Home » NCERT Solutions » NCERT Solutions class 12 Maths Exercise 10.2 (Ex 10.2) Chapter 10 Vector Algebra

NCERT Solutions class 12 Maths Exercise 10.2 (Ex 10.2) Chapter 10 Vector Algebra

NCERT Solutions for Class 12 Maths Exercise 10.2 Chapter 10 Vector Algebra – FREE PDF Download

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.2 (Ex 10.2) 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 10.2 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 10.2) Exercise 10.2



1. Compute the magnitude of the following vectors:

 

 

Ans. Given:  

 ∣∣a∣∣=x2+y2+z2−−−−−−−−−−√=(1)2+(1)2+(1)2−−−−−−−−−−−−−−√=1+1+1−−−−−−−√=3–√|a→|=x2+y2+z2=(1)2+(1)2+(1)2=1+1+1=3

And 

 ∣∣∣b∣∣∣=x2+y2+z2−−−−−−−−−−√=(2)2+(7)2+(3)2−−−−−−−−−−−−−−−−−√=4+49+9−−−−−−−−√=62−−√|b→|=x2+y2+z2=(2)2+(−7)2+(−3)2=4+49+9=62

Also 

 ∣∣c∣∣=x2+y2+z2−−−−−−−−−−√=(13)2+(13)2+(13)2−−−−−−−−−−−−−−−−−−−−√=13+13+13−−−−−−−−−√=1|c→|=x2+y2+z2=(13)2+(13)2+(−13)2=13+13+13=1


2. Write two different vectors having same magnitude.

 

 

Ans. Let  and  

Clearly,  [ Coefficients of  and  are same in vectors  and  coefficients of  in  and  are unequal as  ]

But 

And 

Such possible answers are infinite.


3. Write two different vectors having same direction.

 

 

Ans. Let  and  =  

  where  = 2 > 0

 Vectors  and  have the same direction.

But 

Such possible vectors are infinite.


4. Find the values of  and  so that the vectors  and  are equal.

 

 

Ans. Given:  

If vectors are equal, then their respective components are equal.

Comparing coefficients of  and  on both sides, we have,

 and 


5. Find the scalar and vector components of the vector with initial point (2, 1) and terminal point  

 

 

Ans. Let  be the vector with initial point A (2, 1) and terminal point B 

 Position vector of point A is (2, 1) =  and position vector of point B is  =  

  = Position vector of point B – Position vector of point A

  = 

 Scalar components of the vectors  are coefficients of  and  in  i.e.,  and 6 and vector components of the vector  are  and 


6. Find the sum of the vectors:   and  

 

 

Ans. Given:  and  

Adding, 


7. Find the unit vector in the direction of the vector  

 

 

Ans. We know that a unit vector in the direction of the vector  is 

aˆ=a∣∣a∣∣=iˆ+jˆ+2kˆ(1)2+(1)2+(2)2=iˆ+jˆ+2kˆ1+1+4=iˆ+jˆ+2kˆ6a^=a→|a→|=i^+j^+2k^(1)2+(1)2+(2)2=i^+j^+2k^1+1+4=i^+j^+2k^6

 aˆ=16iˆ+16jˆ+26kˆa^=16i^+16j^+26k^


8. Find the unit vector in the direction of the vector  where P and Q are the points (1, 2, 3) and (4, 5, 6) respectively.

 

 

Ans.  Given: Points P (1, 2, 3) and Q (4, 5, 6) 

 Position vector of point P =  =  and position vector of Q =  = , where O is the origin.

  = Position vector of Q – Position vector of P =   

   = 

Therefore, the unit vector in the direction of vector  = 

3iˆ+3jˆ+3kˆ(3)2+(3)2+(3)23i^+3j^+3k^(3)2+(3)2+(3)2=


9. For given vectors  and  find the unit vector in the direction of  

 

 

Ans. Given: Vectors  and  

  = 

Therefore, ∣∣∣a+b∣∣∣=iˆ+0jˆ+kˆ(1)2+(0)2+(1)2|a→+b→|=i^+0j^+k^(1)2+(0)2+(1)2

 


10. Find the vector in the direction of vector  which has magnitude 8 units.

 

 

Ans. Let  

A vector in the direction of vector  which has magnitude 8 units = 

8.a∣∣a∣∣=8(5iˆjˆ+2kˆ)(5)2+(1)2+(2)28.a→|a→|=8(5i^−j^+2k^)(5)2+(−1)2+(2)2

    = 


11. Show that the vectors  and  are collinear.

 

 

Ans. Let  and  =  =  

  =  where 

Therefore, vectors 2iˆ3jˆ+4kˆ2i^−3j^+4k^ and 4iˆ+6jˆ8kˆ−4i^+6j^−8k^ are collinear.


