text{C}\left( \text{6,0,0} \right)$. 

Ans: For the given triangle $\text{ABC}$. Let $\text{AD}$, $\text{BE}$ and $\text{CF}$ are the medians:

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We know that, median divides the line segment into two equal parts, so $\text{D}$ is the midpoint of $\text{BC}$, therefore,

$\text{Coordinates of point D = }\left( \frac{\text{0+6}}{\text{2}},\text{ }\frac{\text{4+0}}{\text{2}},\text{ }\frac{\text{0+0}}{\text{2}} \right)$

$\text{Coordinates of point D = }\left( \text{3,2,0} \right)$

$\text{AD = }\sqrt{{{\left( \text{0-3} \right)}^{\text{2}}}\text{+}{{\left( \text{0-2} \right)}^{\text{2}}}\text{+}{{\left( \text{6-0} \right)}^{\text{2}}}}$

$\text{AD = }\sqrt{\text{9+4+36}}$

$\text{AD = }\sqrt{\text{49}}$

$\text{AD = 7}$

Similarly, $\text{E}$ is the midpoint of $\text{AC}$,

$\text{Coordinates of point E = }\left( \frac{\text{0+6}}{\text{2}},\text{ }\frac{\text{0+0}}{\text{2}},\text{ }\frac{\text{0+6}}{\text{2}} \right)$

$\text{Coordinates of point E = }\left( \text{3,0,3} \right)$

$\text{AC = }\sqrt{{{\left( \text{3-0} \right)}^{\text{2}}}\text{+}{{\left( \text{0-4} \right)}^{\text{2}}}\text{+}{{\left( \text{3-0} \right)}^{\text{2}}}}$

$\text{AC = }\sqrt{\text{9+16+9}}$

$\text{AC = }\sqrt{\text{34}}$

Similarly, $\text{F}$ is the midpoint of $\text{AB}$,

$\text{Coordinates of point F = }\left( \frac{\text{0+0}}{\text{2}},\text{ }\frac{\text{0+4}}{\text{2}},\text{ }\frac{\text{6+0}}{\text{2}} \right)$

$\text{Coordinates of point F = }\left( \text{0,2,3} \right)$

$\text{CF = }\sqrt{{{\left( \text{6-0} \right)}^{\text{2}}}\text{+}{{\left( \text{0-2} \right)}^{\text{2}}}\text{+}{{\left( \text{0-3} \right)}^{\text{2}}}}$

$\text{CF = }\sqrt{\text{36+4+9}}$

$\text{CF = }\sqrt{\text{49}}$

$\text{CF = 7}$

Therefore, the lengths of the medians of the triangle $\text{ABC}$ we obtain are, $\text{7, }\sqrt{\text{34}}\text{, 7}$

3. If the origin is the centroid of the triangle $\text{PQR}$ with vertices $\text{P}\left( \text{2a,2,6} \right)$, $\text{Q}\left( \text{-4,3b,-10} \right)$ and $\text{R}\left( \text{8,14,2c} \right)$, then find the values of $\text{a}$, $\text{b}$ and $\text{c}$

Ans: The given triangle $\text{PQR}$

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We know that the coordinates of the centroid of triangle with the vertices $\left( {{\text{x}}_{1}},{{\text{y}}_{1}},{{\text{z}}_{1}} \right)$, $\left( {{\text{x}}_{2}},{{\text{y}}_{2}},{{\text{z}}_{2}} \right)$ and $\left( {{\text{x}}_{3}},{{\text{y}}_{3}},{{\text{z}}_{3}} \right)$ are,

$\frac{{{\text{x}}_{\text{1}}}\text{+}{{\text{x}}_{\text{2}}}\text{+}{{\text{x}}_{\text{3}}}}{\text{3}}=\frac{{{\text{y}}_{\text{1}}}\text{+}{{\text{y}}_{\text{2}}}\text{+}{{\text{y}}_{\text{3}}}}{\text{3}}=\frac{{{\text{z}}_{\text{1}}}\text{+}{{\text{z}}_{\text{2}}}\text{+}{{\text{z}}_{\text{3}}}}{\text{3}}$

