Introduction to Three Dimensional Geometry

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Introduction to 3D Geometry Class 11 Maths Chapter 12 deals with the study of essential topics related to the fundamentals of geometry in three dimensions. Class 11 Maths Chapter 12 explains the key concepts of how to use the cartesian coordinate system, compute the distance between points using the distance formula and the section formula, and discover the coordinates of a point in a given space. The concepts taught in Class 11 Chapter 12 Introduction to Three Dimensional Geometry will also help students to gain a thorough understanding of this concept and its practical applications. The expert faculty of Toppr produced these solutions to assist students with their first-term exam preparations.

Introduction to 3D Geometry Class 11 Chapter 12 also covers topics such as Coordinate Axes and Coordinate Planes in Three Dimensional Space, identifying coordinates of a point in space, calculating the distance between two points, the section formula, and establishing point collinearity using the section formula. Students will quickly learn about these concepts if they practice Class 11 Chapter 12 Maths NCERT Solutions of Introduction to 3D Geometry. All of these solutions have been created with the new CBSE pattern in mind, ensuring that students have a thorough understanding of their exams.

Chapter 12 Maths Class 11 Questions and Answers can help you get good grades in tests and properly prepare for all of the important concepts. Several examples provided in the solutions will assist students with a better understanding of 3D geometry. Introduction to 3D Geometry Class 11 NCERT Solutions Chapter 12 is available below in pdf format, and a few solutions are also included in the exercises. These solutions explain the topics covered with examples so that students can easily relate to the notion being discussed.

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EXERCISE 12.1

Question 1

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

Solution

Verified by Toppr

If a point is on the $x$-axis then only $x$-coordinate will have non-zero constant value and other coordinates will be zero.

Hence, $y$-coordinates and $z$-coordinates are zero.

Question 2

If a point is in the $XZ$-plane. What can you say about its $y$-coordinate?

Solution

Verified by Toppr

If a point is in the $XZ$ plane then its $y$-coordinate is zero

Question 3

Name the octants in which the following points lie:

$(1,2,3),(4,−2,3)(4,−2,−5),(4,2,−5),(−4,2,−5),(−4,2,5),(−3,−1,6),(2,−4,−7)$

$(1,2,3),(4,−2,3)(4,−2,−5),(4,2,−5),(−4,2,−5),(−4,2,5),(−3,−1,6),(2,−4,−7)$

Solution

Verified by Toppr

The $x$-coordinate $y$-coordinate and$z$-coordinate of point $(1,2,3)$ are all positive.

Therefore this point lies in octant I

The $x$-coordinate $y$-coordinate and $z$-coordinate of point $(4,−2,3)$ are positive

negative and positive respectively. Therefore this point lies in octant IV

The $x$-coordinate $y$-coordinate and $z$-coordinate of point $(4,−2,−5)$ are positive

negative and negative respectively. Therefore this point lies in octant VIII

The $x$-coordinate, $y$-coordinate and $z$-coordinate of point $(4,2,−5)$ are positive

positive and negative respectively. Therefore this point lies in octant V

The $x$-coordinate, $y$-coordinate and $z$-coordinate of point $(−4,2,−5)$ are negative

positive, and negative respectively. Therefore this point lies in octant VI

The $x$-coordinate, $y$-coordinate and $z$-coordinate of point $(−4,2,5)$ are negative

positive and positive respectively. Therefore this point lies in octant II

The $x$-coordinate $y$-coordinate and $z$-coordinate of point $(−3,−1,6)$ are negative

negative and positive respectively. Therefore this point lies in octant III

The $x$-coordinate, $y$-coordinate and $z$-coordinate of point $(2,−4,−7)$ are positive

negative and negative respectively. Therefore this point lies in octant VIII

Therefore this point lies in octant I

The $x$-coordinate $y$-coordinate and $z$-coordinate of point $(4,−2,3)$ are positive

negative and positive respectively. Therefore this point lies in octant IV

The $x$-coordinate $y$-coordinate and $z$-coordinate of point $(4,−2,−5)$ are positive

negative and negative respectively. Therefore this point lies in octant VIII

The $x$-coordinate, $y$-coordinate and $z$-coordinate of point $(4,2,−5)$ are positive

positive and negative respectively. Therefore this point lies in octant V

The $x$-coordinate, $y$-coordinate and $z$-coordinate of point $(−4,2,−5)$ are negative