12. Find the direction cosines of the vector  

 

 

Ans. The given vector is  

  = iˆ+2jˆ+3kˆ(1)2+(2)2+(3)2=i^+2j^+3k^(1)2+(2)2+(3)2= 

We know that the direction cosines of a vector  are coefficients of  in  i.e., 


13. Find the direction cosines of the vector joining the points A and B directed from A to B.

 

 

Ans. Given: Points A and B 

 Position vector of point A =  = 

And Position vector of point B =  = 

 Vector  =    =    = 

Now  = 

 A unit vector along 

Therefore, the direction cosines of vector  = 


14. Show that the vector  is equally inclined to the axes OX, OY and OZ.

 

 

Ans. Let , then ∣∣a∣∣=(1)2+(1)2+(1)2−−−−−−−−−−−−−−√=3–√|a→|=(1)2+(1)2+(1)2=3 

Let us find angle  (say) between vector  and OX 

  =  = 

 

Similarly angle  (say) between vector  and OY  is 

And angle  (say) between vector  and OZ  is 

  =  = 

 Vector  is equally inclined to OX, OY and OZ.


15. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are  and  respectively, in the ratio 2 : 1 (i) internally (ii) externally.

 

 

Ans. Position vector of P is  and Position vector of Q is  

(i) Position vector of point R dividing PQ internally (i.e., R lies within the segment PQ) in the ratio 2 : 1 =  = PR : QR is 

(ii) Position vector of point R dividing PQ externally (i.e., R lies outside the segment PQ and to the right of point Q because ratio 2 : 1 > 1 i.e., PR : QR = 2 : 1) is 

3iˆ+3kˆ−3i^+3k^


16. Find the position vector of the mid-point of the vector joining the points P (2, 3, 4) and Q 

 

 

Ans. Given: Point P (2, 3, 4) and Q 

 Position vector of point P is 

And Position vector of point Q is 

And Position vector of mid-point R of PQ is a+b2a→+b→2


17. Show that the points A, B and C with position vectors  and  respectively form the vertices of a right angled triangle.

 

 

Ans. Given: Position vector of point A is  

Position vector of point B is  and

Position vector of point C is  where O is the origin.

Now  = Position vector of point B – Position vector of point A

 = 

 ……….(i)

 = Position vector of point C – Position vector of point B

 = 

 ……….(ii)

 = Position vector of point C – Position vector of point A

 = 

 ……….(iii)

Adding eq. (i) and (ii),

 +  =   =  = [By eq. (iii)]

Now From eq. (i),AB = 

From eq. (ii),BC = 

From eq. (iii),AC = 

Here, we can observe that (Longest side BC)2 =  AB2 + AC2

Also, AB−→−.CA−→−=(+1)(2)+(3)(1)+(5)(1)AB→.CA→=(+1)(2)+(−3)(−1)+(−5)(1) = 2 + 3 – 5 = 0

Thus, AB−→−CA−→−AB→⊥CA→ A=90∠A=90∘

Therefore, A B, C are the vertices of a right angled triangle.


18. In triangle ABC (Fig. below), which of the following is not true:

 

 

(A)  

(B) 

(C)  

 

Ans. We know by Triangle law of Addition of vectors that    

  

Therefore option (C) is not true because in option (C),


19. If  and  are two collinear vectors, then which of the following are incorrect:

 

(A)  =  for some scalar  

(B)  

(C) The respective components of  and  are proportional.

(D) Both the vectors  and  have same direction, but different magnitudes.

 

Ans. Option (D) is not true because two collinear vectors can have different directions and also different magnitudes. 

The option (A) and option (C) are true by definition of collinear vectors.

Option (B) is a particular case of option (A) taking