# 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

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  = .

## 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   Þ   or

### 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.

Note:

(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

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:

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

Properties:

Þ

(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.

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.

Properties:

• 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.

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)

…(ii)

(1) – (2) Þ

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

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.

Properties:

• 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

 ·                ·          ·
•  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

where

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

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

Þ         , which is not possible

as

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 :

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 ,

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          or          Þ          Þ          and             …(i)

Now if we take  then from (i),

…(ii)

and                           …(iii)

\ Solving (ii) and (iii);

and

Þ

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

(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 = (ii)  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 .

Further

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.

Q2. If  and  are reciprocal vectors, then:

(A)                                             (B)

(C)                                              (D)      None of these

Sol.  If  and  are reciprocal, then

and

Q3. If  and , then:

(A)                  (B)

(C)            (D)

Sol.

Also,

Thus

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

(A)                                                      (B)

(C)                                                      (D)

Sol.

= 1

Q5. Projection of  on  is equal to:

(A)      3                                                   (B)       -3

(C)      9                                                  (D)      -9

Sol. Projection of  on  is

Thus required projection

=

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

Sol.

Thus projection of  on  is zero.

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

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

(A)                                       (B)

(C)                                        (D)

Sol.

= x(x + 1) + x – 1 + a

+ 2x + a – 1

We must have

+ 2x + a – 1 > 0

4 – 4(a – 1) < 0

a > 2

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.

It’s area vector,

Thus,

Q10. If   where , then:

(A)                            (B)

(C)                            (D)

Sol.

Taking cross with  in first equation, we get

and

Also,

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

(A)                                         (B)

(C)                                         (D)

Sol.

Taking dot with  on both sides, we get

If  be the angle between  and  then

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

(A)      2                                                   (B)       4

(C)      3                                                   (D)      6

Sol.

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

= 9

= 3

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

(A)                                                   (B)

(C)                                           (D)      None of these

Sol.

Thus  are coplaner.

Hence

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

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,

Displacement,

Work done =

= 2 – 12 – 5 = - 15 units.

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

,

and

= 2()

= 4

= 4

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

Sol.

If vector  and  are coplanar, then

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

(A)                                        (B)

(C) Zero                                              (D) None of these

Sol.

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

(A)                                           (B)

(C)      Zero                                              (D)      None of these

Sol.

Q20.  is always equal to:

(A)                                                       (B)

(C)                                                    (D)

Sol.

Similarly,

and

Q21. Value of  is always equal to:

(A)                                     (B)

(C)                                     (D)      None of these

Sol.

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

(A)                                              (B)

(C)                                              (D)      None of these

Sol.

= 0

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

(A)                                     (B)

(C)                                     (D)

Sol.

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.

Now,  and

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)

For

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,

Also

Required distance

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 =

and

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

(A)

(B)

(C)

(D)      None of these

Sol.  Let

Also,

and

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

(A)                                (B)

(C)                                (D)

Sol. We must have

and

Q30. The plane  will contain the line , provided:

(A)                               (B)

(C)                                (D)

Sol. We must have

and

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

(A)                                   (B)

(C)                                   (D)

Sol. We have

and

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

Sol.

=

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

(A)                                                       (B)

(C)                                                       (D)

Sol.

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

(A)                                                    (B)

(C)      Zero                                              (D)      None of these

Sol.

Similarly,

and,

Let

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

Sol.

Q36. Let  be three vectors such that ,Then:

(A)                                      (B)

(C)                                      (D)

Sol.

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

(A)                                                     (B)

(C)                                                    (D)

Sol.

Now,

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

Sol.

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.

Now,

Similarly,

and

as

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

Now,

(2 – 2 – 1) + (2 – 1 – 2)

or 2

If , then

where

If  then

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 .

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

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.

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

Sol.

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

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

(A)      a2                                                             (B)         2a2

(C)      3a2                                                           (D)        none of these

Sol.

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

= 3|a2| – |a2|(cos2a + cos2b + cos2g)

= 2|a2| = 2a2

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

(A)      3                                                   (B)       0

(B)       1                                                   (D)      none of there

Sol.  = 1 – 1 = 0.

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)

Sol.

Þ

Þ   Þ Þ

Þ Þ Þ 2

ÞÞ 2

Þ   .

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.

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

Q50. If , and  ,then  is  equal to

(A)      320                                            (B)       320

(C)      - 320                                           (D)      -320

Sol.

and process gives on

= –320.

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

by

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, ¥).

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

Þ

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

Þ

Þ                                 … (i)

also,                                (as, )                                   … (ii)

from (i) and (ii),

Hence  least value of .

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         Þ   cosx = x

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

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

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.

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

Now

Similarly,

and

Þ    =

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

Þ

=

Also,

Þ

Thus angles are

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

Equation of EF :

Equation of AC :

For point G we must have,

Þ       Þ

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.

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

Thus

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

Þ

Q62. If  and  then

(A)                                           (B)

(C)                                           (D)      none of these

Sol:     Also , and

Þ  are mutually perpendicular vectors.

Þ   and   Þ

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

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

To read more, Buy study materials of Vectors comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.

### Course Features

• Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution