Orthogonal System of Vectors


Table of Content


In the orthogonal system of vectors we choose these vectors as three mutually perpendicular unit vectors denoted by , and  directed along the 

Components of the vector

positive directions of X, Y and Z axes respectively.

Corresponding to any point P(x, y z) we can associate a vector w.r.t. a fixed orthogonal system and then this vector is the position vector (p.v.) of that point. i.e. p.v. of P = 

Distance of P from O =  

x, y, z are called the components of the vector

If a vector makes angles a, b, g with the positive directions of X, Y and Z axes respectively, then cosa, cosb, cosg are called the direction cosines (d.c.’s) of .

cosa =  cosb; cosg = 

So that cos2a + cos2b + cos2g = 1

Unit vector in the direction of is

 = .

Direction cosines


Section Formula

Internal Division 

Let A and B be two points with position vectors  and  respectively, and C be a point dividing AB internally in the ratio m : n. Then the position vector of C is given by .

Proof: Let O be the origin. The , let  be the position vector of C which divides AB internally in the ratio m : n then,


Þ  n.

Þ n(P.V. of  – P.V. of ) = m(P.V. of  – P.V. of )


Þ   Þ  

Þ   or   

Internal Division

External Division 

Let A and B be two points with position vectors  and  respectively and let C be a point dividing  externally in the ratio m : n. Then the position vector of  is given by. External Division


(i) If C is the mid–point of AB, then P.V. of C is .

(ii) We have,. Hence  is in the form of .

where,  and .Thus, position vector of any point  on  can always be taken as  where l + m = 1.

(iii)  If circumcentre is origin and vertices of a triangle have position vectors , then the position vector of orthocentre will be .

Illustration 4: ABC is a triangle. A line is drawn parallel to BC to meet AB and AC in D and E respectively.  Prove that the median through A bisects DE.

Solution:  Take the vertex A of the triangle ABC as the origin. Let  be the p.v. of B and C. The mid point of BC has the p.v. =  The equation of the median is . Let D divide AB in the ratio 1:m

Þ p.v. of .Let E divide AC in the ratio 1:l

Þ p.v. E =  Þl = m. p.v. of the mid-point of DE = which lies on the median. Hence the median bisects DE.

Isector of the Angle Between Two Vectors

Consider two non–zero, non–collinear vectors  and . The bisector of the angle between the two vectors  and  is k 

where k Î R+.

Illustration 5: If the vector  bisects the angle between  and , where  is a unit vector then find .

Key concept: Bisector of the angle between the two vectors  and  is k  where k Î R+.

Solution:  According to the given conditions l = 

Þ         3 = 3l

(3l + 1) – (2 + 9l) + (15l – 2)


Þ         9 = (3l + 1)2 + (2 + 9l)2 + (15l – 2)2 

Þ         315l2 – 18l = 0 Þl = 0, .

If l = 0,  (not acceptable)

For l = 

Scalar (or Dot) Product of Two Vectors

The scalar product of and , written as , is defined to be the cosq where q is the angle between the vectors and  i.e.  = abcosq.


Geometrical Interpretation:

Scalar (or Dot) Product of Two Vectors

 is  the  product  of  length of  one  vector  and  length of the projection of the  other  vector  in the  direction of former.

. Projection of  in direction of  =.Projection of  in direction of 



(acosq)b  = (projection of ) b  = (projection of ) a

(Distributive law)

 Ûare perpendicular to each other Þ



If  then

If  are non-zero, the angle between them is given by 

Illustration 6: Prove by vector method that (a1b1 + a2b2 + a3b3)2 £ (a12+ a22 + a32) (b12 + b22 + b32).

Solution: Let  = a1 + a2 + a3 and  = .

Now × = a1 b1 + a2b2 + a3b, also ×= || || cosq£ || ||.

