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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.