positive, and negative respectively. Therefore this point lies in octant VI

The $x$-coordinate, $y$-coordinate and $z$-coordinate of point $(−4,2,5)$ are negative

positive and positive respectively. Therefore this point lies in octant II

The $x$-coordinate $y$-coordinate and $z$-coordinate of point $(−3,−1,6)$ are negative

negative and positive respectively. Therefore this point lies in octant III

The $x$-coordinate, $y$-coordinate and $z$-coordinate of point $(2,−4,−7)$ are positive

negative and negative respectively. Therefore this point lies in octant VIII

Question 4

Fill in the blanks:

(i) The $x$-axis and $y$-axis taken together determine a plane known as______

(ii) The coordinates of points in the $XY$-plane are of the form_____

(iii) Coordinate planes divide the space into_____ octants

(i) The $x$-axis and $y$-axis taken together determine a plane known as______

(ii) The coordinates of points in the $XY$-plane are of the form_____

(iii) Coordinate planes divide the space into_____ octants

Solution

Verified by Toppr

(i) The $x$-axis and $y$-axis taken together determine a plane known as $XY$-plane

(ii) The coordinates of points in the $XY$-plane are of the form $(x,y,0)$, where $x$ is $x$-coordinate and $y$ is $y$-coordinate

(iii) Coordinate planes divide the space into $8$ octants.

(ii) The coordinates of points in the $XY$-plane are of the form $(x,y,0)$, where $x$ is $x$-coordinate and $y$ is $y$-coordinate

(iii) Coordinate planes divide the space into $8$ octants.

EXERCISE 12.2

Question 1

Find the distance between the following pairs of points:

(i) $(2,3,5)$ and $(4,3,1)$

(ii)$(−3,7,2)$and $((2,4,−1)$

(iii) $(−1,3,−4)$ and $(1,−3,4)$

(iv) $(2,−1,3)$ and $(−2,1,3)$

(i) $(2,3,5)$ and $(4,3,1)$

(ii)$(−3,7,2)$and $((2,4,−1)$

(iii) $(−1,3,−4)$ and $(1,−3,4)$

(iv) $(2,−1,3)$ and $(−2,1,3)$

Solution

Verified by Toppr

The distance between points $P(x_{1},y_{1},z_{1})$ and $Q(x_{2},y_{2},z_{2})$ is given by

$PQ=$ $(x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{2}+(z_{2}−z_{1})_{2} $