For triangle $\text{PQR}$, the coordinates will be,

$\text{ }\!\!\Delta\!\!\text{ PQR = }\frac{\text{2a-4+8}}{\text{3}}=\frac{\text{2+3b+14}}{\text{3}}=\frac{\text{6-10+2c}}{\text{3}}$

$\text{ }\!\!\Delta\!\!\text{ PQR = }\frac{\text{2a+4}}{\text{3}}=\frac{\text{3b+16}}{\text{3}}=\frac{\text{2c-4}}{\text{3}}$

Now, we are given that centroid is the origin,

$\frac{\text{2a+4}}{\text{3}}\text{ = 0}$

$\frac{\text{3b+16}}{\text{3}}\text{ = 0}$

$\frac{\text{2c-4}}{\text{3}}\text{ = 0}$

$\text{a = -2}$,

$\text{b = -}\frac{\text{16}}{\text{3}}$

$\text{c = 2}$

Therefore, we obtain the values as $\text{a = -2}$, $\text{b = -}\frac{\text{16}}{\text{3}}$ and $\text{c = 2}$.

4. Find the coordinates of the point on $\text{y-axis}$ which are at a distance of $\text{5}\sqrt{\text{2}}$ from the point $\text{P}\left( \text{3,-2,5} \right)$

Ans: For the point to be on $\text{x-axis}$ the $\text{y-coordinate}$ and $\text{z-coordinate}$ become zero.

Let the point on $\text{y-axis}$ at a distance of $\text{5}\sqrt{\text{2}}$ from point $\text{P}\left( \text{3,-2,5} \right)$ be  $\text{A}\left( \text{0,b,0} \right)$,

We have, $\text{AP = 5}\sqrt{\text{2}}$

Using distance formula,

$\text{A}{{\text{P}}^{\text{2}}}\text{ = 50}$

${{\left( \text{3-0} \right)}^{\text{2}}}\text{+}{{\left( \text{-2-b} \right)}^{\text{2}}}\text{+}{{\left( \text{5-0} \right)}^{\text{2}}}\text{ = 50}$

$\text{9+4+}{{\text{b}}^{\text{2}}}\text{+4b+25 = 50}$

${{\text{b}}^{\text{2}}}\text{+4b-12 = 0}$

${{\text{b}}^{\text{2}}}\text{+6b-2b-12 = 0}$

$\left( \text{b+6} \right)\left( \text{b-2} \right)\text{ = 0}$

$\text{b = -6}$ or $\text{b = 2}$

The coordinate of the points is $\left( \text{0,2,0} \right)$ and $\left( \text{0,-6,0} \right)$

5. A point $\text{R}$ with $\text{x-coordinate}$ $\text{4}$ lies on the line segment joining the points $\text{P}\left( \text{2,-3,4} \right)$ and $\text{Q}\left( \text{8,0,10} \right)$. Find the coordinates of the point $\text{R}$.

Ans: Let $\text{R}$ is point which divides the line segment $\text{PQ}$in the ratio $\text{k:1}$.

Using section formula,

$\text{ }\left( \frac{\text{k}\left( \text{8} \right)\text{+2}}{\text{k+1}},\text{ }\frac{\text{k}\left( 0 \right)\text{-3}}{\text{k+1}},\text{ }\frac{\text{k}\left( \text{10} \right)+4}{\text{k+1}} \right)\text{ = }\left( \frac{\text{8k+2}}{\text{k+1}},\text{ }\frac{-3}{\text{k+1}},\text{ }\frac{\text{10k+4}}{\text{k+1}} \right)$

The value of $\text{x-coordinate}$ of the point $\text{R}$ is $\text{4}$,

$\frac{\text{8k+2}}{\text{k+1}}\text{ = 4}$

$\text{8k+2 = 4k+4}$

$\text{4k = 2}$

$\text{k = }\frac{\text{1}}{\text{2}}$

So, the coordinates of the point $\text{R}$ are,

$\left( \text{4,}\frac{\text{-3}}{\frac{\text{1}}{\text{2}}\text{+1}}\text{,}\frac{\text{10}\left( \frac{\text{1}}{\text{2}} \right)\text{+4}}{\frac{\text{1}}{\text{2}}\text{+1}} \right)\text{ = }\left( \text{4,-2,6} \right)$

6. If $\text{A}$ and $\text{B}$ be the points $\left( \text{3,4,5} \right)$ and $\left( \text{-1,3,-7} \right)$ respectively, find the equation of the set of points $\text{P}$ such that $\text{P}{{\text{A}}^{\text{2}}}\text{+P}{{\text{B}}^{\text{2}}}\text{ = }{{\text{k}}^{\text{2}}}$, where $\text{k}$ is a constant.

