Maths NCERT Exemplar Solutions Class 11th Chapter Twelve: Overview, Questions, Preparation

Maths NCERT Exemplar Solutions Class 11th Chapter Twelve 2025 ( Maths NCERT Exemplar Solutions Class 11th Chapter Twelve )

Raj Pandey
Updated on Aug 6, 2025 10:57 IST

By Raj Pandey

Table of contents
  • Introduction to Three Dimensional Geometry Short Answer Type Questions
  • Introduction to Three Dimensional Geometry Long Answer Type Questions
  • Introduction to Three Dimensional Geometry Objective Type Questions
  • Introduction to Three Dimensional Geometry Fill in the blank Type Questions
  • Introduction to Three Dimensional Geometry Matching Type Questions
  • JEE Mains 2021
Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Logo

Introduction to Three Dimensional Geometry Short Answer Type Questions

Locate the following points:

i. (1, –1, 3),

ii. (–1, 2, 4)

iii. (–2, –4, –7),

iv. (–4, 2, –5).

Sol.

( i ) L o c a t i o n     o f     P ( 1 , − 1 , 3 ) = ( x , − y , z ) = I Voctant     ( i i ) L o c a t i o n     o f     Q ( − 1 , 2 , 4 ) = ( − x , y , z ) = I Ioctant     ( i i i ) L o c a t i o n     o f     R ( − 2 , − 4 , − 7 ) = ( − x , − y , − z ) = V I Ioctant     ( i v ) L o c a t i o n     o f     S ( − 4 , 2 , − 5 ) = ( − x , y , − z ) = V I    octant

Name the octant in which each of the following points lies:

i. (1, 2, 3),

ii. (4, –2, 3),

iii. (4, –2, –5),

iv. (4, 2, –5),

v. (–4, 2, 5),

vi. (–3, –1, 6),

vii. (2, –4, –7),

viii. (–4, 2, –5).

Sol:

( i )Point ( 1 , 2 , 3 )     l i e s     i n     IOctant     ( i i )Point ( 4 , − 2 , 3 )     l i e s     i n     I VOctant     ( i i i )Point ( 4 , − 2 , − 5 )     l i e s     i n     V I I I  Octant   ( i v )Point ( 4 , 2 , − 5 )     l i e s     i n     VOctant     ( v )Point ( − 4 , 2 , 5 )     l i e s     i n     I IOctant     ( v i )Point ( − 3 , − 1 , 6 )     l i e s     i n     I I IOctant     ( v i i )Point ( 2 , − 4 , − 7 )     l i e s     i n     V I I IOctant     ( v i i i )Point ( − 4 , 2 , − 5 )     l i e s     i n     V IOctant    

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Commonly asked questions
Q:  

Show that if x 2 + y 2 = 1 , then the point   ( x , y , 1 − x 2 − y 2 ) is at a distance of 1 unit from the origin.

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This is a short answer type question as classified in NCERT Exemplar

Given    is  (x,y,1−x2−y2)∴  Distance between  the  origin  and  the   point is=(x−0)2+(y−0)2+(1−x2−y2−0)2                                                                                                     =1=1Hence  proved.

Q:  

Name the octant in which each of the following points lies:

i. (1, 2, 3),

ii. (4, –2, 3),

iii. (4, –2, –5),

iv. (4, 2, –5),

v. (–4, 2, 5),

vi. (–3, –1, 6),

vii. (2, –4, –7),

viii. (–4, 2, –5).

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A: 

This is a short answer type question as classified in NCERT Exemplar

(i)Point(1,2,3)  lies  in  IOctant  (ii)Point(4,−2,3)  lies  in  IVOctant  (iii)Point(4,−2,−5)  lies  in  VIII Octant (iv)Point(4,2,−5)  lies  in  VOctant  (v)Point(−4,2,5)  lies  in  IIOctant  (vi)Point(−3,−1,6)  lies  in  IIIOctant  (vii)Point (2,−4,−7)  lies  in  VIIIOctant  (viii)Point(−4,2,−5)  lies  in  VIOctant

Q:  

Let   A , B , C be the feet of perpendiculars from a point   P on the x -, y -, and z -axes, respectively. Find the coordinates of   A , B , C in each of the following where the point   P is:

i. A = (3, 4, 2),

ii. (–5, 3, 7),

iii. (4, –3, –5).

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A: 

This is a short answer type question as classified in NCERT Exemplar

The  coordinates  of  A,   B  and  C  are (i)A (3, 0, 0),   B (0, 4, 0)  and  C (0, 0, 2) (ii)A (−5, 0, 0),   B (0, 3, 0)  and  C (0, 0, 7) (iii)A (4, 0, 0),   B (0, −3, 0)  and  C (0, 0, −5)

Q:  

Let   A , B , C be the feet of perpendiculars from a point   P on the x y -, y z -, and z x -planes, respectively. Find the coordinates of   A , B , C in each of the following where the point   P is:

i. (3, 4, 5),

ii. (–5, 3, 7)

iii. (4, –3, –5).

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This is a short answer type question as classified in NCERT Exemplar

T h e     c o o r d i n a t e s     o f     A ,     B     a n d     C     a r e ( i ) A ( 3 , 4 , 0 ) ,     B ( 0 , 4 , 5 )     a n d     C ( 3 , 0 , 5 ) ( i i ) A ( − 5 , 3 , 0 ) ,     B ( 0 , 3 , 7 )     a n d     C ( − 5 , 0 , 7 ) ( i i i ) A ( 4 , − 3 , 0 ) ,     B ( 0 , − 3 , − 5 )     a n d     C ( 4 , 0 , − 5 )

Q:  

How far apart are the points (2, 0, 0) and (–3, 0, 0)?

A: 

This is a short answer type question as classified in NCERT Exemplar

Given  points  are   (2, 0, 0),   and   (−3, 0, 0)∴  Distance between  the  given  = (2+3)2+ (0−0)2+ (0−0)2=25=5Hence,   the  required  =5.