(i) Distance between points $(2,3,5)$ and $(4,3,1)$

$=$ $(4−2)_{2}+(3−3)_{2}+(1−5)_{2} $

$=$ $(2)_{2}+(0)_{2}+(−4)_{2} $

$=$ $4+16 $

$=$ $20 $

$=$ $25 $

(ii) Distance between points $(−3,7,2)$ and $(2,4,−1)$

$=$ $(2+3)_{2}+(4−7)_{2}+(−1−2)_{2} $

$=$ $(5)_{2}+(−3)_{2}+(−3)_{2} $

$=$ $25+9+9 $

$=$ $43 $

(iii) Distance between points $(−1,3,−4)$ and $(1,−3,4)$

$=$ $(1+1)_{2}+(−3−3)_{2}+(4+4)_{2} $

$=$ $(2)_{2}+(−6)_{2}+(8)_{2} $

$=$ $4+36+64 =104 $

$=226 $

(iv) Distance between points $(2,−1,3)$ and $(−2,1,3)$

$=$ $(−2−2)_{2}+(1+1)_{2}+(3−3)_{2} $

$=$ $(−4)_{2}+(2)_{2}+(0)_{2} $

$=$ $16+4 $

$=$ $20 $

$=$ $25 $

$PQ=$ $(x_{2}−x_{1})_{2}+(y_{2}−y_{1})_{2}+(z_{2}−z_{1})_{2} $

(i) Distance between points $(2,3,5)$ and $(4,3,1)$

$=$ $(4−2)_{2}+(3−3)_{2}+(1−5)_{2} $

$=$ $(2)_{2}+(0)_{2}+(−4)_{2} $

$=$ $4+16 $

$=$ $20 $

$=$ $25 $

(ii) Distance between points $(−3,7,2)$ and $(2,4,−1)$

$=$ $(2+3)_{2}+(4−7)_{2}+(−1−2)_{2} $

$=$ $(5)_{2}+(−3)_{2}+(−3)_{2} $

$=$ $25+9+9 $

$=$ $43 $

(iii) Distance between points $(−1,3,−4)$ and $(1,−3,4)$

$=$ $(1+1)_{2}+(−3−3)_{2}+(4+4)_{2} $

$=$ $(2)_{2}+(−6)_{2}+(8)_{2} $

$=$ $4+36+64 =104 $

$=226 $

(iv) Distance between points $(2,−1,3)$ and $(−2,1,3)$

$=$ $(−2−2)_{2}+(1+1)_{2}+(3−3)_{2} $

$=$ $(−4)_{2}+(2)_{2}+(0)_{2} $

$=$ $16+4 $

$=$ $20 $

$=$ $25 $

Question 2

Show that the points $(−2,3,5),(1,2,3)$ and $(7,0,−1)$ are collinear.

Solution

Verified by Toppr

Let points $(−2,3,5),(1,2,3)$ and $(7,0,−1)$ be denoted by $P,Q$ and $R$ respectively.

Points $P,Q$ and $R$ are collinear if they lie on a line.

$PQ=(1+2)_{2}+(2−3)_{2}+(3−5)_{2} $

$=(3)_{2}+(−1)_{2}+(−2)_{2} $

$=9+1+4 $

$=14 $

$QR=(7−1)_{2}+(0−2)_{2}+(−1−3)_{2} $

$=(6)_{2}+(−2)_{2}+(−4)_{2} $

$=36+4+16 $

$=56 $

$=214 $

$PR=(7+2)_{2}+(0−3)_{2}+(−1−5)_{2} $

$=(9)_{2}+(−3)_{2}+(−6)_{2} $

$=81+9+36 $

$=126 $

$=314 $

Here $PQ+QR=14 +214 $

$=314 $

$=PR$

$⇒PQ+QR=PR$

Hence points $P(−2,3,5),Q(1,2,3)$ and $R(7,0,−1)$ are collinear

Points $P,Q$ and $R$ are collinear if they lie on a line.

$PQ=(1+2)_{2}+(2−3)_{2}+(3−5)_{2} $

$=(3)_{2}+(−1)_{2}+(−2)_{2} $

$=9+1+4 $

$=14 $

$QR=(7−1)_{2}+(0−2)_{2}+(−1−3)_{2} $

$=(6)_{2}+(−2)_{2}+(−4)_{2} $

$=36+4+16 $

$=56 $

$=214 $

$PR=(7+2)_{2}+(0−3)_{2}+(−1−5)_{2} $

$=(9)_{2}+(−3)_{2}+(−6)_{2} $

$=81+9+36 $

$=126 $

$=314 $

Here $PQ+QR=14 +214 $

$=314 $

$=PR$

$⇒PQ+QR=PR$

Hence points $P(−2,3,5),Q(1,2,3)$ and $R(7,0,−1)$ are collinear

Question 3

Verify the following

(i) $(0,7,−10),(1,6,−6)$ and $(4,9,−6)$ are the vertices of an isosceles triangle

(ii) $(0,7,10),(−1,6,6)$ and $(−4,9,6)$ are the vertices of a right angled triangle

(iii) $(−1,2,1),(1,−2,5),(4,−7,8)$ and $(2,−3,4)$ are the vertices of a parallelogram

(i) $(0,7,−10),(1,6,−6)$ and $(4,9,−6)$ are the vertices of an isosceles triangle

(ii) $(0,7,10),(−1,6,6)$ and $(−4,9,6)$ are the vertices of a right angled triangle

(iii) $(−1,2,1),(1,−2,5),(4,−7,8)$ and $(2,−3,4)$ are the vertices of a parallelogram

Solution

Verified by Toppr

(i) Let point$(0,7,−10),(1,6,−6)$ and $(4,9,−6)$ be denoted by $A,B$ and $C$ respectively