Þ (×)2£ ||2 ||Þ (a1b1 + a2b2 + a3b3)2£

Illustration 7: If || = 3, || = 1, || = 4 and , find the value of  

Solution:  We know, 

Þ 0 =  (Given )

Þ0 = 

Þ = – = –13

 Illustration 8: In a DABC, prove by vector method that

cos 2A + cos 2B + cos 2C ³ –3/2  

Solution:  As we know;                                    …(i)

and                                        … (ii)

Now using (i), we get

Þ 3R2 + 2R2 (cos 2A + cos2B + cos 2C) ³ 0

Þ cos 2A + cos 2B + cos 2C ³ –3/2

Vector (Or Cross) Product of Two Vectors

The vector product of two vectors and , denoted by , is defined as the vector , where q is the angle between the vectors and and  is a unit vector perpendicular to both and (i.e., perpendicular to the plane of and ).The sense of  is obtained by the right hand thumb rule i.e. and  form a right-handed screw.

Vector (Or Cross) Product of Two Vectors

If we curl the fingers of our right hand from to through the smaller angle (keeping the initial point of  and same), the thumb points in the direction of . In this case, ,and  (or ), in that order are said to form a right handed system. It is evident that = absinq.


  • Þ

  •   (non-commutative)

  •  (Distributive)

  •  Û are collinear (if none of is a zero vector)

  • If  then  


  • Any vector perpendicular to the plane of  is l () where l is a real number. Unit vector perpendicular to  is ± 

 denotes the area of the parallelogram OACB, whereas area of DOAB = 

Area is also treated as a vector with its direction in the proper sense.

Area of the parallelogram

Illustration 9:   and  are unit vectors and || = 4. If angle between  and  is cos–1 and , then show that can be written as   also find the value of l. 

Key concept:     If , then it can be written as =0, that means vector  and (-2) are collinear.

Solution:  Since vector  and (-2) are collinear, vector can be written asÞ

Þ  Þ16 + 4 – 4(4) (1)  = l2(1)

Þ  l2 = 16 Þl± 4.

Illustration 10: If  are three vectors such that ,  , then show that  = 1,  

Solution: Given              …(i)


(1) – (2) Þ


Now either  is perpendicular to  which is not possible

Or  = 0 Þ.

Also  Þ.  

Þ   .

Equation the coefficients of , we get .

Thus  and .

Scalar Triple Product

It is defined for three vectors in that order as the scalar  which can also be written simply as . It denotes the volume of the parallelopiped formed by taking a, b, c as the co-terminus edges.

i.e. V = magnitude of 

Scalar Triple Product

The value  of  scalar triple  product  depends on the  cyclic  order  of  the  vectors and  is  independent of the  position of  the  dot  and  cross. These  may be  interchange at  pleasure. However and  anti-cyclic  permutation of the vectors  changes  the value of  triple product  in sign but  not  a magnitude.


  • If  are given as  etc., then 

  •  i.e. position of dot and cross can be interchanged without altering the product. Hence it is also represented by 




Position of dot and cross

  •  in that order form a right handed system if 

Illustration 11:   Show that .

Key concept:     In such type of questions first we reduce each term as a product of two or three vector, by substituting a group of vectors by a single vector and then solve.

Solution:  Let 

Now L.H.S. = 


= 0 = R.H.S.

Vector Triple Product

The vector product of two vectors, one of which is itself the vector product of two vectors, is a vector  quantity called  vector  triple product.

It is defined for three vectors as the vector . This vector being perpendicular to , is coplanar with  i.e. 

Take the scalar product of this equation with a. We get

0 =  Þ

If we choose the coordinate axes in such a way that

, it is easy to show that l = 1. Hence

In general,  ( Vector  triple product  is  not  associative ) .

, if some or all of are zero vectors or  are collinear.

Illustration 12: Let  be three mutually perpendicular vectors of the same magnitude. If the vector  satisfy the equation

 then find  

Solution: Here  -


or l2


l2 { 

let  then ,and .