Q:  

Find the distance from the origin to (6, 6, 7).

A: 

This is a short answer type question as classified in NCERT Exemplar

Coordinates  of  the  origin  are  (0,0,0)∴  Distance from  (0,0,0)  to  (6,6,7)=(6−0)2+(6−0)2+(7−0)2                                                                           =36+36+49=121=11 units.Hence,  the  required  =11.

Q:  

Three consecutive vertices of a parallelogram A B C D  are A ( 6,–2,4 ) , B ( 2,4,–8 ) , and C ( –2,2,4 ) . Find the coordinates of the fourth vertex.

[Hint: Diagonals of a parallelogram have the same mid-point.]

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Let  the  coordinates  of  the  fourth  vertex  be  (a,b,c)We  know  that  the  diagonals  of  a    parallelogram bisect  each  other.∴  Midpoint  of  diagonal  AC=(6−22,−2+22,4+42)=(2,0,4)and  the  midpoint  of  diagonal  BD=(a+22,b+42,c−82)∴        a+22=2     ⇒a=2           b+42=0     ⇒b=−4           c−82=4     ⇒c=16Hence,  the  required  coordinates  are  (2,−4,16).

Q:  

Show that the triangle   A B C with vertices A ( 0 , 4 , 1 ) , B ( 2,3,–1 ) , C ( 4 , 5 , 0 ) and  is right angled.

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G i v e n     v e r t i c e s     a r e     A ( 0 , 4 , 1 ) , B ( 2 , 3 , − 1 )     a n d     C ( 4 , 5 , 0 )               A B = ( 2 − 0 ) 2 + ( 3 − 4 ) 2 + ( − 1 − 1 ) 2 = 4 + 1 + 4 = 9 = 3               B C = ( 4 − 2 ) 2 + ( 5 − 3 ) 2 + ( 0 + 1 ) 2 = 4 + 4 + 1 = 9 = 3               A C = ( 4 − 0 ) 2 + ( 5 − 4 ) 2 + ( 0 − 1 ) 2 = 1 6 + 1 + 1 = 1 8 ?     ( 3 ) 2 + ( 3 ) 2 = ( 1 8 ) 2 S o ,     A B 2 + B C 2 = A C 2 . H e n c e ,     Δ A B C     i s     a     r i g h t     a n g l e d     t r i a n g l e .

Q:  

Find the third vertex of a triangle whose centroid is the origin and two vertices are ( 2 , 4 , 6 )  and ( 0,–2,–5 ) .

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Let  the  coordinates  of  the  third  vertex  i.e.,  A  be  (a,b,c)  Since the  centroid  is  at  origin  i.e,(0,0,0)∴        0=a+2+02       ⇒a=−2           0=b+4−22       ⇒b=−2and   0=c+6−52       ⇒c=−1Hence,  the  required  coordinates  are  (−2,−2,−1).

Q:  

Three vertices of a parallelogram A B C D  are A ( 1 , 2 , 3 ) , B ( –1,–2,–1 ) , and C ( 2 , 3 , 2 ) . Find the fourth vertex .

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Let  the  coordinates  of  D  be  (a,b,c)We  know  that  the  diagonals  of  a   parallelogram bisect   each  other.∴  Mid  of  AC  i.e.,  O=(1+22,2+32,3+22)=(32,52,52)     Mid  of  BD  i.e.,  O=(a−12,b−22,c−12)Equating  the  corresponding  coordinate,  we  have                       a−12=32      ⇒a=4                       b−22=52      ⇒b=7and                c−12=52      ⇒c=6Hence,  the  coordinates  of  D  (4,7,6).

Q:  

Find the coordinates of the points which trisect the line segment joining the points A ( 2,1,–3 )  and B ( 5,–8,3 ) .


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Let  C  and  D  be  the  points  which  divides  the  given  line  AB into   three  equal  parts.Here,  AC:CB=1:2Let  (x1,y1,z1)  be  the  coordinates  of  C∴         x1=1×5+2×21+2=3     and    y1=1×−8+2×11+2=−2            z1=1×3+2×−31+2=−1     So,  C=(3,−2,−1)Now  D  is  mid point of  CBLet  (x2,y2,z2)  be  the  coordinates  of  D∴         x2=3+52=4     and    y2=−8−22=−5            z2=3−12=1           So,  D=(4,−5,1)Hence,  the  required  coordinates  are  C(3,−2,−1)  and  D(4,−5,1).

Q:  

If the origin is the centroid of a triangle A B C  having vertices A ( a , 1 , 3 ) , B ( –2,b,–5 ) , and C ( 4 , 7 , c ) , find the values of a , b , c .

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C o o r d i n a t e s     o f     t h e     c e n t r o i d     G = ( 0 , 0 , 0 ) ∴                   0 = x 1 + x 2 + x 3 3                 ⇒ 0 = a − 2 + 4 3             ⇒ a = − 2                       0 = y 1 + y 2 + y 3 3                 ⇒ 0 = 1 + b + 7 3               ⇒ b = − 8 a n d       0 = z 1 + z 2 + z 3 3                   ⇒ 0 = 3 − 5 + c 3                 ⇒ c = 2 H e n c e ,     t h e     r e q u i r e d     v a l u e s     a r e     a = − 2 ,     b = − 8     a n d     c = 2 .

Q:  

Locate the following points:

i. (1, –1, 3),

ii. (–1, 2, 4)

iii. (–2, –4, –7),

iv. (–4, 2, –5).

A: 

This is a short answer type question as classified in NCERT Exemplar

(i)Location  of  P (1, −1, 3)= (x, −y, z)=IVoctant   (ii)Location  of  Q (−1, 2, 4)= (−x, y, z)=IIoctant   (iii)Location  of  R (−2, −4, −7)= (−x, −y, −z)=VIIoctant   (iv)Location  of  S (−4, 2, −5)= (−x, y, −z)=VI  octant

Q:  

Show that the points A   ( 1,–1,3 ) , B   ( 2,–4,5 ) , and   ( 5,–13,11 ) are collinear.

