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 

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