$AB=(1−0)_{2}+(6−7)_{2}+(−6+10)_{2} $

$=(1)_{2}+(−1)_{2}+(4)_{2} $

$=1+1+16 $

$=18 $

$⇒AB=32 $

$BC=(4−1)_{2}+(9−6)_{2}+(−6+6)_{2} $

=$(3)_{2}+(3)_{2} $

=$9+9 =18 $

$⇒BC=32 $

$CA=(0−4)_{2}+(7−9)_{2}+(−10+6)_{2} $

=$(−4)_{2}+(−2)_{2}+(−4)_{2} $

=$16+4+16 =36 =6$

Here $AB=BC$ $=$ $CA$

Thus the given points are the vertices of an isosceles triangle

(ii) Let $(0,7,10),(−1,6,6)$ and $(−4,9,6)$ be denoted by $A,B$ and $C$ respectively

$AB=(−1−0)_{2}+(6−7)_{2}+(6−10)_{2} $

$=(−1)_{2}+(−1)_{2}+(−4)_{2} $

$=1+1+16 =18 $

$=32 $

$BC=(−4+1)_{2}+(9−6)_{2}+(6−6)_{2} $

$=(−3)_{2}+(3)_{2}+(0)_{2} $

$=9+9 =18 $

$=32 $

$CA=(0+4)_{2}+(7−9)_{2}+(10−6)_{2} $

$=(4)_{2}+(−2)_{2}+(4)_{2} $

$=16+4+16 $

$=36 $

$=6$

Now $AB_{2}+BC_{2}=(32 )_{2}+(32 )_{2}=18+18=36=AC_{2}$

Therefore by pythagoras theorem $ABC$ is a right triangle

Hence the given points are the vertices of a right-angled triangle

(iii) Let $(−1,2,1),(1,−2,5),(4,−7,8)$ and $(2,−3,4)$ be denoted by $A,B,C$ and $D$ respectively

$AB=(1+1)_{2}+(−2−2)_{2}+(5−1)_{2} $

$=4+16+16 $

$AB=36 $

$AB=6$

$BC=(4−1)_{2}+(−7+2)_{2}+(8−5)_{2} $

$=9+25+9 =43 $

$CD=(2−4)_{2}+(−3+7)_{2}+(4−8)_{2} $

$=4+16+16 $

$=36 $

$CD=6$

$DA=(−1−2)_{2}+(2+3)_{2}+(1−4)_{2} $

$DA=9+25+9 =43 $

Here $AB=CD=6$, $BC=AD=43 $

Hence the opposite sides of quadrilateral $ABCD$ whose vertices are taken in order are equal

Therefore $ABCD$ is a parallelogram

Hence the given points are the vertices of a parallelogram

$AB=(1−0)_{2}+(6−7)_{2}+(−6+10)_{2} $

$=(1)_{2}+(−1)_{2}+(4)_{2} $

$=1+1+16 $

$=18 $

$⇒AB=32 $

$BC=(4−1)_{2}+(9−6)_{2}+(−6+6)_{2} $

=$(3)_{2}+(3)_{2} $

=$9+9 =18 $

$⇒BC=32 $

$CA=(0−4)_{2}+(7−9)_{2}+(−10+6)_{2} $

=$(−4)_{2}+(−2)_{2}+(−4)_{2} $

=$16+4+16 =36 =6$

Here $AB=BC$ $=$ $CA$

Thus the given points are the vertices of an isosceles triangle

(ii) Let $(0,7,10),(−1,6,6)$ and $(−4,9,6)$ be denoted by $A,B$ and $C$ respectively

$AB=(−1−0)_{2}+(6−7)_{2}+(6−10)_{2} $

$=(−1)_{2}+(−1)_{2}+(−4)_{2} $

$=1+1+16 =18 $

$=32 $

$BC=(−4+1)_{2}+(9−6)_{2}+(6−6)_{2} $

$=(−3)_{2}+(3)_{2}+(0)_{2} $

$=9+9 =18 $

$=32 $

$CA=(0+4)_{2}+(7−9)_{2}+(10−6)_{2} $

$=(4)_{2}+(−2)_{2}+(4)_{2} $

$=16+4+16 $

$=36 $

$=6$

Now $AB_{2}+BC_{2}=(32 )_{2}+(32 )_{2}=18+18=36=AC_{2}$

Therefore by pythagoras theorem $ABC$ is a right triangle

Hence the given points are the vertices of a right-angled triangle

(iii) Let $(−1,2,1),(1,−2,5),(4,−7,8)$ and $(2,−3,4)$ be denoted by $A,B,C$ and $D$ respectively