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Given  points  are  A(1,−1,3),B(2,−4,5)  and  C(5,−13,11)       AB=(2−1)2+(−4+1)2+(5−3)2=1+9+4=14       BC=(5−2)2+(−13+4)2+(11−5)2=9+81+36=126=314       AC=(5−1)2+(−13+1)2+(11−3)2=16+144+64=224=414Here,  we  observe  that  14+314=414So,  AB+BC=AC.Hence,  the  given   points are  collinear.

Q:  

Let A ( 2,2,–3 ) , B ( 5 , 6 , 9 ) , and C ( 2 , 7 , 9 )  be the vertices of a triangle. The internal bisector of angle A  meets B C  at the point D . Find the coordinates of D .

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Given  that  AD  is  the   internalbisector   of  ∠A∴    ABAC=BDDC        AB=(5−2)2+(6−2)2+(9+3)2=9+16+144=169=13        AC=(2−2)2+(7−2)2+(9+3)2=0+25+144=169=13∴    ABAC=BDDC=1313        ⇒BD=DC⇒D  is  the  mid point of  BC∴  Coordinates  of  D=(5+22,6+72,9+92)=(72,132,9)Hence,  the  required  coordinates  are  (72,132,9).

Q:  

If for some q, q, r ∈ R, not at all have same sign, one of the roots of the equation ( p 2 + q 2 ) x 2 − 2 q ( p + r ) x + q 2 + r 2 = 0  is also a root of the equation x2 + 2x – 8 = 0, then q 2 + r 2 p 2  is equal to………….

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A: 

Let a and b are the roots of ( p 2 + q 2 ) x 2 − 2 q ( p + r ) x + q 2 + r 2 = 0  

  ∴ α + β > 0     a n d     α β > 0     Also, it has a common root with x2 + 2x – 8 = 0

∴  The common root between above two equations is 4.

  ⇒ 1 6 ( p 2 + q 2 ) − 8 q ( p + r ) + q 2 + r 2 = 0 ⇒ ( 1 6 p 2 − 8 p q + q 2 ) + ( 1 6 q 2 − 8 q r + r 2 ) = 0  

      ⇒ ( 4 p − q ) 2 + ( 4 q − r ) 2 = 0 ⇒ q = 4 p     a n d     r = 1 6 p ∴ q 2 + r 2 p 2 = 1 6 p 2 + 2 5 6 p 2 p 2 = 2 7 2  

Q:  

The number of 5-digit natural numbers, such that the product of their digits is 36, is……

A: 

Factors of 36 = 22.32.1

Five-digit combinations can be 

(1, 2, 3, 3), (1, 4, 3, 1), (1, 9, 2, 1), (1, 4, 9, 11), (1, 2, 3, 6, 1), (1, 6, 1, 1)

i.e., total numbers  5 !       5 ! 2 !       2 ! + 5 ! 2 !       2 ! + 5 ! 2 !       2 ! + 5 ! 3 ! + 5 ! 2 ! + 5 ! 3 !     2 ! = ( 3 0 × 3 ) + 2 0 + 6 0 + 1 0 = 1 8 0 .  

Q:  

The series of positive multiples of 3 is divided into sets : { 3 } , { 6 ,     9 ,     1 2 } , { 1 5 ,     1 8 ,     2 1 ,     2 4 ,     2 7 } , . . . . . .  Then the sum of the elements in the 11th set is equal to…………..

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A: 

Given series { 3 × 1 } , { 3 × 2 , 3 × 3 , 3 × 4 } , { 3 × 5 , 3 × 6 , 3 × 7 , 3 × 8 , 3 × 9 } . . . . . . . . .  

∴  11th set will have 1 + (10)2 = 21 terms

Also up to 10th set total 3 × k type terms will be 1 + 3 + 5 + ……… +19 = 100 terms


∴ S e t     1 1 = { 3 × 1 0 1 , 3 × 1 0 2 , . . . . . . 3 × 1 2 1 }
∴ Sum of elements = 3 × (101 + 102 + ….+121)

= 3 × 2 2 2 × 2 1 2 = 6 9 9 3 .   

Q:  

The number of distinct real roots of the equation x5 ( x 3 − x 2 − x + 1 ) + x ( 3 x 3 − 4 x 2 − 2 x + 4 ) − 1 = 0  is …………..

A: 

x 8 − x 7 − x 6 + x 5 + 3 x 4 − 4 x 3 − 2 x 2 + 4 x − 1 = 0  

⇒ x 7 ( x − 1 ) − x 5 ( x − 1 ) + 3 x 3 ( x − 1 ) − x ( x 2 − 1 ) + 2 x ( 1 − x ) + ( x − 1 ) = 0  

⇒ ( x − 1 ) ( x 2 − 1 ) ( x 5 + 3 x − 1 ) = 0 ∴ x = ± 1  are roots of above equation and x5 + 3x – 1 is a monotonic term hence vanishes at exactly one value of x other then 1 or -1.

 3 real roots.