Þl2 {3 Þ3 

Hence  .


Collinear and Coplanar Vectors

1. Methods to Prove Collinearity

  • Two vectors and are collinear if there exists kÎR such that .

  • If are collinear.

  • Three points A(), B(), C()are collinear if there exists kÎR such that  that is .

  • If then A, B, C are collinear.

  • A(), B(), C() are collinear if there exists scalars l, m, n, (not all zero) such that where l + m + n = 0.

All the above methods are equivalent and any of them can be utilized to prove the collinearity. (of 2 vectors or 3 points)

Illustration 13:  Let  be three non–zero vectors such that any two of them are non–collinear. If  is collinear with  and  is collinear with , then prove that .

Key concept:  Two vectors and are collinear if there exists kÎR such that .

Solution:  It is given that  is collinear with 

Þfor some scalar l               …(i)

Also   is collinear with 

Þ for some scalar m             …(ii)

from (i) and (ii)

Þ(1 + 2m)  + (3 – ml)  = 0

Þ1 + 2m = 0  and  3 – ml = 0  { and  are non–collinear vectors}

Þ         m = – 1/2 and l = – 6

Substituting the values of l and m in (i) & (ii), we get

2. Methods to Prove Coplanarity

  • Three vectors are coplanar if there exists l, mÎR such that  i.e., one can be expressed as a linear combination of the other two.

  • If  are coplanar (necessary and sufficient condition).

  • Four points A, B, C and Dlie in the same plane if there exist l, mÎR such that  i.e. .

  • If = 0 then A, B, C, D are coplanar.

  • A, B, C, D are coplanar if there exists scalars k, l, m, n (not all zero), such that where k + l + m + n = o.

Again all the above methods are equivalent. Choose the best amongst them depending on convenience.

Illustration 14: Prove that if cos a¹ 1, cosb¹ 1 and cos g¹ 1, then the vectors  can never be coplanar.

Key concept: If three vectorsare coplanar then .

Solution: Suppose that  are coplanar. 

Þ         ( R2® R2 – R1 and R3® R3 – R1 )


or         cosa (cosb – 1)(cosg – 1) – (1 – cosa)(cosg – 1) – (1 – cosa)(cosb – 1) = 0

dividing through out by (1 – cosa) (1 – cosb)(1 – cosg); we get


or         –1 + 


Þ         , which is not possible


Hence they can not be coplanar

Applications of Vectors

 1. Vector Equation of a Straight Line

Following are the two most useful forms of the equation of a line.

(i) Line passing through a given point Aand parallel to a vector 

Vector Equation of a Straight Line

where is the p.v. of any general point P on the line and l is any real number. The vector equation of a straight line passing through the origin and parallel to a vector  is  = n .

(ii) Line passing through two given points Aand B

For each particular value of l, we get a particular point on the line. Each of the above equations can be written easily in Cartesian form also.

For example, in case (i), writing


Line passing through two given points

we get 

Þ x = a1 + lb1, y = a2 + lb2, z = a3 + lb3.

Illustration 15: Given vectors  where O is the centre of circle circumscribed about DABC, then find vector .

Solution:  Here,     

Þ and 

Now          [as radii of circle]




Þ          and             …(i)

Centre of circle circumscribed about DABC

Now if we take  then from (i),


and                           …(iii)

 \ Solving (ii) and (iii);



2. Shortest Distacne Between Two Lines

Two lines in space can be parallel, intersecting or neither (called skew lines). Let be two lines.

(i) They intersect if .

(ii) They are parallel if are collinear. Parallel lines are of the form  Perpendicular distance between them is constant and is equal to          .

(iii) For skew lines, shortest distance between them (along common perpendicular) is given by .


3. Equation of a Plane in Vector Form 

Following are the four useful ways of specifying a plane.

(i) A plane at a perpendicular distance d from the origin and normal to a given direction has the equation 

or  (is a unit vector).


