NCERT Solutions class 12 Maths Exercise 10.3 (Ex 10.3) Chapter 10 Vector Algebra

NCERT Solutions for Class 12 Maths Exercise 10.3 Chapter 10 Vector Algebra – FREE PDF Download

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.3 (Ex 10.3) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 10 Vector Algebra Exercise 10.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra (Ex 10.3) Exercise 10.3



1. Find the angle between two vectors   and   with magnitude   and 2 respectively having 
 
Ans. Given:   and  Let   be the angle between the vector   and 
We know that 
 
 = 
 
 

2. Find the angle between the vectors   and   

 

Ans. Given: Let   and     and   
Also 
= Product of coefficients of   + Product of coefficients of   + Product of coefficients 

Let   be the angle between the vector   and 
We know that 
 
 = 
 
 

3. Find the projection of the vector   on the vector   

 

Ans. Let   and  Projection of vector   and   = 


If projection of vector   and   is zero, then vector   is perpendicular to 

4. Find the projection of the vector iˆ+3jˆ+7kˆi^+3j^+7k^  on the vector   

 

Ans. Let   and  Projection of vector   and   = 


5. Show that each of the given three vectors is a unit vector:

   
Also show that they are mutually perpendicular to each other.
 

Ans. Let     ……….(i)    ……….(ii)
  ……….(iii)
 


 Each of the three given vectors   is a unit vector.
From eq. (i) and (ii),
  

   and   are perpendicular to each other.
From eq. (ii) and eq. (iii),
  

   and   are perpendicular to each other.
From eq. (i) and (iii),
  

   and   are perpendicular to each other.
Hence,   are mutually perpendicular vectors.

6. Find   and   if   and   

 

Ans. Given:   and     ……….(i)  
 
      ……….(ii)
Putting   in eq. (ii),

 
 
 
 
Putting   in eq (i),

7. Evaluate the product   

 

Ans. Given:   = 


8. Find the magnitude of two vectors   and   having the same magnitude such that the angle between them is   and their scalar product is   

 

Ans. Given:  , angle   (say) between   and   is   and their scalar (i.e., dot) product =        
Putting   and   we have 
 
 
 
   = 
   and 

9. Find   if for a unit vector     

 

Ans. Given:   is a unit vector      ……….(i)(xa)(x+a)=12(x→−a→)(x→+a→)=12
 
 
 
 
Putting   from eq. (i), 
       ∣∣x∣∣=13−−√|x→|=13

10. If   and   are such that   is perpendicular to   then find the value of   

 

Ans. Given:   and  Now   =

 
Again, 
Since,   is perpendicular to   therefore,  .
 
      
       

11. Show that   is perpendicular to   for any two non-zero vectors   and   

 

Ans. Let  , where   and  Let 
Now 



Putting,   and  ,
 = 0
 
Therefore, vectors   and   are perpendicular ot each other.

12. If   and  , then what can be concluded about the vector   

 

Ans. Given:       
Again     
    
 0 = 0 for all (any vector  )
Therefore,   can be any vector.

13. If   are unit vectors such that   find the value of 

 

Ans. Since,   are unit vectors.Therefore,   and      ……….(i)
Also given 
 
    
 
Putting the values from eq. (i), we get
 
 
 

14. If either vector   or   then  . But the converse need not be true. Justify your answer with an example.

 

Ans. Case I: Vector  . Therefore by definition of zero vector,    ……….(i)  
  [From eq. (i)]
 
Case II: Vector  . Therefore by definition of zero vector,    ……….(ii)
 
  [From eq. (ii)]
 
But the converse is not true.
Justification: Let 
Therefore, 
Therefore, 
Again let 
 
Therefore, 
But 
Hence, here  , but   and  .

15. If the vertices A, B, C of a triangle ABC are   and (0, 1, 2) respectively, then find   

 

Ans. Vertices A, B, C of a triangle are A (1, 2,3 ), B  and C (0, 1, 2) respectively.  Position vector of point A = 
Position vector of point B = 
Position vector of point C = 
Now   = Position vector of point A – Position vector of point B

  ……….(i)
And   = Position vector of point C – Position vector of point B

  ……….(ii)
Let   be the angle between the vectors   and  .
 
  [Using eq. (i) and (ii)]
 
 

16. Show that the points A (1, 2, 7), B (2, 6, 3) and C  are collinear.

 

Ans. Vertices A, B, C of a triangle are A (1, 2, 7), B (2, 6, 3) and C  respectively.  Position vector of point A = 
Position vector of point B = 
Position vector of point C = 
Now   = Position vector of point B – Position vector of point A

  ……….(i)
And   = Position vector of point C – Position vector of point A

 ……….(ii)
   = 2.    [Using eq. (i)]
 Vectors   and   are collinear and parallel. 
Thus, points A, B and C are collinear.
And also vectors   and   have a common point A and hence can’t be parallel.

17. Show that the vectors   and   form the vertices of a right angled triangle.

 

Ans. Let the given position vectors be A, B, C.  Position vector of point A is   Position vector of point B is   and Position vector of point C is 
   = Position vector of B – Position vector of A

 =    ……….(i)
 = Position vector of C – Position vector of B

 =       ……….(ii)
 = Position vector of C – Position vector of A

 =        ……….(iii)
Adding eq. (i) and (ii),
 +   =   +   =   =   [Using eq. (iii)]
Therefore, by Triangle law of addition of vectors, points A, B, C are the vertices of a triangle ABC.
Now from eq. (i) and (ii),
.  =   = 
Again from eq. (ii) and (iii),
.  =   = 2 + 3 – 5 = 0
   is perpendicular to  .
 Angle C is   Therefore   is a right angled at C.
Thus, A, B, C are the vertices of a right angled triangle.

18. If   is a non-zero vector of magnitude   and   is a non-zero scalar, then   is a unit vector if:

(A)    
(B)    
(C)    
(D)   
 

Ans. Given:   is a non-zero vector of magnitude   
Also given   and   is a unit vector.
 
 
 
 
Therefore, option (D) is correct.

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