Q:  

If the coefficients of x and x2 in the expansion of (1 + x)p (1 – x)q, p, q ≤  15, are -3 and -5 respectively, then the coefficient of x3 is equal to…………………

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A: 

Coefficient of x in ( 1 + x ) p ( 1 − x ) q = − p C 0 q C 1 + p C 1 q C 0 = − 3 ⇒ p − q = 3  

              Coefficient of x2 in ( 1 + x ) p ( 1 − x ) q = p C 0 q C 2 − p C 1 q C 1 − p C 2 q C 0 = − 5  

⇒ q ( q − 1 ) 2 − p q + p ( p − 1 ) 2 = − 5 ⇒ q ( q − 1 ) 2 − ( q − 3 ) q + ( q − 3 ) ( q − 4 ) 2 = − 5 ⇒ q = 1 1 , p = 8  

              Coefficient of x3 in ( 1 + x ) 8 ( 1 − x ) 1 1 = − 1 1 C 3 + 8 C 1 1 1 C 2 − 8 C 2 1 1 C 1 + 8 C 3 = 2 3  

Q:  

If n ( 2 n + 1 ) ∫ 0 1 ( 1 − x n ) 2 n d x = 1 1 7 7 ∫ 0 1 ( 1 − x n ) 2 n + 1 dx, then n  N is equal to

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A: 

∫ 0 1 1 . ( 1 − x n ) 2 n + 1 d x using by parts we get

( 2 n 2 + n + 1 ) ∫ 0 1 ( 1 − x n ) 2 n + 1 d x = 1 1 7 7 ∫ 0 1 ( 1 − x n ) 2 n + 1 d x

⇒ 2 n 2 + n + 1 = 1 1 7 7 ⇒ n = 2 4     o r     − 4 9 2 ∴ n = 2 4

Q:  

Let a cure y = y(x) pass through the point (3, 3) and the area of the origin under this curve, above the x-axis and between the abscissae 3 and x (>3) be  ( y x ) 3 . If the curve also passes through the point ( α ,     6 1 0 )  in the first quadrant, then a is equal to………….

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A: 

∫ 3 x f ( x ) d x = ( f ( x ) x ) 3 ⇒ x 3 ∫ 3 x f ( x ) d x = f 3 ( x ) , differentiating w.r.to x

x 3 f ( x ) + 3 x 2 f 3 ( x ) x 3 = 3 f 2 ( x ) f ' ( x ) ⇒ 3 y 2 d y d x = x 3 y = 3 y 3 x ⇒ 3 x y d y d x = x 4 + 3 y 2  

After solving we get  y 2 = x 4 3 + c x 2  also curve passes through (3, 3) Þ c = -2


∴ y 2 = x 4 3 − 2 x 2
which passes through ( α , 6 1 0 ) ∴ α 4 − 6 α 2 3 = 3 6 0 ⇒ α = 6  

Q:  

The equations of the sides AB, BC and CA of a triangle ABC are 2x + y = 0, x + py = 15a and x – y = 3 respectively. If its orthocenter is (2, a), − 1 2 < a < 2 ,  then p is equal to…………

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A: 

Slope of AH = a + 2 1 slope of BC = − 1 p ∴ p = a + 2 ∴ C ( 1 8 p − 3 0 p + 1 , 1 5 p − 3 3 p + 1 )  

slope of HC =  1 6 p − p 2 − 3 1 1 6 p − 3 2  

slope of BC × slope of HC = -1 Þ p = 3 or 5

hence p = 3 is only possible value.

Q:  

Let the function f(x) = 2x2 – loge x, x  > 0, be decreasing in (0, a) and increasing in (a, 4). A tangent to the parabola y2 = 4ax at a point P on it passes through the point (8a, 8a – 1) but does not pass through the point ( − 1 a ,     0 ) . If the equation of the normal at P is x α + γ β = 1 ,  then a + b is equal to…………….

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A: 

f ' ( x ) = 4 x 2 − 1 x so f (x) is decreasing in ( 0 , 1 2 ) a n d     ( 1 2 , ∞ ) ⇒ a = 1 2  

Tangent at y2 = 2x is y = mx + 1 2 m it is passing through (4, 3) therefore we get m = 1 2 o r     1 4  

So tangent may be  y = 1 2 x + 1     o r     y = 1 4 x + 2       b u t     y = 1 2 x + 1  passes through (-2, 0) so rejected.

Equation of normal  x 9 + y 3 6 = 1  

Q:  

Let Q and R be two points on the line x + 1 2 = y + 2 3 = z − 1 2 at a distance 2 6  from the point P(4, 2, 7). Then the surface of the area of the triangle PQR is……………

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A: 

Let PT perpendicular to QR

x + 1 2 = y + 2 3 = z − 1 2 = λ ⇒ T ( 2 λ − 1 , 3 λ − 2 , 2 λ + 1 ) therefore

2 ( 2 λ − 5 ) + 3 ( 3 λ − 4 ) + 2 ( 2 λ − 6 ) = 0 ⇒ λ = 2

T ( 3 , 4 , 5 ) ∴ P T = 1 + 4 + 4 = 3 ∴ Q T = 2 6 − 9 = 1 7

∴ Δ P Q R = 1 2 × 2 1 7 × 3 = 3 1 7  

Therefore square of  a r ( Δ P Q R ) = 153.

Try these practice questions

Q1:

Let f : R →R be a continuous function such that f(3x) – f(x) =. If f(8) = 7, then f(14) is equal to:

Q2:

Let O be the origin and A be the point z1 = 1 + 2i. If B is the point z2, Re(z2) < 0, such that OAB is a right angled isosceles triangle with OB as hypotenuses, then which of the following is NOT true?

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Q3:

If the system of linear equations

8x + y + 4z = -=-2

x + y + z = 0

λ x - 3y = μ  

has infinitely many solutions, then the distance of the point  ( λ , μ − 1 2 )  from the plane 8x + y + 4z + 2 = 0 is:

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Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Logo

Introduction to Three Dimensional Geometry Long Answer Type Questions

1. Show that the three points A ( 2 , 3 , 4 )  ,  B ( 1 , 2 , 3 )  , and  C ( 4 , 1 , 1 0 )  are collinear and find the ratio in which  C  divides  A B  .