Equation of a Plane in Vector Form


(ii) A plane passing through the point Aand normal to has the equation .

A plane passing through the point

(iii)  Parameteric equation of the plane passing through Aand parallel to the plane of vectors is given by Þ.

(iv) Parameteric equation of the plane passing through A, B C(A, B, C non-collinear) is given by  Þ.

In Cartesian form, the equation of the plane assumes the form Ax + By + Cz = D. The vector normal to this plane is and the perpendicular distance of the plane from the origin is .

Angle Between a Line and a Plane:

The angle between a line and a plane is the complement of the angle between the line and the normal to the plane.

Angle Between Two Planes:

It is equal to the angle between their normal unit vectors .  i.e. cosq = 

4. Some Miscellaneous Result


(i) Volume of the tetrahedron ABCD = 


Volume of the tetrahedron


(ii)  Area of the quadrilateral with diagonals 
 = .


Area of the quadrilateral with diagonals

5. Reciprocal System of Vectors

 If  are three non-coplanar vectors, then a system of vectors defined by is called the reciprocal system of vectors because .


The scalar product of any vector of one system with a vector of other system which does not correspond to it is zero i.e.

If is a reciprocal system to  then  is also reciprocal system to .

Illustration 16: Show that the points and  are equi-distant from the plane  r × (5i + 2j – 7k) + 9 = 0 and are on the opposite sides of it.

Solution: The given plane is  = -9

Length of the perpendicular from to it is

Length of the perpendicular from 


Thus the length of the two perpendiculars are equal in magnitude but opposite in sign. Hence they are located on opposite sides of the plane.

Solved Objective Problems

Q1.  Any four non-zero vector will always be:

(A)  Linearly dependent                          (B)  Linearly independent

(C)  Either ‘A’ or ‘B’                                  (D)  None of these

Sol.  Four or more than four non zero vectors are always linearly dependent.

         Hence (A) is correct answer.

Q2. If  and  are reciprocal vectors, then:

(A)                                             (B)       

(C)                                              (D)      None of these

Sol.  If  and  are reciprocal, then





Hence (C) is correct answer.

Q3. If  and , then:

(A)                  (B)       

(C)            (D)      




Hence (C) is correct answer.

Q4. If three unit vectors  satisfy , then angle between  and  is:

(A)                                                      (B)       

(C)                                                      (D)      


  = 1

Hence (B) is correct answer.

Q5. Projection of  on  is equal to:

(A)      3                                                   (B)       -3

(C)      9                                                  (D)      -9

Sol. Projection of  on  is 

Thus required projection


Hence (B) is correct answer.

Q6. If  and  are two non-collinear unit vectors, then projection of  on  is equal to:

 (A)      2                                                   (B)       -2

(C)      1                                                   (D)      None of these


Thus projection of  on  is zero.

Hence (D) is correct answer.

Q7. ABCD is a parallelogram with  and . Area of this parallelogram is equal to:

(A)       sq. units                              (B)        sq. units

(C)       sq. units                                (D)       sq. units

Sol. Area vector of parallelogram

*      Area of the parallelogram

sq. units

Hence (B) is correct answer.


Q8. If  and   always make an acute angle with each other for every value of ,  then:

(A)                                       (B)       

(C)                                        (D)      


= x(x + 1) + x – 1 + a

+ 2x + a – 1

We must have

*+ 2x + a – 1 > 0 

 4 – 4(a – 1) < 0

 a > 2

Hence (B) is correct answer.

Q9. Let  be three non zero vectors such that  Then , where   is equal to:

(A)      1                                                   (B)       2

(C)      -1                                                  (D)      -2

Sol. Clearly  and  represents the sides of a triangle.

                        Sides of a triangle

It’s area vector,



Hence (D) is correct answer.

