Ans. Let  be the unit vector in XY-plane such that XOP =  

Therefore,     OP = 1   ……….(i)

By Triangle Law of Addition of vectors,

In ,  = 

[Unit vector along OX is  and that is along OY is ]

   [Dividing and multiplying by OP in R.H.S.]

  [Using eq. (i)]

 

Ans. Given points are P and Q 

 Position vector of point P =  = 

And Position vector of point Q =  = 

Now   = Position vector of Q – Position vector of P

= 

=  = 

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

And magnitude of vector  = 

Ans. Let the initial point of departure is origin (0, 0) and the girl walks a distance OA = 4 km towards west. 

Through the point A, draw a line AQ parallel to a line OP, which is  East of North, i.e., in East-North quadrant making an angle of  with North.

Again, let the girl walks a distance AB = 3 km along this direction 

  =    ……….(i)  [ Vector  is along OX’]

Now, draw BM perpendicular to axis.

In  by Triangle Law of Addition of vectors,

Dividing and multiplying by AB in R.H.S.,

 = 

      ……….(ii)

  Girl’s displacement from her initial point O of departure to final point B,

 =  = 

 

 

Ans. Given:  

 Either the vectors  are collinear or form the sides of a triangle.

Case I: Vectors  are collinear.

Let  and 

Then 

Also,  = AC = AB + BC = 

Case II: Vectors  form a triangle.

Here also by Triangle Law of vectors, 

But   [ Each side of a triangle is less than sum of the other two sides]

   is true only when vectors  and  are collinear vectors.

Ans. Since  is a unit vector, 

Therefore, 

       

Squaring both sides,  

 

 

Ans. Given: Vectors  and  

Let vector  be the resultant vector of  and 

   =  + 

= 

 Required vector pf magnitude 5 units and parallel (or collinear) to resultant vector  is

= 

= 

= 

  

= 

Ans. Given: Vectors  and  

Let 

= 

= 

= 

  A unit vector parallel to the vector  is

= 

= 

Ans. Given: Points A B and C (11, 3, 7). 

  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

=  =  = 

  

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

=  =  = 

  

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

=  =  = 

  

Now  =  = 

Therefore, points A, B, C are either collinear or are the vertices of a triangle ABC.

Again AB + BC =  = AC

Now to find ratio in which B divides AC

Let the point B divides AC in the ratio 

Therefore, using section formula, Position vector of point B  is 

   = 

 

 

 

Comparing coefficients of  both sides, we get

 −6λ=−4,−6λ=−4,     3λ=2,3λ=2,      −2λ−2=7λ−8−2λ−2=7λ−8

 

Therefore, required ratio =  =  : 1 = 2 : 3

Ans. Since position vector of point R dividing the join of P and Q externally in the ratio 1 : 2 =  is given by  

       

Again position vector of the middle point of the line segment RQ

=  (Position vector of point R – Position vector of point Q)

=  =  =  = Position vector of point P (given)

Therefore, P is the middle point of the line segment RQ.

Ans. Let ABCD is a parallelogram. 

Given: The vectors representing two adjacent sides of this parallelogram say,

 and 

Now vectors along the diagonals  and  of the parallelogram are

 and 

   =  = 

And  =  = 

Therefore, Unit vectors parallel to (or along) diagonals are

 and 

    and      and 

   and 

Now Area of parallelogram =  = 

=  = 

 

=  =  =  sq. units

Ans. Let  be the direction cosines of a vector equally inclined to axes OX, OY and OZ respectively. 

  A unit vector along the given vector is

  and 

       ……….(i)

Let the given vector (for which unit vector is ) make equal angle (given)  (say) with OX OY and OZ

 The given vector is in positive octant OXYZ and hence  is acute. ……….(ii)

Now angle  between  and   

 

 

     ……….(iii)

Similarly, angle  between  and ,       ……….(iv)

And angle  between  and ,       ……….(v)

Putting the values of  in eq. (i), we get

  

 

 

But  [  is acute and hence  is positive]

Therefore, required vectors  are  and 

Ans. Given: Vectors  and  

We know that the cross-product of two vectors,  is a vector perpendicular to both  and 

Hence, vector  which is also perpendicular to both  and  is  where  or some other scalar.

Therefore, 

= λ[iˆ(28+4)−jˆ(7−6)+kˆ(−2−12)]λ[i^(28+4)−j^(7−6)+k^(−2−12)]

           ………..(i)

 

Now given  and 

 

 

 

 

 

Putting  in eq. (i), we get

 

Ans. Let ,  and  

Now  (say) = 

   a unit vector along  is

 = 

= 

= 

   

     …..(i)

Also given Dot product of  and  is 1.

 . = 1

 

 

 

Squaring both sides,

 

 

    

Ans. Given:  are mutually perpendicular vectors of equal magnitude. 

        ……….(i)

And  (say)        ……….(ii)

Let vector  make angles  with vectors  respectively.

 

= 

=   [From eq. (i)]

  =  =       ……….(iii)

We know that 

= 

Putting the values from eq. (i) and (ii),

=  = 

 

Now  = 

= 

 

Similarly,  and 

  

Therefore,  is equally inclined to the vectors  and 

Ans. We know that  =  

  

=      ……….(i)

Now if  and  are perpendicular     

Putting  in 

= ,

 

=       ……….(ii)

 

=  [Putting value of  in eq. (i)]

 

 

But  (given)

Therefore, vectors  and  are perpendicular to each other.

Ans. Given:  

 

 

[ and  being lengths of vectors are always  0]

Therefore, option (B) is correct.

Ans. Given:  and  are unit vectors. 

  and 

Now squaring both sides of , we have,

 

 

  , where  is the given angle between vectors  and .

Putting , we have,  

 

  = 

 

=  = 

Therefore, option (D) is correct.

Ans.  

Also  = 1 – 1 + 1 = 1

Therefore, option (C) is correct.

Ans. Given:  

 

And this equation is true only for option (B) namely , since 

Therefore, option (B) is correct.