Sol:

G i v e n      points   a r e     A ( 2 , 3 , 4 ) ,     B ( − 1 , 2 , − 3 )     a n d     C ( − 4 , 1 , − 1 0 )                 A B = ( 2 + 1 ) 2 + ( 3 − 2 ) 2 + ( 4 + 3 ) 2 = 9 + 1 + 4 9 = 5 9                 B C = ( − 1 + 4 ) 2 + ( 2 − 1 ) 2 + ( − 3 + 1 0 ) 2 = 9 + 1 + 4 9 = 5 9                 A C = ( 2 + 4 ) 2 + ( 3 − 1 ) 2 + ( 4 + 1 0 ) 2 = 3 6 + 4 + 1 9 6 = 2 3 6 = 2 5 9 ∴         A B + B C = A C         5 9 + 5 9 = 2 5 9 H e n c e ,     A , B     a n d     C     a r e     c o l l i n e a r     a n d     A C : B C = 2 5 9 : 5 9 = 2 : 1 H e n c e ,     C     d i v i d e s     A B     i s     2 : 1     e x t e r n a l l y .

2. Prove that the points ( 0 , 1 , 7 )  ,  ( 2 , 1 , 9 )  , and  ( 6 , 5 , 1 3 )  are collinear. Find the ratio in which the first point divides the join of the other two.
L e t     t h e     g i v e n    points     a r e     A ( 0 , − 1 , − 7 ) ,     B ( 2 , 1 , − 9 ) ,     C ( 6 , 5 , − 1 3 )                         A B = ( 2 − 0 ) 2 + ( 1 + 1 ) 2 + ( − 9 + 7 ) 2 = 4 + 4 + 4 = 1 2 = 2 3                         B C = ( 6 − 2 ) 2 + ( 5 − 1 ) 2 + ( − 1 3 + 9 ) 2 = 1 6 + 1 6 + 1 6 = 4 8 = 4 3                         A C = ( 6 − 0 ) 2 + ( 5 + 1 ) 2 + ( − 1 3 + 7 ) 2 = 3 6 + 3 6 + 3 6 = 1 0 8 = 6 3                           2 3 + 4 3 = 6 3 i . e . ,                     A B + B C = A C ∴                                 A B : A C = 2 3 : 6 3 = 1 : 3 H e n c e ,    point     A     d i v i d e s     B     a n d     C     i n     1 : 3     e x t e r n a l l y .
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Commonly asked questions
Q:  

Show that the three points A(2,3,4) , B(1,2,3) , and C(4,1,10) are collinear and find the ratio in which C divides AB .

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A: 

This is a short answer type question as classified in NCERT Exemplar

Given   points are  A(2,3,4),  B(−1,2,−3)  and  C(−4,1,−10)        AB=(2+1)2+(3−2)2+(4+3)2=9+1+49=59        BC=(−1+4)2+(2−1)2+(−3+10)2=9+1+49=59        AC=(2+4)2+(3−1)2+(4+10)2=36+4+196=236=259∴    AB+BC=AC    59+59=259Hence,  A,B  and  C  are  collinear  and  AC:BC=259:59=2:1Hence,  C  divides  AB  is  2:1  externally.

Q:  

Prove that the points (0,1,7) , (2,1,9) , and (6,5,13) are collinear. Find the ratio in which the first point divides the join of the other two.

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A: 

This is a long answer type question as classified in NCERT Exemplar

Let  the  given  points  are  A(0,−1,−7),  B(2,1,−9),  C(6,5,−13)            AB=(2−0)2+(1+1)2+(−9+7)2=4+4+4=12=23            BC=(6−2)2+(5−1)2+(−13+9)2=16+16+16=48=43            AC=(6−0)2+(5+1)2+(−13+7)2=36+36+36=108=63             23+43=63i.e.,          AB+BC=AC∴                AB:AC=23:63=1:3Hence,  point  A  divides  B  and  C  in  1:3  externally.

Q:  

What are the coordinates of the vertices of a cube whose edge is 2 units, one of whose vertices coincides with the origin and the three edges passing through the origin, coincides with the positive direction of the axes through the origin?

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A: 

This is a long answer type question as classified in NCERT Exemplar

G i v e n     t h a t     e a c h     e d g e     o f     t h e     c u b o i d     i s     2     u n i t s . ∴     C o o r d i n a t e s     o f     t h e     v e r t i c e s     a r e             A ( 2 , 0 , 0 ) ,     B ( 2 , 2 , 0 ) ,     C ( 0 , 2 , 0 ) ,     D ( 0 , 2 , 2 ) ,     E ( 0 , 0 , 2 ) ,     F ( 2 , 0 , 2 ) ,     G ( 2 , 2 , 2 )     a n d     O ( 0 , 0 , 0 ) .

Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Logo

Introduction to Three Dimensional Geometry Objective Type Questions

1. The distance of point P ( 3 , 4 , 5 )  from the y z -plane is

(a) 3 units

(b) 4 units

(c) 5 units

(d) 550

Sol.

G i v e n    point     i s     P ( 3 , 4 , 5 ) ∴    Distance o f     P     f r o m     y z p l a n e = ( 0 − 3 ) 2 + ( 4 − 4 ) 2 + ( 5 − 5 ) 2 = 9 = 3   u n i t s H e n c e ,     t h e     c o r r e c t     o p t i o n     i s     ( a ) .

2. What is the length of the foot of the perpendicular drawn from the point P ( 3 , 4 , 5 )  on the y -axis?

(a) 4 1  

(b) 3 4  

(c) 5

(d) None of these

Sol. O n     y - a x i s ,     x = 0     a n d     z = 0     g i v e n     point     i s     P ( 3 , 4 , 5 ) ∴     T h e     point     A     i s     ( 0 , 4 , 0 ) ∴     P A = ( 0 − 3 ) 2 + ( 4 − 4 ) 2 + ( 0 − 5 ) 2 = 9 + 0 + 2 5 = 3 4 H e n c e ,     t h e     c o r r e c t     o p t i o n     i s     ( b ) .
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Commonly asked questions
Q:  

The distance of point P ( 3 , 4 , 5 )  from the y z -plane is

(a) 3 units

(b) 4 units

(c) 5 units

(d) 550

A: 

This is an Objective Type Questions as classified in NCERT Exemplar

Given  point  is  P (3, 4, 5)∴  Distance of  P  from  yzplane= (0−3)2+ (4−4)2+ (5−5)2=9=3 unitsHence,   the  correct  option  is   (a).

