NCERT Solutions For Class 11 Maths Chapter 11 Introduction To Three Dimensional Geometry - 2025-26

Exercise 11.1: This exercise introduces the concept of three-dimensional space and the coordinate system used to represent points in three dimensions. Students will learn about the distance formula in three-dimensional space and how to find the coordinates of a point dividing a line segment in a given ratio.

Exercise 11.2: In this exercise, students will learn about the direction ratios of a line, the angle between two lines, and the equation of a plane in different forms. They will also practice finding the distance between a point and a plane.

Exercise 11.3: This exercise focuses on the concept of the angle between two planes and the shortest distance between two skew lines. Students will learn how to find the angle between two planes and the shortest distance between two skew lines.

Miscellaneous Exercise: This exercise includes a mix of questions covering all the concepts taught in the chapter. Students will have to apply their knowledge of three-dimensional geometry to solve various problems and answer questions. They will also practice finding the direction ratios of a line, the equation of a plane, the angle between two lines, and the distance between a point and a plane.

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

Exercise 11.1

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 11.2

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 11.3

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 $\text{D}\left( \text{x,y,z} \right)$.

(Image Will Be Updated Soon)

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:

(Image Will Be Updated Soon)

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 $\text{PQR}$

(Image Will Be Updated Soon)

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 11 - Introduction to Three Dimensional Geometry

Let us consider a room with a rectangular floor. 

(Image to be added soon)

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.

(Image to be added soon)

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. 

(Image to be added soon)

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 11.1 - 4 Questions with Solutions

EXERCISE 11.2 - 5 Questions with Solutions

MISCELLANEOUS EXERCISE - 4 Questions with Solutions.

Important Study Material Links for Chapter 12: Limits and Derivatives

• Introduction to Three Dimensional Geometry Important Questions

• Introduction to Three Dimensional Geometry Important Formulas

• Introduction to Three Dimensional Geometry Revision Notes

• Introduction to Three Dimensional Geometry NCERT Exemplar Solutions

• Introduction to Three Dimensional Geometry RD Sharma Solutions

• Introduction to Three Dimensional Geometry RS Aggarwal Solutions

NCERT Class 11 Maths Solutions Chapter-wise Links - Download the FREE PDF

Additional Study Materials for CBSE Class 11 Maths