Ans: Let the coordinates of point $\text{P}$ be $\left( \text{x,y,z} \right)$.

Using distance formula we get,

$\text{P}{{\text{A}}^{\text{2}}}\text{=}{{\left( \text{x-3} \right)}^{\text{2}}}\text{+}{{\left( \text{y-4} \right)}^{\text{2}}}\text{+}{{\left( \text{z-5} \right)}^{\text{2}}}$

$\text{P}{{\text{A}}^{\text{2}}}\text{ = }{{\text{x}}^{\text{2}}}\text{+9-6x+}{{\text{y}}^{\text{2}}}\text{+16-8y+}{{\text{z}}^{\text{2}}}\text{+25-10z}$

$\text{P}{{\text{A}}^{\text{2}}}\text{ = }{{\text{x}}^{\text{2}}}\text{-6x+}{{\text{y}}^{\text{2}}}\text{-8y+}{{\text{z}}^{\text{2}}}\text{-10z+50}$

Similarly,

$\text{P}{{\text{B}}^{\text{2}}}\text{ = }{{\left( \text{x-1} \right)}^{\text{2}}}\text{+}{{\left( \text{y-3} \right)}^{\text{2}}}\text{+}{{\left( \text{z-7} \right)}^{\text{2}}}$

$\text{P}{{\text{A}}^{\text{2}}}\text{ = }{{\text{x}}^{\text{2}}}\text{-2x+}{{\text{y}}^{\text{2}}}\text{-6y+}{{\text{z}}^{\text{2}}}\text{-14z+59}$

We are given that,

$\text{P}{{\text{A}}^{\text{2}}}\text{+P}{{\text{B}}^{\text{2}}}\text{ = }{{\text{k}}^{\text{2}}}$

So,

$\left( {{\text{x}}^{\text{2}}}\text{-6x+}{{\text{y}}^{\text{2}}}\text{-8y+}{{\text{z}}^{\text{2}}}\text{-10z+50} \right)\text{+}\left( {{\text{x}}^{\text{2}}}\text{-2x+}{{\text{y}}^{\text{2}}}\text{-6y+}{{\text{z}}^{\text{2}}}\text{-14z+59} \right)\text{ = }{{\text{k}}^{2}}$

$\text{2}{{\text{x}}^{\text{2}}}\text{+2}{{\text{y}}^{\text{2}}}\text{+2}{{\text{z}}^{\text{2}}}\text{-4x-14y+14z+109 =}\ {{\text{k}}^{\text{2}}}$

$\text{2}\left( {{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}\text{+}{{\text{z}}^{\text{2}}}\text{-2x-7y+7z} \right)\text{ = }{{\text{k}}^{\text{2}}}\text{-109}$

${{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}\text{+}{{\text{z}}^{\text{2}}}\text{-2x-7y+7z = }\frac{{{\text{k}}^{\text{2}}}\text{-109}}{\text{2}}$

Therefore, the equation is as follows ${{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}\text{+}{{\text{z}}^{\text{2}}}\text{-2x-7y+2z = }\frac{{{\text{k}}^{\text{2}}}\text{-109}}{\text{2}}$

NCERT Solutions for Class 11 Maths Chapter 12 - Introduction to Three Dimensional Geometry

Let us consider a room with a rectangular floor. 