$AB=(1+1)_{2}+(−2−2)_{2}+(5−1)_{2} $

$=4+16+16 $

$AB=36 $

$AB=6$

$BC=(4−1)_{2}+(−7+2)_{2}+(8−5)_{2} $

$=9+25+9 =43 $

$CD=(2−4)_{2}+(−3+7)_{2}+(4−8)_{2} $

$=4+16+16 $

$=36 $

$CD=6$

$DA=(−1−2)_{2}+(2+3)_{2}+(1−4)_{2} $

$DA=9+25+9 =43 $

Here $AB=CD=6$, $BC=AD=43 $

Hence the opposite sides of quadrilateral $ABCD$ whose vertices are taken in order are equal

Therefore $ABCD$ is a parallelogram

Hence the given points are the vertices of a parallelogram

Question 4

Find the equation of the set of points which are equidistant from the points $(1,2,3)$ and $(3,2,−1)$

Solution

Verified by Toppr

Let $P(x,y,z)$ be the point that is equidistant from points $A(1,2,3)$ and $B(3,2,−1)$ .

i.e. $PA=PB$

$⇒PA_{2}=PB_{2}$

$⇒(x−1)_{2}+(y−2)_{2}+(z−3)_{2}=(x−3)_{2}+(y−2)_{2}+(z+1)_{2}$

$⇒x_{2}−2x+1+y_{2}−4y+4+z_{2}−6z+9=x_{2}−6x+9+y_{2}−4y+4+z_{2}+2z+1$

$⇒−2x−6z+14=−6x+2z+14$

$⇒−2x−6z+6x−2z=0$

$⇒4x−8z=0$

$⇒x−2z=0$

Thus the required equation is $x−2z=0$.

i.e. $PA=PB$

$⇒PA_{2}=PB_{2}$

$⇒(x−1)_{2}+(y−2)_{2}+(z−3)_{2}=(x−3)_{2}+(y−2)_{2}+(z+1)_{2}$

$⇒x_{2}−2x+1+y_{2}−4y+4+z_{2}−6z+9=x_{2}−6x+9+y_{2}−4y+4+z_{2}+2z+1$

$⇒−2x−6z+14=−6x+2z+14$

$⇒−2x−6z+6x−2z=0$

$⇒4x−8z=0$

$⇒x−2z=0$

Thus the required equation is $x−2z=0$.

Question 5

Find the equation of the set of points $P$, the sum of whose distances from $A(4,0,0)$ and $B(−4,0,0)$ is equal to $10$.

Solution

Verified by Toppr

Let the coordinates of $P$ be $(x,y,z)$.

The coordinates of points $A$ and $B$ are $(4,0,0)$ and $(−4,0,0)$ respectively.

It is given that $PA+PB=10$

$⇒(x−4)_{2}+y_{2}+z_{2} +(x+4)_{2}+y_{2}+z_{2} =10$

$⇒(x−4)_{2}+y_{2}+z_{2} =10−(x+4)_{2}+y_{2}+z_{2} $

On squaring both sides, we obtain

$⇒(x−4)_{2}+y_{2}+z_{2}=100+(x+4)_{2}+y_{2}+z_{2}−20(x+4)_{2}+y_{2}+z_{2} $

$⇒x_{2}−8x+16+y_{2}+z_{2}=100−20x_{2}+8x+16+y_{2}+z_{2} +x_{2}+8x+16+y_{2}+z_{2}$

$⇒20x_{2}+8x+16+y_{2}+z_{2} =100+16x$

$⇒5x_{2}+8x+16+y_{2}+z_{2} =(25+4x)$

On squaring both sides again, we obtain

$25(x_{2}+8x+16+y_{2}+z_{2})=625+16x_{2}+200x$

$⇒25x_{2}+200x+400+25y_{2}+25z_{2}=625+16x_{2}+200x$

$⇒9x_{2}+25y_{2}+25z_{2}−225=0$

Thus the required equation is $9x_{2}+25y_{2}+25z_{2}−225=0$.

The coordinates of points $A$ and $B$ are $(4,0,0)$ and $(−4,0,0)$ respectively.

