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NCERT Solutions class 12 Maths Exercise 10.4 (Ex 10.4) Chapter 10 Vector Algebra

NCERT Solutions for Class 12 Maths Exercise 10.4 Chapter 10 Vector Algebra – FREE PDF Download

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4 (Ex 10.4) 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 10.4 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 10.4) Exercise 10.4



1. Find  if  and  

 

Ans. Given:  and  

Expanding along first row,

 = 

 


2. Find a unit vector perpendicular to each of the vectors  and  where  and  

 

 

Ans. Given:  and  

On Adding   =  +  = 

On Subtracting   =    = 

Therefore, 

Expanding along first row = 

 

  = 

Therefore, a unit vector perpendicular to both   and  is

 =  = 

±23iˆ23jˆ23kˆ±23i^∓23j^∓23k^


3. If a unit vector  makes an angle  with   with  and an acute angle  with  then find  and hence, the components of .

 

 

Ans. Let  be a unit vector.  ……….(i) 

   

Squaring both sides,   ……….(ii)

Given: Angle between vectors  and iˆi^ is 

    

  ……….(iii)

Again, given Angel between vectors  and jˆj^ is 

    

  ……….(iv)

Again, given Angel between vectors  and kˆk^ is  where  is acute angle.

    

  ……….(v)

Putting the values of  and  in eq. (ii),

   

    

Since  is acute angle, therefore  is positive and hence   

From eq. (v), 

Putting values of  and  in eq. (i), 

 Components of  are coefficients of  in 

  and angle 


4. Show that (ab)×(a+b)=2(a×b)(a→−b→)×(a→+b→)=2(a→×b→) 

 

 

Ans. L.H.S. =  =    

 = 2(a×b)2(a→×b→) = R.H.S.


5. Find  and  if  

 

 

Ans. Given:  

 

Expanding along first row,

 = 

Comparing the coefficients of  on both sides, we have

  ……….(i)

  ……….(ii)

And   ……….(iii)

From eq. (ii),     

From eq. (iii),     

Putting the values of  and  in eq. (i),

    0 = 0

Therefore,  and λ=3.λ=3.


6. Given that  and  What can you conclude about the vectors  and  

 

 

Ans. Given:    

   or   or   

   or   or vector  is perpendicular to  …..(i)

Again, given      

   or   or   

     or   or vector  and  are collinear or parallel. …..(ii)

Since, vectors  &  are perpendicular to each other as well as parallel are not possible. ..(iii)

Therefore, from eq. (i), (ii) and (iii),  either   or 

  and 


7. Let the vectors  be given as  then show that  

 

 

Ans. Given: Vector  and  

 

Now L.H.S. = 

 + 

[By Property of Determinants]

 = R.H.S.


8. If either  and  then  Is the converse true? Justify your answer with an example.

 

 

Ans. Given: Either  or  

  or ……….(i)

 

   [Using eq. (i)]

  [By definition of zero vector]

But the converse is not true.

Let   

  is a non-zero vector.

Let 

 

  is a non-zero vector.

But 

Taking 2 common from R3 =   [ R2 and R3 are identical]


9. Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).

 

 

Ans. Vertices of  are A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5). 

 Position vector of point A = (1, 1, 2) = 

Position vector of point B = (2, 3, 5) = 

Position vector of point C = (1, 5, 5) = 

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

=   

 

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

=   

 

  x  = 

 = 6i3j+4k−6i→−3j→+4k→

Now Area of triangle ABC = 

 sq. units


10. Find the area of the parallelogram whose adjacent sides are determined by the vectors   and  

 

 

Ans. Given: Vectors representing two adjacent sides of a parallelogram are 

 and  

  =  = 

Now Area of parallelogram = 

 sq. units


11. Let the vectors  and  such that  then  is a unit vector, if the angle between  and  is:

 

(A)  

(B)  

(C)  

(D)  

 

Ans. Given:  and  is a unit vector. 

    , where  is the angle between  and 
 
 
 
 
 
Therefore, option (B) is correct.


12. Area of a rectangle having vertices A, B, C and D with position vectors  and  respectively is:

 

(A)  

(B) 1

(C) 2

(D) 4

 

Ans. Given: ABCD is a rectangle.
Now  = Position vector of point B – Position vector of point A
=  


 AB = 
And   = Position vector of point D – Position vector of point A
=  


 AD = 
 Area of rectangle ABCD = Length x Breadth = AB x AD = 2 x 1 = 2 sq. units
Therefore, option (C) is correct.