Q10. If   where , then:

(A)                            (B)  

(C)                            (D)  


Taking cross with  in first equation, we get



Hence (A) is correct answer.

Q11.  Let   be unit vectors such that  Then angle between   and   is :

(A)                                         (B)       

(C)                                         (D)      


Taking dot with  on both sides, we get

If  be the angle between  and  then


Hence (B) is correct answer.


Q12. Let  be pair wise mutually perpendicular vectors, such that . Then length of   is equal to:

(A)      2                                                   (B)       4

(C)      3                                                   (D)      6


= 1 + 4 + 4 + 0 + 0 + 0

= 9

 = 3

Hence (C) is correct answer.


Q13. Let   be three unit vectors such that  Then  is equal to:

(A)                                                   (B)       

(C)                                           (D)      None of these


Thus  are coplaner.


Hence (D) is correct answer.

Q14. ABCD is a parallelogram  and  are the midpoints of side BC and CD respectively. If , then   is equal to:

(A)      1/2                                                 (B)       1

(C)      3/2                                                (D)      2

Sol. Let P.V. of A,B,D be  and  respectively.

Then P.V. of C = 

Also,  P.V. of 


and, P.V. of 

Hence (C) is correct answer.


Q15. Two constant force  and  act on a particle. If particle is displaced from a point A with position vector   to the point B with position vector  Then work done in the process is equal to:

(A)      15 units                                           (B)       10 units

(C)      -15 units                                          (D)      -10 units

Sol. Total force, 


Work done = 

= 2 – 12 – 5 = - 15 units.

Hence (C) is correct answer.

Q16. A, B, C and D are any four points in the space. If  , where  is the area of triangle ABC, then  is equal to:

(A)      2                                                   (B)       1/2

(C)      4                                                   (D)      1/4

Sol. Let P.V. of A, B, C and D be  and 

* ,



= 2()


= 4

= 4

Hence (C) is correct answer.

Q17.  The position vector of the points A, B, C and D are  and . It is known that these points are coplanar, then  is equal to:

(A)                                               (B)       

(C)                                               (D)      None of these


If vector  and  are coplanar, then

Hence (A) is correct answer.

Q18. Let  be any three vectors. Then  is always equal to:

(A)                                        (B)  

(C) Zero                                              (D) None of these


Hence (B) is correct answer.

Q19.  Let  be any three vectors. Then  is also equal to:

(A)                                           (B)       

(C)      Zero                                              (D)      None of these


Hence (A) is correct answer.


Q20.  is always equal to:

(A)                                                       (B)       

(C)                                                    (D)      




Hence (C) is correct answer.

Q21. Value of  is always equal to:

(A)                                     (B)       

(C)                                     (D)      None of these


Hence (A) is correct answer.

Q22. For any four vectors  the expression  is always equal to:

(A)                                              (B)       

(C)                                              (D)      None of these


 = 0

Hence (D) is correct answer.


Q23.  For any four vectors  and is always equal to:

(A)                                     (B)       

(C)                                     (D)      


Hence (B) is correct answer.

Q24. In the parallelogram ABCD if the internal bisectors of the angle  and  intersect at the point P, then  is equal to:

(A)                                                      (B)       

(C)                                                      (D)      

Sol. Let P.V. of B, A and C be  and  respectively.

                        Parallelogram ABCD

Now,  and  

Hence (D) is correct answer.

Q25. If the vector  bisects the angle between  and , where  is a unit vector, then:

(A)               (B)       

(C)             (D)      

Sol. We must have

For  (not acceptable)


Hence (D) is correct answer.


Q26. Distance of  from the plane  is:

(A)                                                (B)       

(C)                                                (D)      None of these

Sol. Let Q() be the foot of altitude drawn from P to the plane  = 0,


* Required distance

Hence (C) is correct answer.