It is given that $PA+PB=10$

$⇒(x−4)_{2}+y_{2}+z_{2} +(x+4)_{2}+y_{2}+z_{2} =10$

$⇒(x−4)_{2}+y_{2}+z_{2} =10−(x+4)_{2}+y_{2}+z_{2} $

On squaring both sides, we obtain

$⇒(x−4)_{2}+y_{2}+z_{2}=100+(x+4)_{2}+y_{2}+z_{2}−20(x+4)_{2}+y_{2}+z_{2} $

$⇒x_{2}−8x+16+y_{2}+z_{2}=100−20x_{2}+8x+16+y_{2}+z_{2} +x_{2}+8x+16+y_{2}+z_{2}$

$⇒20x_{2}+8x+16+y_{2}+z_{2} =100+16x$

$⇒5x_{2}+8x+16+y_{2}+z_{2} =(25+4x)$

On squaring both sides again, we obtain

$25(x_{2}+8x+16+y_{2}+z_{2})=625+16x_{2}+200x$

$⇒25x_{2}+200x+400+25y_{2}+25z_{2}=625+16x_{2}+200x$

$⇒9x_{2}+25y_{2}+25z_{2}−225=0$

Thus the required equation is $9x_{2}+25y_{2}+25z_{2}−225=0$.

EXERCISE 12.3

Question 1

Find the coordinates of the point which divides the line segment joining the points $(−2,3,5)$ and $(1,−4,6)$ in the ratio

(i) $2:3$ internally (ii) $2:3$ externally.

(i) $2:3$ internally (ii) $2:3$ externally.

Solution

Verified by Toppr

(i) The coordinates of point R that divides the line segment joining
points $P(x_{1},y_{1},z_{1})$ and $Q(x_{2},y_{2},z_{2})$ internally in the
ratio $m:n$ are

$(m+nmx_{2}+nx_{1} ,m+nmy_{2}+ny_{1} ,m+nmz_{2}+nz_{1} )$

Let $R(x,y,z)$ be the points that divides the line segment joining points $(−2,3,5)$ and $(1,−4,6)$ internally in the ratio $2:3$

$x=2+32(1)+3(−2) ,y=2+32(−4)+3(3) $ and $z=2+32(6)+3(5) $

i.e., $x=5−4 ,y=51 ,$ and $z=527 $

Thus, the coordinates of the required point are $(5−4 ,51 ,527 )$

(ii) The coordinates of point R that divides the line segment joining points $P(x_{1},y_{1},z_{1})$ and $Q(x_{2},y_{2},z_{2})$ externally in the ratio $m:n$ are

$(m−nmx_{2}+nx_{1} ,m−nmy_{2}+ny_{1} ,m−nmz_{2}+nz_{1} )$

Let $R(x,y,z)$ be the point that divides the line segment joining points $(−2,3,5)$ and $(1,−4,6)$ externally in the ratio $2:3$

$x=2−32(1)−3(2) ,y=2−32(−4)−3(3) ,$ and $z=2−32(6)−3(5) $

i.e., $x=−8,y=17,$ and $z=3$

Thus, the coordinates of the required point are $(−8,17,3).$

$(m+nmx_{2}+nx_{1} ,m+nmy_{2}+ny_{1} ,m+nmz_{2}+nz_{1} )$

Let $R(x,y,z)$ be the points that divides the line segment joining points $(−2,3,5)$ and $(1,−4,6)$ internally in the ratio $2:3$

$x=2+32(1)+3(−2) ,y=2+32(−4)+3(3) $ and $z=2+32(6)+3(5) $

i.e., $x=5−4 ,y=51 ,$ and $z=527 $

Thus, the coordinates of the required point are $(5−4 ,51 ,527 )$

(ii) The coordinates of point R that divides the line segment joining points $P(x_{1},y_{1},z_{1})$ and $Q(x_{2},y_{2},z_{2})$ externally in the ratio $m:n$ are

$(m−nmx_{2}+nx_{1} ,m−nmy_{2}+ny_{1} ,m−nmz_{2}+nz_{1} )$

Let $R(x,y,z)$ be the point that divides the line segment joining points $(−2,3,5)$ and $(1,−4,6)$ externally in the ratio $2:3$

$x=2−32(1)−3(2) ,y=2−32(−4)−3(3) ,$ and $z=2−32(6)−3(5) $

i.e., $x=−8,y=17,$ and $z=3$

Thus, the coordinates of the required point are $(−8,17,3).$

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Practice more questions

**Introduction**

- A cartesian coordinate system in three dimensions is made up of three mutually perpendicular lines known as the x, y, and z-axes. They are both measured in the same length unit.
- The three planes determined by the pair of axes are the coordinate planes. These planes are known as the x-y, y-z, and x-z planes.
- The three coordinate planes split space into eight equal portions called octants.