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Let $OX$ and $OY$ be two adjacent edges of the floor, which we take as x-axis & y-axis. Then for locating a point $p$ exactly on the floor we draw perpendiculars $PM$ & $PN$ respectively on $OX$ or $OY$ and measure their lengths. The distances $OM ( = PN)$ and $ON ( = PM)$ are called the x- and y-coordinates respectively. $OP$ is the distance of point $P$ from the meeting point $O$ of $OX$ & $OY$, which we call the origin -- this is what is called the 2D Cartesian Coordinate system. 

But how can we exactly locate a point like a marking $Q$ made on the dome of a fan hanging from the ceiling fan? For this, we have to draw a line from $P$, perpendicular to the floor, to the marking $Q$ and measure its height. This means we require another line $OZ$ which will be the line pointing to the two adjacent walls where the bottom edges are $OX$ & $OY$. The distance $OZ (=PQ)$ is the $3^{rd}$ coordinate of the point $Q$. The lines $OX, OY$ & $OZ$ are the axes of what we call the 3D Cartesian Coordinate system.

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Here the origin $O$ is the meeting point of the two adjacent walls and the floor. This 3D Cartesian Coordinate system is necessary to locate a point in space-like a balloon floating in the air above the ground, the location of an airplane flying in the sky, or a satellite orbiting the earth.

Coordinates of a Point in Space

Let the origin be $O$ & let the mutually perpendicular lines be $OX, OY$ and $OZ$, taken as X-axis, Y-axis & Z-axis respectively. 

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The planes, $YOZ, ZOX$ & $XOY$ are respectively called a y-z plane, z-x plane & xy- plane. These panes known as coordinate planes, divide the space into eight parts, called octants.

Let $p$ be a point in space. Through $P$, we draw planes parallel to the coordinate planes and meet the axes $OX, OY$ & $OZ$ at points $A, B$ & $C$ respectively. We complete the parallelepiped whose coterminous edges are $OA, OB$ & $OC$.

Let $OA = x, OB = y$ & $OC =z$

Then we say that the coordinates of $P$ are $(x, y, z)$. As can be seen from the fi. $x, y, z$ are a distance of point $P$ respectively from $yz, xz$ & $xy$ -plane.

Note:

1. Any point on the $y-z$ plane will have its x-coordinate equal to zero. Similarly, a point on the $z-x$ plane will have $y = 0$ & a point on the $xy$ plane will have $z = 0$.

2. Coordinates of the origin are $O(0, 0, 0)$.

Distance Formula in 3D

In 2D coordinate geometry, the distance between two pints $P(x_1,y_1)$ & $Q(x_2,y_2)$ is given by

$PQ = \sqrt{(x_2 -x_1)^2+ (y_2 -y_1)^2}$  similar in a 3D coordinate system, the distance between two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ is given by $AB = \sqrt{(x_2 -x_1)^2+ (y_2 -y_1)^2+(z_2 -z_1)^2}$

Note: Distance of a point $P(x, y, z)$ from the origin $O(0, 0, 0)$ is 

$OP = \sqrt{(x -0)^2+ (y-0)^2+(z-0)^2}$ 

$OP = \sqrt{x^2 + y^2 + z^2}$

Illustrated Example

Example1: Find the distance between the points $A( -2, 1, -3)$ & $B(4, 3, -6)$.

Sol. 

Required Distance

$AB = \sqrt {((4-(-1))^2 + (3-1)^2 +(6-(-3))^2}$

$AB = \sqrt{6^ 2+2^2+(-3)^2} = \sqrt{36+4+9} = \sqrt49 = 7\text{units}$

Section Formula

In the 2D coordinate system, if a point $R(\bar{x},\bar{y})$ divides the two joins of the points $P(x_1+y_1)$ & 

$Q (x_2 + y_2)$ in the ratio m:n, then

$\bar{x}=\dfrac{mx_2+nx_1}{m + n}, \bar{y}=\dfrac{my_2+ny_1}{m + n}, \bar{z}=\dfrac{mz_2+nz_1}{m + n}$

Note: 

1. The coordinates of the midpoint of the join $P(x_1, y_1, z_1)$ & $Q(z_2, y_2, z_2)$ is given by 

(here $m =n$) $\bar{x}=\dfrac{x_2+x_1}{2}, \bar{y}=\dfrac{y_2+y_1}{2}, \bar{z}=\dfrac{z_2+z_1}{2}$ 