Q27. Distance of  from the line  is:

(A)                      (B)       

(C)                      (D)      None of these

Sol. (A) Let Q() be the foot of altitude drawn from P() to the line = 


Q28. Let  and  be unit vector that are mutually perpendicular, then for any arbitrary :




(D)      None of these

Sol.  Let 



Hence (A) is correct answer.

Q29. The line  will not meet the plane , provided:

(A)                                (B)       

(C)                                (D)      

Sol. We must have


Hence (C) is correct answer.

Q30. The plane  will contain the line , provided:

(A)                               (B)       

(C)                                (D)      

Sol. We must have


Hence (C) is correct answer.

Q31. If the projection of point  on the plane  is the point , then:

(A)                                   (B)       

(C)                                   (D)      

Sol. We have


Hence (B) is correct answer.

Q32. Let  and  be unit vectors that are perpendicular to each other, then  will always be equal to:

(A)      1                                                   (B)       Zero

(C)      -1                                                  (D)      None of these



Hence (A) is correct answer.

Q33. If , then the vector  is always equal to:

(A)                                                       (B)       

(C)                                                       (D)      


Hence (D) is correct answer.

Q34.  For any two vectors  and , the expression  is always equal to:

(A)                                                    (B)       

(C)      Zero                                              (D)      None of these





Hence (B) is correct answer.

Q35. Let   and  be three non zero vectors such that , angle between  and  is  and  is perpendicular to  and , then 

 where  is equal to:

(A)      1/2                                                 (B)       1/4

(C)      1                                                   (D)      2


Hence (A) is correct answer.

Q36. Let  be three vectors such that ,Then:

(A)                                      (B)       

(C)                                      (D)      



Hence (A) is correct answer.

Q37. Let  and  be unit vectors such that , then the value of  is equal to:

                        (A)                                                     (B)       

                        (C)                                                    (D)      



Hence (C) is correct answer.

Q38. Let  be three unit vectors such that  If the angle between  and  is , then , where  is equal to:

(A)                                                    (B)       

(C)                                                    (D)      None of these


Hence (B) is correct answer.

Q39. Let P is any arbitrary point on the circumcircle of a given equilateral triangle of side length  units then,  is always equal to:

(A)                                                   (B)       

(C)                                                      (D)      

Sol.  Let P.V. of P, A, B and C are  and  respectively and O() be the circumcentre of the equilateral triangle ABC.





Hence (A) is correct answer.

Q40. . A vector coplanar with  and , whose projection on  is of magnitude  is:

(A)                                      (B)       

(C)                                      (D)      

Sol. Let the required vector be 

Then,  and 


(2 – 2 – 1) + (2 – 1 – 2)

 or 2

If , then


If  then 

Hence (A) is correct answer.

Q41. If a + bm + cm, m =1, 2, 3, are pairwise perpendicular unit vectors, then  is equal to

(A)      0                                                   (B)   1 or –1

(C)      3 or  -3                                          (D)      4 or –4

Sol.  =1 Þ= ± 1 .

Hence (B) is correct answer.


Q42. If  are three non-coplanar unit vectors, then is equal to

(A)                              (B) 

(C)                                                 (D) 

Sol.  = projection of   in  the  direction of .

Hence the given vector is 

Hence (D) is correct answer.

Q43. If sec2Aandare coplanar, then cot2A + cot2B + cot2C is

(A)      equal to 1                                         (B)       equal to 2

(C)      equal to 0                                        (D)      not defined

Sol. The vectors are co-planar

Þ   = 0

Þ   cot2A + cotB + cot2 C + 1 = 0 which is not possible.

Hence (C) is correct answer.

Q44. If a, b, c are three non - coplanar vectors and p, q, r are vectors defined by the relations r =  then the value of expression (a + b).p + (b + c).q + (c + a).r is equal to

(A)      0                                                   (B)       1

(C)      2                                                   (D)      3


Hence the given  scalar expression  = 1 + 1 + 1 = 3.

Hence (D) is correct answer.


