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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
= .
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
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.
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 = ,
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 ||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
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.
(non-commutative)
(Distributive)
Û are collinear (if none of is a zero vector)
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 .
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.
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.
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 .
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 , which is not possible
as
Hence they can not be coplanar
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),
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.
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:
(C) (D)
Sol.
Also,
Thus
Q4. If three unit vectors satisfy , then angle between and is:
= 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
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
Q8. If and always make an acute angle with each other for every value of , then:
= 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:
Taking cross with in first equation, we get
Q11. Let be unit vectors such that , , , Then angle between and is :
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
= 1 + 4 + 4 + 0 + 0 + 0
= 9
= 3
Q13. Let be three unit vectors such that Then is equal to:
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
= 2()
= 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:
If vector and are coplanar, then
Q18. Let be any three vectors. Then is always equal to:
(C) Zero (D) None of these
Q19. Let be any three vectors. Then is also equal to:
Q20. is always equal to:
Similarly,
Q21. Value of is always equal to:
Q22. For any four vectors the expression is always equal to:
= 0
Q23. For any four vectors and is always equal to:
Q24. In the parallelogram ABCD if the internal bisectors of the angle and intersect at the point P, then is equal to:
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:
Sol. We must have
For (not acceptable)
For
Q26. Distance of from the plane is:
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:
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 :
(A)
(B)
(C)
(D) None of these
Sol. Let
Q29. The line will not meet the plane , provided:
Q30. The plane will contain the line , provided:
Q31. If the projection of point on the plane is the point , then:
Sol. We have
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
Q33. If , then the vector is always equal to:
Q34. For any two vectors and , the expression is always equal to:
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
Q36. Let be three vectors such that ,, Then:
Q37. Let and be unit vectors such that , then the value of is equal to:
Now,
Q38. Let be three unit vectors such that If the angle between and is , then , where is equal to:
Q39. Let P is any arbitrary point on the circumcircle of a given equilateral triangle of side length units then, is always equal to:
Sol. Let P.V. of P, A, B and C are and respectively and O() be the circumcentre of the equilateral triangle ABC.
Q40. . A vector coplanar with and , whose projection on is of magnitude is:
Sol. Let the required vector be
Then, and
(2 – 2 – 1) + (2 – 1 – 2)
or 2
If , then
Q41. If am + 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
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 + cot2 B + 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
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
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, M1 , N1 be the mid points of AB, BC, A1B1 and B1C1 respectively, then
Þ Þ Þ
Þ Þ Þ 2
ÞÞ 2
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
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
(as, a.b = |a| |b| cosq)
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 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
(C) (D) none of these
Sol: We have . Taking dot by vector , we get
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:
Sol: Let P.V. of P, A, B and C are and respectively and O() be the circumcentre of the equilateral triangle ABC.
Now
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
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
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
Sol: (D) Let P.V. of A.B. and D be . Then
Equation of EF :
Equation of AC :
For point G we must have,
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
Q61. If the lines and intersect (t and s are scalars) then.
Sol: For the point of intersection of the lines
Q62. If and then
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)
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
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