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# Vector[1]

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### Vector[1]

1. 1. VECTORSA vector is a quantity with both magnitude and direction, there aretwo operations defined on vectors and these both have a very directgeometric interpretation. We draw a vector as a line with an arrow,for now I will call the end without the arrow the start of the vectorand the end with the arrow the end of the vector. Vector addition: to add two vectors we take the start of the second vector and move it to the end of the first vector. The addition of these two vectors is the vector from the start of the first vector to the end of the second vector. Scalar multiplication changes the length of a vector without changing its direction. That is we scale it by the multiplying factor. So scalar multiplication involves multiplying a scalar (single number) by a vector to give another number.
2. 2. Vectors are represented in a natural way by directed linesegments, the length of the segment measuring themagnitude of the vector and the arrowhead indicating thedirection of the vector. The arrowhead is usually placed atone end (or near the end) of the line segment; this end iscalled the head of the vector and the other end is called thetail of the vector. Such vectors are called free vectorsbecause they are distinguished only by their length anddirection, and not by their position in space.The vector with the same magnitude as u but oppositedirection, the negative of u, is labelled -u.The zero vector (a vector with zero length and no direction) iswritten 0. Position vectorsThe line segment , where A and B are points in space, iscalled the position vector of B relative to A. It is also knownas the vector from A to B as well as the displacement of Bfrom A.
3. 3. The vector from B to A has the same magnitude as thevector from A to B but opposite direction. That is, is thenegative of .The zero vector is the position vector of a point A withrespect to itself; that is, 0 = . The length of the zero vectoris 0 but it has no direction. (TYPES OF VECTORS) Zero vector , Proper vector (non-zero vector), Unit vector, Equal vector, Collinear vector(two or more vectors have same or parallel supports, irrespective of their magnitudes & directions), Coplanar vectors(three or more vectors lie in the same plane or parallel to the same plane ). Equality of vectorsGiven points A and B, we may think of the position vector v = as the vector that acts on the point A to get B.If the line segments AB and DC have thesame direction and the same length, thenABCD is a parallelogram and the positionvectors and are equal; we write thisas = .Using a head-to-tail procedure, a vector v can be added toitself to give the vector v + v.
4. 4. We have v = and v = . It’s natural to write + =as 2v. The vector 2v has length 2|v| and is in the samedirection as v. The notion of multiplication of a vector by apositive integer is then generalised to define the vector sv forall scalars s, as follows.For example, -3v has three times the magnitude of v butpoints in the opposite direction; v (also written ) hasmagnitude and has the same direction as v.Note that -1v has the same magnitude as v but has theopposite direction, and so is the same vector as the negativeof v, that is, -1v = -v. When are vectors parallel?
5. 5. Two vectors u and v are said to be parallel if they have eitherthe same direction or opposite direction. This means thateach is a scalar multiple of the other: for some non-zeroscalar s, v = su and so u = v.Since the zero vector can be written 0 = 0v, the zero vector isconsidered to be parallel to every other vector v. Unit vectors and how to get themIt is often useful to be able to find a unit vector (a vector oflength 1) in the direction of a given non-zero vector v. Acommon notation used for this unit vector is .For example, if v has length 3, then = v has length 1 andis in the same direction as v.In general, the unit vector in the same direction as v isSince = , we also have v = |v| . Introduction
6. 6. Because non-zero vectors have direction as well asmagnitude, adding vectors involves more than simply addingnumbers. The sum of two vectors is another vector, and sothe definition of addition must give a process for determiningboth the magnitude and the direction of the sum vector.There are two equivalent procedures for addition of vectors,called the parallelogram rule and the triangle rule.The parallelogram rule for additionSuppose u and v are two vectors (in the plane or in space).Translate them so that they are tail-to-tail at point O.From the head of each vector, draw a copy of the othervector to complete a parallelogram OAPB. In thisparallelogram, u = = and v = = .
7. 7. The triangle rule for additionAnother way to define addition of two vectors is by a head-to-tail construction that creates two sides of a triangle. Thethird side of the triangle determines the sum of the twovectors, as shown below.Place the tail of the vector v at the head of the vector u. Thatis, u = and v = .Now construct the vector to complete the third side of thetriangle OAP.
8. 8. This method is equivalent to the parallelogram law ofaddition, as can be easily seen by drawing a copy of v tail-to-tail with u, to obtain the same parallelogram as before.Using position vector notation, the triangle rule of addition iswritten as follows: for any three points X, Y , Z,Both the triangle and the parallelogram rules of addition areprocedures that are independent of the order of the vectors;that is, using either rule, it is always true that u + v = v + u forall vectors u and v. This is known as the commutative law ofaddition. There are other rules like this one, and they arediscussed in the component Vector Algebra. Adding a vector to its negativeTo add u to -u, place the tail of -u = at the head of u = .
9. 9. Then u + (-u) = = 0, the zero vector. By the commutativelaw,Subtraction of two vectors is just a special case of addition.The vector u - v is defined to beUsing the parallelogram law of addition, u - v = u + (-v) =+ = .The vector v - u can be found in the same way.