Q45. The value of  |a  ´ |2 + |a ´ |2 + |a ´|2is

(A)      a2                                                             (B)         2a2

(C)      3a2                                                           (D)        none of these


Þ |a|2 sin2a + |a|2 sin2b + |a|2 sin2g

= 3|a2| – |a2|(cos2a + cos2b + cos2g)

= 2|a2| = 2a2

Hence (B) is correct answer.

Q46. If   are non-coplanar vectors  then  is equal to

(A)      3                                                   (B)       0

(B)       1                                                   (D)      none of there

Sol.  = 1 – 1 = 0.

Hence (B) is correct answer.


Q47. Consider  DABC  and  DA1B1C1  in such a  way  that   and  M, N, M , N1 be  the  mid points of  AB, BC, A1B1 and B1C1 respectively,  then

(A)                                       (B)       

(C)                                        (D)      




Þ   Þ Þ

Þ Þ Þ 2

ÞÞ 2

Þ   .

Hence (D) is correct answer.


Q48. Let , where x1, x2, x3Î {-3, -2, -1, 0, 1, 2}. Number of possible vectors  such that  are mutually perpendicular, is

(A)      25                                                 (B)       28

(C)      22                                                 (D)      none  of these

Sol.   Þ x1 + x2 + x3 = 0

Thus  we  have to obtain  the  number  of  integral solution of this  equation.

Coefficient of  x° | ( x-3 +x-2 +x-1 + x0 + x + x2)3

= x°

=  x9 |(1 – x6)3 ( 1– x)-3 =11C9  – 3.5C3  = 25.

Hence (A) is correct answer.

Q49. Let a, b, c, be distinct and non-negative. If the vectors ai + aj + ck, i + k, and  ci + cj + bk lie in a plane, then c is 

(A)      A.M.  of  a and  b                              (B)       G.M.  of  a and  b

(C)      H.M  of  a and  b                               (D)      equal  to zero.

Sol.  = 0

C2® C2 – C1

–1(ab – c2) = 0 Þ c2 = ab

Hence (B) is correct answer.


Q50. If , and  ,then  is  equal to

(A)      320                                            (B)       320 

(C)      - 320                                           (D)      -320 


 and process gives on

= –320.

Hence (C) is correct answer.

Q51. If  is the vector whose initial point divides the joining of  and  in the ratio k:1 and terminal point is origin. Also  then the interval in which k lies

(A)      (–¥, –6] È [–1/6, ¥)                          (B)       (–¥, –6] È [1/6, ¥)

(C)      (–¥, 6] È [–1/6, ¥)                            (D)      (¥, 6] È [–1/6, ¥)

Sol. The point that divides  and  in the ratio of k : 1 is given


Also      ÞÞ

On squaring both sides, we get 

Or 12k2 + 74k + 12 ³ 0Þ (6k + 1) (k + 6) ³ 0

Hence k Î (–¥, –6] È [–1/6, ¥).

Hence (A) is correct answer.

Q52. If 'a' is real constant and A, B, C are variable angles and,  then the least value of is :

(A)      10                                                  (B)       11

(C)      12                                                  (D)      13

Sol: The given relation can be re–written as


(as, a.b = |a| |b| cosq)


Þ                                 … (i)

also,                                (as, )                                   … (ii)

from (i) and (ii), 

Hence  least value of .

Hence (C) is correct answer.

Q53. The vector  and  are collinear for

(A)      unique value of x , 0 < x < p/6              

(B)       unique value of x , p/6 < x < p/3

(C)      no value of x                                                         

(D)      infinity many value og x, 0 < x < p/2

Sol: Since  and  are collinear, for some l, we can write .


Þ         Þ   cosx = x

Here we will get only one unique value of x which belongs to 

Hence (B) is correct answer.