**Distance between Two Points**

Let P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) be two points referring to a system of rectangular axes OX, OY and OZ. Distance between PQ is:

*√ [(x _{2 }- x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]*

**Section Formula**

Let the two given points be P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}). The coordinates of the Point R(x, y, z) if:

- If point R (x, y, z) divide PQ in the given ratio
*m : n*internally:

*x = (mx _{2} + nx_{1}) / m + n, y = (my_{2} + ny_{1}) / m + n, z = (mz_{2} + nz_{1}) / m + n*

- If point R (x, y, z) divide PQ in the given ratio
*m : n*externally:

*x = (mx _{2} - nx_{1}) / m - n, y = (my_{2} - ny_{1}) / m - n, z = (mz_{2} - nz_{1}) / m - n*

- Coordinates of the midpoint of the line segment, if R is the midpoint of PQ:

*x = (x _{1} + x_{2}) / 2, y = (y_{1} + y_{2}) / 2, z = (z_{1} + z_{2}) / 2*

- Coordinates of the centroid of the triangle:

*( (x _{1} + x_{2} + x_{3}) / 3, (y_{1} + y_{2} + y_{3}) / 3, (z_{1} + z_{2} + z_{3}) / 3 )*

Related Chapters

- Chapter 1 : SetsChapter 2 : Relations and FunctionsChapter 3 : Trigonometric FunctionsChapter 4 : Principle of Mathematical InductionChapter 5 : Complex Numbers and Quadratic EquationsChapter 6 : Linear InequalitiesChapter 7 : Permutations and CombinationsChapter 8 : Binomial TheoremChapter 9 : Sequences and SeriesChapter 10 : Straight LinesChapter 11 : Conic SectionsChapter 13 : Limits and DerivativesChapter 14 : Mathematical ReasoningChapter 15 : StatisticsChapter 16 : Probability

**Q1. ****How many exercises are there in the Introduction to Three Dimensional Geometry Class 11 NCERT Solutions****?**

**Answer:** Class 11 Maths Chapter 12 Introduction To 3D Geometry has 14 questions, 9 of which are easy, 3 of which are somewhat straightforward, and 3 of which are extended answer-type questions. The solutions to the exercise-specific problems are also accessible in PDF format, with the goal of assisting students in performing well in the yearly exam.

**Q2. What are the key subtopics in Introduction to 3D Geometry Class 11 Maths Chapter 12 that could be tested?**

**Answer:** NCERT Solutions Class 11 Maths Chapter 12 begins with a brief overview of the 3D Geometry concepts. The NCERT Solutions Class 11 Maths Chapter 12 focuses on Coordinate Axes and Coordinate Planes in Three Dimensional Space, identifying coordinates of a point in space, calculating the distance between two points, the section formula, the formula for finding the coordinates of the mid-point of the line segment, and to find the coordinates of the centroid of the triangle. Students can now study and stay up to date on the latest CBSE syllabus by using the NCERT Solutions, which are available in PDF format.

**Q3. Is it necessary to practice all of the questions in Introduction to 3D Geometry Class 11 NCERT Solutions?**

**Answer:** Introduction to 3D Geometry Class 11 NCERT Solutions are well-designed to help students study 3D geometry concepts easily. It thoroughly covers all essential principles, allowing students to grasp the concepts. All of the exercises, examples, and practice problems have been meticulously designed to help students gain a rapid and effective comprehension of the various topics.

Related Chapters

- Chapter 1 : SetsChapter 2 : Relations and FunctionsChapter 3 : Trigonometric FunctionsChapter 4 : Principle of Mathematical InductionChapter 5 : Complex Numbers and Quadratic EquationsChapter 6 : Linear InequalitiesChapter 7 : Permutations and CombinationsChapter 8 : Binomial TheoremChapter 9 : Sequences and SeriesChapter 10 : Straight LinesChapter 11 : Conic SectionsChapter 13 : Limits and DerivativesChapter 14 : Mathematical ReasoningChapter 15 : StatisticsChapter 16 : Probability