2. If a point $R(\bar{x}, \bar{y}, \bar{z})$ divides the join of $P(x_1, y_1, z_1)$ & $Q(x_2, y_2, z_2)$ externally, then 

$\bar{x}=\dfrac{mx_2+nx_1}{m + n}, \bar{y}=\dfrac{my_2+ny_1}{m + n}, \bar{z}=\dfrac{mz_2+nz_1}{m + n}$

Illustrated Example

Example 2: Find the coordinates of the point which divides the join of the points $P(5, 4, 2)$ & 

$Q( -1, -2, 4)$ in ratio $2:3$ .

Sol. 

If $R(\bar{x}, \bar{y}, \bar{z})$ be the required point, then $\bar{x}=\dfrac{2(-1)+3 \times 5}{2+3}, \bar{y}=\dfrac{2(-2)+3 > 4}{2+3}, \bar{z}=\dfrac{2 \times 4+3 \times 2}{2+3}$ $   = \dfrac{13}{5} = \dfrac{8}{5} = \dfrac{14}{5}$ $\therefore$ Required points $\left(\dfrac{13}{5}, \dfrac{8}{5},\dfrac{14}{5}\right)$

We Cover All Exercises in the Chapter Given Below:- 

EXERCISE 12.1 - 4 Questions with Solutions

EXERCISE 12.2 - 5 Questions with Solutions

EXERCISE 12.3 - 5 Questions with Solutions.

Summary

1. In 3D, the coordinates axes of a rectangular Cartesian coordinates system are three mutually perpendicular lines. The axes are named the $x, y$ and $z$ axes. 

2. The coordinate planes are the three planes ascertained by the pair of axes. These planes are called $x-y, y-z$ and $x-z$ planes and they divide the space into eight regions known as octants.

3. In space, the coordinates of a point $P$ are the perpendicular distances from $P$ on three mutually perpendicular coordinates planes $XY, YZ, XZ$ respectively. The coordinates of a point $P$ are written in the triplet $(x,y,z)$. 

4. The coordinates of a point are also the distances from the origin of the feet of the perpendiculars from the point on the respective coordinate axes. 

5. The distance between two points $A(x_1,y_1, z_1)$ and $B(x_2,y_2, z_2)$ is demonstrated by
   $AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$

6. The coordinates of any point on:

1. x-axis are of the form $(x, 0, 0)$

2. y-axis are of the form $(0, y, 0)$

3. z-axis are of the form $(0, 0, z)$

4. xy-plane are of the form $(x, y, 0)$

5. yz-plane are of the form $(0, y, z)$

6. xz- plane are of the form $(x, 0, z)$

7. The distance of a point $P(x,y,z)$ from the origin $O(0,0,0$ is given by
   $OP = \sqrt{(x-0)^2 + (y-0)^2 + (z-0)^2}$
   $OP = \sqrt{(x)^2 + (y)^2 + (z)^2}$

8. The coordinates of the midpoint of the join $P(x_1,y_1,z_1)$ & $Q(x_2,y_2,z_2)$ is given by
   (here $m =n$) $\bar{x}=\dfrac{x_1+x_2}{2}, \bar{y}=\dfrac{y_1+y_2}{2}, \bar{z}=\dfrac{z_1+z_2}{2}$ 

9. The coordinates of the centroid of the triangle whose vertices are $(x_1,y_1,z_1), (x_2,y_2,z_2)$ and $(x_3,y_3,z_3)$ are $\dfrac{x_1+x_2+x_3}{3}, \dfrac{y_1+y_2+y_3}{3}, \dfrac{z_1+z_2+z_3}{3}$.

10. In the 2D coordinate system, if a point $R(\bar{x}, \bar{y})$ divides the join of the points $P(x_1 + y_1)$ & $Q (x_2 + y_2)$ internally in the ratio $m:n$, then $\bar{x} = \dfrac{mx_2+mx_1}{m+n}, \bar{y} = \dfrac{my_2+my_1}{m+n}, \bar{z} = \dfrac{mz_2+mz_1}{m+n}$.