Q:  

What is the length of the foot of the perpendicular drawn from the point P ( 3 , 4 , 5 )  on the y -axis?

(a) 4 1  

(b) 3 4  

(c) 5

(d) None of these

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A: 

This is an Objective Type Questions as classified in NCERT Exemplar

On  y-axis,   x=0  and  z=0  given  point  is  P (3, 4, 5) ∴  The  point  A  is   (0, 4, 0) ∴     P A = ( 0 − 3 ) 2 + ( 4 − 4 ) 2 + ( 0 − 5 ) 2 = 9 + 0 + 2 5 = 3 4 H e n c e ,     t h e     c o r r e c t     o p t i o n     i s     ( b ) .

Q:  

Distance of the point ( 3 , 4 , 5 )  from the origin ( 0 , 0 , 0 )  is

(a) 5 0    

(b)   3

(c)   4

(d)    5

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A: 

This is an Objective Type Questions as classified in NCERT Exemplar

Given  points  A (3, 4, 5)  and  the  given  O (0, 0, 0)∴  OA= (3−0)2+ (4−0)2+ (5−0)2=9+16+25=50Hence,   the  correct  option  is   (a).

Q:  

If the distance between the points (a,0,1) and (0,1,2) is 27 , then the value of a is

(a) 5

(b) ± 5

(c) –5

(d) None of these

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A: 

This is an Objective Type Questions as classified in NCERT Exemplar

Let  the  given  points  be  A(a,0,1)  and  B(0,1,2)∴  AB=(a−0)2+(0−1)2+(1−2)2  27=a2+1+1Squaring  both  sides,  we  get      27=a2+2     ⇒a2=25     ∴a=±5Hence,  the  correct  option  is  (b).

Q:  

  x -axis is the intersection of two planes

(a) x y  and xz

(b) yz and zx

(c) xy and yz

(d) None of these

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A: 

This is an Objective Type Questions as classified in NCERT Exemplar

We  know  that  on  the  xy  and  xzplanes,   the  line  of intersection   is  xaxis.Hence,   the  correct  option  is   (a).

Q:  

Equation of y -axis is considered as

(a) x = 0 , y = 0  

(b)  y = 0 , z = 0

(c)    

(d)   None of these

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A: 

This is an Objective Type Questions as classified in NCERT Exemplar

On  y-axis,   x=0  and  z=0 H e n c e ,     t h e     c o r r e c t     o p t i o n     i s     ( c ) .

Q:  

The point (2,3,4) lies in the

(a) First octant

(b) Seventh octant

(c) Second octant

(d) Eighth octant

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A: 

This is an Objective Type Questions as classified in NCERT Exemplar

The  point   (−2, −3, −4)  lies  in  seventh  octant. H e n c e ,     t h e     c o r r e c t     o p t i o n     i s     ( b ) .

Q:  

A plane is parallel to yz-plane so it is perpendicular to:

(a) x -axis

(b) y-axis

(c) z-axis

(d) None of these

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A: 

This is an Objective Type Questions as classified in NCERT Exemplar

Any  plane  parallel  to  yz-plane  is  perpendicular  to  x-axis. H e n c e ,     t h e     c o r r e c t     o p t i o n     i s     ( a ) .

Q:  

The locus of a point for which y = 0 , z = 0  is

(a) Equation of x-axis

(b) Equation of y-axis

(c) Equation of z-axis

(d) None of these

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A: 

This is an Objective Type Questions as classified in NCERT Exemplar

We  know  that  one  equation  of  xaxis,   y=0,   z=0Hence,   the  locus  of  the   point is  equation  of  xaxis.Hence,   the  correct  option  is   (a).

Q:  

The locus of a point for which x = 0  is

(a) x y   -plane

(b) y z -plane

(c) z x -plane

(d) None of these

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A: 

This is an Objective Type Questions as classified in NCERT Exemplar

On  the  yzplane,   x=0Hence,   the  locus  of  the  point  is  yzplane.Hence,   the  correct  option  is   (b).

Q:  

If a parallelepiped is formed by planes drawn through the points ( 5 , 8 , 1 0 )  and   ( 3 , 6 , 8 ) parallel to the coordinate planes, then the length of the diagonal of the parallelepiped is

(a) 2 3    

(b) 3 2

(c) 2  

(d) 3

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A: 

This is an Objective Type Questions as classified in NCERT Exemplar

Given  points  are  A (5, 8, 10)  and  B (3, 6, 8)∴      AB= (5−3)2+ (8−6)2+ (10−8)2=4+4+4=12=23Hence,   the  correct  option  is   (a).

Q:  

 L is the foot of the perpendicular drawn from a point P(3,4,5) on the xy -plane. The coordinates of point L are

(a) ( 3 , 0 , 0 )          

(b) ( 0 , 4 , 5 )       

(c) ( 3 , 0 , 5 )     

(d)  None of these

Read more
A: 

This is an Objective Type Questions as classified in NCERT Exemplar

We  know  that  on  xyplane,   z=0So,   the  coordinates  of  the   point L  are   (3, 4, 0).Hence,   the  correct  option  is   (d).

Q:  

 L is the foot of the perpendicular drawn from a point (3,4,5) on the x -axis. The coordinates of L are

(a) ( 3 , 0 , 0 )  

(b) ( 0 , 4 , 0 )  

(c) ( 0 , 0 , 5 )      

(d) None of these

Read more
A: 

This is an Objective Type Questions as classified in NCERT Exemplar

We  know  that  on  x-axis,   y=0  and  z=0. S o ,     t h e     r e q u i r e d     c o o r d i n a t e s     a r e     ( 3 , 0 , 0 ) . H e n c e ,     t h e     c o r r e c t     o p t i o n     i s     ( a ) .

Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Logo

Introduction to Three Dimensional Geometry Fill in the blank Type Questions

1. The three axes O X , O Y , O Z  determine ____.

Sol:

T h e     t h r e e     a x e s     O X ,     O Y     a n d     O Z      determine   t h r e e     c o o r d i n a t e     p l a n e s . H e n c e ,     t h e     f i l l e r     v a l u e     i s     t h r e e     c o o r d i n a t e     p l a n e s .

2. The three planes determine a rectangular parallelepiped which has ____ rectangular faces.

Sol:

T h r e e     p a i r s . H e n c e ,     t h e     v a l u e     o f     t h e     f i l l e r     i s     t h r e e     p a i r s .

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Commonly asked questions
Q:  

The three axes OX,OY,OZ determine ____.

A: 

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

The  three  axes  OX,   OY  and  OZ   determine three  coordinate  planes.Hence,   the  filler  value  is  three  coordinate  planes.

Q:  

A line is parallel to xy -plane if all the points on the line have equal ____.

A: 

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

z-coordinates. Hence,   the  value  of  the  filler  is  z-coordinates.

Q:  

If the midpoints of the sides of a triangle AB,BC,CA are D(1,2,3),E(3,0,1),F(1,1,4) , then the centroid of the triangle ABC is ____.

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A: 

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

L e t     t h e     v e r t i c e s     o f     t h e     Δ A B C     b e A ( x 1 , y 1 , z 1 ) ,     B ( x 2 , y 2 , z 2 )     a n d     C ( x 3 , y 3 , z 3 ) D  is  the  mid-point  of  AB ∴                         1 = x 1 + x 2 2         ⇒ x 1 + x 2 = 2                                                       … ( i )                             2 = y 1 + y 2 2         ⇒ y 1 + y 2 = 4                                                     … ( i i ) a n d     − 3 = z 1 + z 2 2         ⇒ z 1 + z 2 = − 6                                                     … ( i i i ) E  is  the  mid-point  of  BC ∴                         3 = x 2 + x 3 2         ⇒ x 2 + x 3 = 6                                                       … ( i v )                             0 = y 2 + y 3 2         ⇒ y 2 + y 3 = 0                                                       … ( v ) a n d             1 = z 2 + z 3 2         ⇒ z 2 + z 3 = 2                                                             … ( v i ) F  is  the  mid-point  of  AC ∴             − 1 = x 1 + x 3 2         ⇒ x 1 + x 3 = − 2                                                       … ( v i i )                           1 = y 1 + y 3 2         ⇒ y 1 + y 3 = 2                                                           … ( v i i i ) a n d     − 4 = z 1 + z 3 2         ⇒ z 1 + z 3 = − 8                                                         … ( i x ) A d d i n g     e q . ( i ) , ( i v )     a n d     ( v i i )     w e     g e t                       2 ( x 1 + x 2 + x 3 ) = 2 + 6 − 2 ∴                                 x 1 + x 2 + x 3 = 3                                                                                                     … ( x ) A d d i n g     e q . ( i i ) , ( v )     a n d     ( v i i i )     w e     g e t                       2 ( y 1 + y 2 + y 3 ) = 4 + 0 + 2 ∴                                 y 1 + y 2 + y 3 = 3                                                                                                     … ( x i ) A d d i n g     e q . ( i i i ) , ( v i )     a n d     ( i x )     w e     g e t                       2 ( z 1 + z 2 + z 3 ) = − 6 + 2 − 8 ∴                                 z 1 + z 2 + z 3 = − 6                                                                                                   … ( x i i ) S u b t r a c t i n g     ( i )     f r o m     e q . ( x )     w e     g e t                                       x 3 = 3 − 2 = 1             ⇒ x 3 = 1 S u b t r a c t i n g     ( i v )     f r o m     e q . ( x )     w e     g e t                                       x 1 = 3 − 6 = − 3             ⇒ x 1 = − 3 S u b t r a c t i n g     ( v i i )     f r o m     e q . ( x )     w e     g e t                                       x 2 = 3 − ( − 2 ) = 5             ⇒ x 2 = 5

Q:  

The three planes determine a rectangular parallelepiped which has ____ rectangular faces.

A: 

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

Three  pairs.Hence,   the  value  of  the  filler  is  three  pairs.

Q:  

The coordinates of a point are the perpendicular distances from the ____ on the respective axes.

A: 

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

Givenpoints  Hence,   the  value  of  the  filler  is  given  .

Q:  

The three coordinate planes divide the space into ____ parts.

A: 

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

E i g h t . H e n c e ,     t h e     v a l u e     o f     t h e     f i l l e r     i s     e i g h t .

Q:  

If a point P lies in the yz-plane, then the coordinates of the point on the yz-plane are of the form ____.

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A: 

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

We  know  that  yzplane,   x=0So,   the  coordinates  of  the  required  point  is   (0, y, z).Hence,   the  value  of  the  filler  is   (0, y, z).

Q:  

The equation of yz-plane is ____.

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A: 

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

The  equation  of  yz-plane  is  x=0. H e n c e ,     t h e     v a l u e     o f     t h e     f i l l e r     i s     x = 0 .

Q:  

If the point P lies on the z -axis, then the coordinates of P are of the form ____.

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A: 

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

On  the  z-axis,   x=0  and  y=0. ∴     T h e     r e q u i r e d     c o o r d i n a t e     i s     i n     t h e     f o r m     o f     ( 0 , 0 , z ) . H e n c e ,     t h e     v a l u e     o f     t h e     f i l l e r     i s     ( 0 , 0 , z ) .

Q:  

The equation of z -axis is ____.

A: 

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

The  equation  of  z-axis  are,   x=0  and  y=0. H e n c e ,     t h e     v a l u e     o f     t h e     f i l l e r     i s     x = 0     a n d     y = 0 .

