Orthogonal System of Vectors

Table of Content

Section Formula
- Internal Division
- External Division
Isector of the Angle Between Two Vectors
Scalar (or Dot) Product of Two Vectors
Vector (Or Cross) Product of Two Vectors
Scalar Triple Product
Vector Triple Product
Collinear and Coplanar Vectors
Applications of Vectors
Solved Objective Problems

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 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 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:

(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 + a₃b₃, also ×= || || cosq£ || ||.

Þ (×)²£ ||² ||²Þ (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

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 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 :

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 and …(i)

Now if we take then from (i),

…(ii)

and …(iii)

\ Solving (ii) and (iii);

and

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

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)

Sol. If and are reciprocal, then

and

Hence (C) is correct answer.

Q3. If and , then:

(A) (B)

Sol.

Also,

Thus

Hence (C) is correct answer.

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

(A) (B)

Sol.

= 1

Hence (B) is correct answer.

Q5. Projection of on is equal to:

(A) 3 (B) -3

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

Sol.

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

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)

Sol.

= 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

Sol. Clearly and represents the sides of a triangle.

It’s area vector,

Thus,

Hence (D) is correct answer.

Q10. If , where , then:

(A) (B)

Sol.

Taking cross with in first equation, we get

and

Also,

Hence (A) is correct answer.

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

(A) (B)

Sol.

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

Sol.

= 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)

Sol.

Thus are coplaner.

Hence

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

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

Sol. Total force,

Displacement,

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

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

and

= 2()

= 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)

Sol.

If vector and are coplanar, then

Hence (A) is correct answer.

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

(A) (B)

Sol.

Hence (B) is correct answer.

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

(A) (B)

Sol.

Hence (A) is correct answer.

Q20. is always equal to:

(A) (B)

Sol.

Similarly,

and

Hence (C) is correct answer.

Q21. Value of is always equal to:

(A) (B)

Sol.

Hence (A) is correct answer.

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

(A) (B)

Sol.

= 0

Hence (D) is correct answer.

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

(A) (B)

Sol.

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)

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

Now, and

Hence (D) is correct answer.

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

(A) (B)

Sol. We must have

For (not acceptable)

For

Hence (D) is correct answer.

Q26. Distance of from the plane is:

(A) (B)

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

Also

Required distance

Hence (C) is correct answer.

Q27. Distance of from the line is:

(A) (B)

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

Hence (A) is correct answer.

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

(A) (B)

Sol. We must have

and

Hence (C) is correct answer.

Q30. The plane will contain the line , provided:

(A) (B)

Sol. We must have

and

Hence (C) is correct answer.

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

(A) (B)

Sol. We have

and

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

Sol.

Hence (A) is correct answer.

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

(A) (B)

Sol.

Hence (D) is correct answer.

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

(A) (B)

Sol.

Similarly,

and,

Let

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

Sol.

Hence (A) is correct answer.

Q36. Let be three vectors such that ,, Then:

(A) (B)

Sol.

Hence (A) is correct answer.

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

(A) (B)

Sol.

Now,

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)

Sol.

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)

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

Hence (A) is correct answer.

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

(A) (B)

Sol. Let the required vector be

Then, and

Now,

(2 – 2 – 1) + (2 – 1 – 2)

or 2

If , then

where

If then

Hence (A) is correct answer.

Q41. If a_m + b_m + c_m, m =1, 2, 3, are pairwise perpendicular unit vectors, then is equal to

(A) 0 (B) 1 or –1

Sol. = =1 Þ= ± 1 .

Hence (B) is correct answer.

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

(A) (B)

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

Sol. The vectors are co-planar

Þ = 0

Þ cot²A + cot²B + 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

Sol.

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

Sol.

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, M₁ , N₁ be the mid points of AB, BC, A₁B₁ and B₁C₁ respectively, then

(A) (B)

Sol.

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

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

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

Sol.

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, ¥)

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

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

(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

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)

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)

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

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)

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

= =

Also, =

= Þ

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)

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

Hence (D) is correct answer.

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

(A) coplanar (B) none coplanar

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

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

Hence (B) is correct answer.

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

(A) (B)

Sol: For the point of intersection of the lines

Hence (B) is correct answer.

Q62. If and then

(A) (B)

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

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 – 1}C₇ = 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)

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.

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.