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# NCERT Solutions for Class 11 Maths Chapter 11 - Introduction To Three Dimensional Geometry

Last updated date: 13th Sep 2024
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## NCERT Solutions for Class 11 Maths Chapter 11 - Introduction to Three Dimensional Geometry

The leading education online platform, Vedantu provides the notes and solutions for Chapter 11 Maths Class 11 NCERT Solutions. This chapter introduces three-dimensional geometry to you which is one of the important topics in Maths. In this chapter, you will learn the cartesian coordinate system, how to calculate the distance between points using distance formula and section formula, and how to find the coordinates of a point in a given space. Each NCERT Solution provided by vedantu is curated by our expert teachers. You can download the pdf of the notes and solutions for this chapter from the official website of Vedantu. You can also register for online Maths tuitions on www.vedantu.com which will help you master the subject and score excellent marks in the exams.

 Class: NCERT Solutions for Class 11 Subject: Class 11 Maths Chapter Name: Chapter 11 - Introduction to Three Dimensional Geometry Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes

## Introduction to Three Dimensional Geometry Chapter at a Glance - Class 11 NCERT Solutions

• In three dimensions, the coordinate axes of a rectangular Cartesian coordinate system are three mutually perpendicular lines. The axes are called the $x, y$ and $z$-axis.

• The three planes determined by the pair of axes are the coordinate planes, called $X Y, Y Z$ and $Z X$ - planes.

• The three coordinate planes divide the space into eight parts known as octarts.

• The coordinates of a point $P$ in three-dimensional geometry is always written in the form of triplet like $(x, y, z)$. Here $x, y$ and $z$ are the distances from the $Y Z, Z X$ and $X Y$ - planes.

• (i) Any point on $x$-axis is of the form $(x, 0,0)$

(ii) Any point on $y$-axis is of the form $(0, y, 0)$

(iii) Any point on $z$-axis is of the form $(0,0, z)$.

• Distance between two points $P\left(x_1, y_1, z_1\right)$ and $Q\left(x_2, y_2, z_2\right)$ is given by

$$P Q=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}$$

• The coordinates of the point $R$ which divides the line segment

• joining two points $P\left(x_1, y_1, z_1\right)$ and $Q\left(x_2, y_2, z_2\right)$ internally and externally in the ratio $m: n$ are given by $\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}, \frac{m z_2+n \Sigma_1}{m+n}\right)$ and $\left(\frac{m x_2-n x_1}{m-n}, \frac{m y_2-n y_1}{m-n}, \frac{m z_2-n z_1}{m-n}\right)$, respectively.

• The coordinates of the mid-point of

• The line segment joining two points $P\left(x_1, y_1, z_1\right)$ and $Q\left(x_2, y_2, z_2\right)$ and $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)$.

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

Competitive Exams after 12th Science

## Exercises under NCERT Class 11 Maths Chapter 11 – Introduction to Three-Dimensional Geometry

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.

 Octants $I$ $II$ $III$ $IV$ $V$ $VI$ $VII$ $VIII$ $x$ $+$ $-$ $-$ $+$ $+$ $-$ $-$ $+$ $y$ $+$ $+$ $-$ $-$ $+$ $+$ $-$ $-$ $z$ $+$ $+$ $+$ $+$ $-$ $-$ $-$ $-$

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.

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.

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.

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

1. 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

EXERCISE 11.3 - 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 $\bar{x} = \dfrac{mx_2+mx_1}{m+n}, \bar{y} = \dfrac{my_2+my_1}{m+n}, \bar{z} = \dfrac{mz_2+mz_1}{m+n}$.

## FAQs on NCERT Solutions for Class 11 Maths Chapter 11 - Introduction To Three Dimensional Geometry

1. What are the topics of Class 11 Maths Chapter 11- Introduction to Three Dimensional Geometry? What are the major learnings from these topics?

Following are the topics of Class 11 Maths Chapter 11- Introduction to Three Dimensional Geometry:

• Introduction: This section talks about concepts such as coordinate axes, coordinate planes in real life, the basics of geometry in three-dimensional space, etc.

• Coordinate Axes and Coordinate Planes in Three Dimensional Space: In this section, students will learn about the rectangular coordinate system, naming of a coordinate plane and different notations in coordinate planes.

• Coordinates of a Point in Space: This section explains in detail about the coordinate system in space with the help of examples.

• Distance between Two Points: Through this topic, students will learn how to calculate the distance between three points in a three-dimensional coordinate system using the distance formula.

• Section Formula: This section explains the section formula for a three-dimensional geometry and the different cases associated with it.

2. Where to find NCERT Solutions for Class 11 Maths Chapter 11 Introduction to Three Dimensional Geometry?

There are plenty of e-learning platforms available that caters to NCERT Solutions for Class 11 Maths. However, students should opt for platforms that are known to provide 100% accurate and verified solutions. For those who are looking for NCERT Solutions for Class 11 Maths, Vedantu is the best choice. At Vedantu, you can find the best NCERT Solutions for Class 11 Maths Chapter 11 Introduction to Three Dimensional Geometry and other chapters. These solutions are prepared by expert Maths teachers. They are well aware of the NCERT guidelines and board pattern. Hence, these solutions are really effective in scoring high marks in exams. Also, these are available at no cost.

3. Why should I download Vedantu’s NCERT Solutions for Class 11 Maths Chapter 11?

NCERT Solutions by Vedantu is a great study material for scoring excellent marks in exams. These solutions include chapter-wise explanations to NCERT exercise problems. Maths is a subject that requires regular practice to get the concepts right. For students facing any difficulties in finding the solutions to the exercise questions of Class 11 Maths Chapter 11, NCERT Solutions offered by Vednatu is the best resource. These solutions help students to find the answers to questions they are not able to solve. This will save a lot of time. Also, these solutions are provided by experts. Hence, students can be assured of the quality of these materials.

4. What is a Cartesian Coordinate System?

A Cartesian coordinate system is defined as a coordinate system that specifies each point uniquely in a plane by a set of numerical coordinates. These numerical coordinates are the signed distances to the point from two fixed perpendicular oriented lines. These are measured in the same unit of length. Each reference line is called a coordinate axis (or, simply axis) of the system. The point where the axes meet is called Origin (0,0). In three dimensional space, a cartesian coordinate system is based on three mutually perpendicular coordinate axes called the x-axis, y-axis and z-axis.

5. Explain the concept of 3D Geometry covered in Chapter 11 of NCERT Solutions for Class 11 Maths.

Chapter 11 of NCERT Class 11 Mathematics is an Introduction to Three Dimensional Geometry. This chapter includes concepts such as finding the point coordinates in a provided space, using section formula and distance formula to calculate the difference between the two points, and a brief about the cartesian coordinate system. Vedantu offers the students with notes, study material, and the solutions to all the exercises of this chapter in a step-by-step manner. These answers are verified by experienced experts.

6. Where can I get the NCERT Solutions for Class 11 Maths Chapter 11?

As the exams approach, most of the students bend towards finding the solutions to the given exercises. Vedantu offers solutions in a step-by-step manner for the Class 11 Maths Chapter 11. These solutions are verified by the subject-matter-experts. You can either refer to it online on the Vedantu website or on the Vedantu app, you can also download it for free for offline reference. These solutions are provided in accordance with the CBSE guidelines too.

7. How do you introduce 3D geometry?

3D Geometry can be defined as a three-dimensional solid figure that has length, width, and height as the measurements and finds application in everyday life such as building constructions, computer graphics, designing, etc. system. Chapter 11 in Class 11 introduces students to this subject and forms a base for further studies related to the same. Vedantu offers complete solutions for this chapter on their website for free.

8. What is 3D geometry,  explain in detail?

3D geometry is the mathematics of three-dimensional shapes that involves three parameters to determine the location of a particular point in space. The 3D shapes have the dimensions of height, breadth and length. It finds real-life applications in building constructions, computer graphics, designing, etc.  The experts at Vedantu have designed the complete study guide that includes topics such as finding the point coordinates in a provided space, using section formula and distance formula to calculate the difference between the two points, and a brief about the cartesian coordinate system.

9. What are the real-life applications of 3D geometry?

3D geometry is an important topic in today’s world. It finds application in Computer graphics, video-game programming, creation of virtual reality, construction, art, interior design, etc. Without the concept of 3D geometry, several things would have been challenging and unresolved. Take a look at Vedantu’s official website or download the Vedantu app for the chapter-wise simplified solutions at free of cost.