Q54. The vectors  have their initial points at (1, 1), the value of l so that the vectors terminate on one straight line is

(A)      0                                                  (B)     3

(C)      6                                                  (D)      9

Sol: Since initial point of  is , their terminal points will be  and . Now given all the vectors terminate on one straight line.  Hence 

Þ l1 = 1 and l = 9

Hence (D) is correct answer.

Q55. Given that  is a perpendicular to  and p is a non-zero scalar, then a vector  satisfying is given by

(A)                                           (B)         

(C)                                        (D)        none of these

Sol: We have . Taking dot by vector , we get

Þ     Þ  

Þ      Þ.

Hence (A) is correct answer.


Q56. Let P is any arbitrary point on the circumcircle of a given equilateral triangle of side length '' units then  is always equal to:

(A)                                                   (B)       

(C)                                                     (D)      

Sol: Let P.V. of P, A, B and C are  and  respectively and O() be the circumcentre of the equilateral triangle ABC.





Þ    = 

Hence (A) is correct answer.


Q57. Let  and  are two non collinear vector such that . The angle of a triangle whose two sides are represented by the vector  and  are

(A)                                           (B)       

(C)                                       (D)      none of these

Sol: Let, clearly  and  are mutually perpendicular as  is coplanar with  and  and  is at right angle to the plane of  and . And   





Thus angles are 

Hence (B) is correct answer.


Q58. E and F are the interior points on the sides BC and CD of a parallelogram ABCD. Let  and . If the line EF meets the diagonal AC in G then  where l is equal to

(A)                                                   (B)       

(C)                                                    (D)      

Sol: (D) Let P.V. of A.B. and D be . Then 

Þ       and 

E and F are the interior points on the sides BC and CD of a parallelogram ABCD

Equation of EF : 

Equation of AC : 

For point G we must have, 

Þ       Þ

Hence (D) is correct answer.


Q59. If and the vector  and are non-coplanar, then the vectors  and are:

(A)      coplanar                                          (B)       none coplanar

(C)      collinear                                          (D)      non collinear

Sol: Given   

  [since X, Y, Z are non-coplanar]

Hence  and  are coplanar.

Hence (A) is correct answer.

Q60. If b and c are any two perpendicular unit vectors and a is any vector, then  

(A)      b                                 (B)    a

(C)      c                                 (D)   b + c

Sol: Consider three non-coplanar vectors b, c and . Any vector a can be written as ……(i).

Taking dot product with  in (i) we get

  .Taking dot product with b in (i)

Taking dot product with c in (i), we get,  a.c = y


Hence (B) is correct answer.


Q61. If the lines  and  intersect (t and s are scalars) then.

(A)                                           (B)       

(C)                                           (D)      none of these

Sol: For the point of intersection of the lines


Hence (B) is correct answer.


Q62. If  and  then

(A)                                           (B)       

(C)                                           (D)      none of these

Sol:     Also , and 

Þ  are mutually perpendicular vectors.

Þ   and   Þ    

Hence (C) is correct answer.

Q63. The position vector of a point P is  where x, y, z Î N and . If  = 10, then the number of possible positions of P is

(A)      30                                                 (B)       72

(C)      66                                                 (D)      9C2

Sol: Given  = 10 Þ x + y + z = 10, x, y, z ³1

The number of possible positions of  P

= coefficient of x10 in (x + x2 + x3 + … )3

= coefficient of x7 in (1 – x)-3

3 + 7 – 1C7 = 9C7 = 9C2 = 36

Hence (D) is correct answer.


Q64. If vectors ax and x make an acute angle with each other, for all x Î R, then a belongs to the interval

(A)                                            (B)       ( 0, 1)

(C)                                             (D)      

Sol: Since vectors make an acute angle with each other so their dot product must be positive i.e. ax2 – 10 ax + 6 > 0 " x Î R

Þ- ax2 + 10ax – 6 < 0 " x Î R  Þ  –a < 0 and 100a2< 24a

Hence (C) is correct answer.

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