Q:  

A line is parallel to x -axis if all the points on the line have equal ____.

A: 

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

y     a n d     z     c o o r d i n a t e s . H e n c e ,     t h e     v a l u e     o f     t h e     f i l l e r     i s     y     a n d     z     c o o r d i n a t e s .

Q:  

x = a represents a plane parallel to ____.

A: 

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

x=a  represents  a  plane  parallel  to  yz-plane.

Q:  

The plane parallel to yz -plane is perpendicular to ____.

A: 

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

The  plane  parallel  to  yz-plane  is  perpendicular  to  x-axis. Hence,   the  value  of  the  filler  is  x-axis.

Q:  

The length of the longest piece of string that can be stretched straight in a rectangular room whose dimensions are 10, 13, and 8 units is ____.

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A: 

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

The  given  dimensions
 are  10,  13  and  8Let  a=10,  b=13  and  c=8∴ Required   length=a2+b2+c2                                        =(10)2+(13)2+(8)2=100+169+64=333Hence,  the  value  of  the  filler  is  333.

Q:  

If the distance between the points (a,2,1) and (1,1,1) is 5, then a ____.

A: 

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

The  given  points  are  (a,2,1)  and  (1,−1,1)∴  Distance=(a−1)2+(2+1)2+(1−1)2                     5=a2+1−2a+9Squaring  both  sides,  we  have                  25=a2+1−2a+9⇒              a2−2a−15=0⇒     a2−5a+3a−15=0⇒a(a−5)+3(a−5)=0⇒            (a+3)(a−5)=0∴    a=−3  or  5Hence,  the  value  of  the  filler  is  5  or  −3.

Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Logo

Introduction to Three Dimensional Geometry Matching Type Questions

Match each item given under Column C1 to its correct answer given under Column C2:

 Sol.  ( a ) I n     x y - p l a n e ,     z - c o o r d i n a t e     i s     z e r o .             H e n c e ,     ( a ) ↔ ( i i i ) . ( b ) T h e     point     ( 2 , 3 , 4 )     l i e s     i n     f i r s t     octant .             H e n c e ,     ( b ) ↔ ( i ) . ( c ) L o c u s     o f     t h e     points     w i t h     x - c o o r d i n a t e     i s     z e r o     i s     y z - p l a n e .             H e n c e ,     ( c ) ↔ ( i i ) . ( d ) A     l i n e     i s     p a r a l l e l     t o     x - a x i s     i f     a n d     o n l y     i f     a l l     t h e     points     o n     t h e     l i n e               h a v e     e q u a l     y     a n d     z - c o o r d i n a t e s .             H e n c e ,     ( d ) ↔ ( v i ) . ( e ) x = 0 ,     y = 0     r e p r e s e n t s     z - a x i s .             H e n c e ,     ( e ) ↔ ( i v ) . ( f ) z = c     r e p r e s e n t     a     p l a n e     p a r a l l e l     t o     x y - p l a n e .             H e n c e ,     ( f ) ↔ ( v ) . ( g ) T h e     p l a n e s     x = a ,     y = b     r e p r e s e n t     t h e     l i n e     p a r a l l e l     t o     z - a x i s .             H e n c e ,     ( g ) ↔ ( v i i i ) . ( h ) C o o r d i n a t e s     o f     a     point     a r e     t h e     distances     f r o m     t h e     o r i g i n     t o     t h e     f e e t     o f     p e r p e n d i c u l a r     f r o m                 t h e     point     o n     t h e     r e s p e c t i v e     a x e s ,             H e n c e ,     ( h ) ↔ ( v i i ) . ( i ) A     b a l l     i s     s o l i d     r e g i o n     i n     t h e     s p a c e     e n c l o s e d     b y     a     s p h e r e .             H e n c e ,     ( i ) ↔ ( x ) . ( j ) T h e     r e g i o n     i n     t h e     p l a n e     e n c l o s e d     b y     a     c i r c l e     i s     k n o w n     a s     a     d i s c .             H e n c e ,     ( i ) ↔ ( i x ) .

 

Q&A Icon
Commonly asked questions
Q:  

Match each item given under Column C1 to its correct answer given under Column C2:

A: 

This is a matching answer type question as classified in NCERT Exemplar

(a)In  xy-plane,  z-coordinate  is  zero.             H e n c e ,     ( a ) ↔ ( i i i ) . (b)The  point  (2,3,4)  lies  in  first  octant.             H e n c e ,     ( b ) ↔ ( i ) . (c)Locus  of  the  points  with  x-coordinate  is  zero  is  yz-plane.             H e n c e ,     ( c ) ↔ ( i i ) . (d)A  line  is  parallel  to  x-axis  if  and  only  if  all  the  points  on  the  line        have  equal  y  and  z-coordinates.             H e n c e ,     ( d ) ↔ ( v i ) . (e)x=0,  y=0  represents  z-axis.             H e n c e ,     ( e ) ↔ ( i v ) . (f)z=c  represent  a  plane  parallel  to  xy-plane.             H e n c e ,     ( f ) ↔ ( v ) . (g)The  planes  x=a,  y=b  represent  the  line  parallel  to  z-axis.             H e n c e ,     ( g ) ↔ ( v i i i ) . (h)Coordinates  of  a  point  are  the  distances  from  the  origin  to  the  feet  of  perpendicular  from         the  point  on  the  respective  axes,             H e n c e ,     ( h ) ↔ ( v i i ) . ( i ) A     b a l l     i s     s o l i d     r e g i o n     i n     t h e     s p a c e     e n c l o s e d     b y     a     s p h e r e .             H e n c e ,     ( i ) ↔ ( x ) . ( j ) T h e     r e g i o n     i n     t h e     p l a n e     e n c l o s e d     b y     a     c i r c l e     i s     k n o w n     a s     a     d i s c .             H e n c e ,     ( i ) ↔ ( i x ) .

Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Logo

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Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Exam

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