10. 10. The vectors u + v and u - v are both diagonals of theparallelogram formed by u and v.Adding the zero vector 0 to any vector v gives the vector vagain. That is,for all vectors v. For if v = , thenand The head-minus-tail rule
11. 11. Given a vector and any point O, the triangle rule foraddition shows us that = + . We can rewrite thisequation asSince points to the head of the vector and pointsto its tail, we call this the “head-minus-tail” rule.The advantage of this rule is that it allows us to rewrite all thevectors occurring in a problem as combinations of vectorsemanating from a given point O (often called the origin). Internal division of a line segmentWith an assigned point O as origin, the position of any pointP is given uniquely by the vector , which is called theposition vector of P relative to O.Let P1 and P2 be any points, and let R be a point on the lineP1P2 such that R divides the line segment P1P2 in the ratio m :n. That is, R is the point such that = . Our task isto find the position vector of R (relative to O) in terms of theposition vectors of P1 and P2.
12. 12. As = , we have n =m and therefore (1)which rearranges to giveWhen m and n are both positive, the vectors andhave the same direction, since = . Thiscorresponds to the situation where R lies between P1 and P2,as shown in the diagram above. R is then said to divide theline segment P1P2 internally in the ratio m : n.As a special case of the general formula (1) we can obtain aformula for the position vector of the midpoint Mbetween two points P1 and P2.In this case, m : n = 1 : 1 and so
13. 13. External division of a line segmentWe look at a situation similar to the one on the previouspage, but allow m and n to have opposite sign. Again, R isthe point such that = , and we want to find theposition vector of R (relative to O) in terms of the positionvectors of P1 and P2.Since is a negative multiple of , the vectorsand have opposite direction.The point R then lies outside the line segment P1P2 (but stillon the line joining P1 and P2). In these cases, R is said todivide the line segment P1P2 externally in the ratio m : n. Theformula is, as in case of an internal division, = , m-n≠ 0 . Coordinate axes and Cartesian coordinatesBy three dimensional space we mean the space we live in. Tofix a point P in three dimensional space requires a system ofaxes and three numbers. First select any point, call it theorigin and mark it as O. All measurements will from now onoriginate from this point O. Next place three mutually
14. 14. perpendicular axes OX, OY , OZ through O. This axis systemis drawn on a page like this:Note that although it may not look it, the angles XOY , XOZ, YOZ are all right-angles!To fix any given point P in three dimensional space, we referit to the axis system. Let us first show the point P and thecoordinate axes.1. Drop a line from P perpendicular to the XOY plane (think ofthis plane as the floor), meeting the XOY plane at a point Q(the foot of the perpendicular).2. Now we are in the XOY plane with a point Q and an axissystem in two dimensions.
15. 15. 3. Drop a perpendicular from Q to OX and OY .4. Transfer the two-dimensional picture from 3 into the threedimensional diagram from 1.(Some of the right angles are marked: you should mark allthe other right angles in the picture.)
16. 16. 5. Measure the lengths OR, OS, QP, and denote them by x, y,z respectively. We call the three numbers (x,y,z), in the ordergiven, the Cartesian coordinates of the point P.Notice that the order in which the numbers are written isimportant: (1, 2, 1) and (2, 1, 1) are the Cartesian coordinatesof different points.To make the diagrams uncluttered, only half of each axis hasbeen drawn. However, if the axes are extended infinitely inboth directions, it can be seen that this axis system createseight octants in space, just as the two dimensional axissystem creates four quadrants in the plane. Points in theeight different octants have Cartesian coordinatescorresponding to all possible combinations of positive and
17. 17. negative values of x, y and z. The figures above illustratepoints in the octant corresponding to positive values of x, yand z. Cartesian form of a vectorWe begin with two dimensions. We have the following pictureillustrating how to construct the Cartesian form of a point Qin the XOY plane. Vectors i and j are vectors of length1 in the directions OX and OY respectively. The vector is xi. The vector is yj. The vector is the sum of and , that is, We now extend this to threedimensions to show how to construct the Cartesian form of apoint P. Define k to be a vector of length 1 in the direction of OZ. We now have the following picture .Draw a perpendicular PT from P to the OZ axis.
18. 18. In the rectangle OQPT,PQ and OT both have length z. Thevector is zk. We know that = xi + yj. The vector ,being the sum of the vectors and , is thereforeThis formula, which expresses in terms of i, j, k, x, y andz, is called the Cartesian representation of the vector inthree dimensions. We call x, y and z the components ofalong the OX, OY and OZ axes respectively.The formulaapplies in all octants, as x, y and z run through all possiblereal values. Length of a vector in terms of componentsThe length of the vector in the XOY plane is .From the three dimensional picture below,
19. 19. “cut out” the triangle OPQ to obtain:It can be seen that the length of the vector is = . Addition and subtraction of vectors in terms of componentsGiven two vectors in Cartesian formthe sum + = is obtained by completing theparallelogram.
20. 20. It can be proved that this is the same as the followingcalculation:That is, the components of a sum are the sums of thecomponents.Notice that the parallelogram OQRP is part of a twodimensional plane sitting within three dimensional space (ina tilted way like the slanting face of the roof of a house). SubtractionThe rule for subtraction works in exactly the same way.Writing for the vector - , the rule above gives
21. 21. The subtraction is illustrated below. Recall that = - = .An example of the use of this rule is the calculation of theCartesian form of the position vector of a point P2 relativeto a point P1. Suppose that in Cartesian form, = x1 i + y1 j +z1k and = x2i + y2j + z2k.Then Multiplication by scalars in terms of componentsIf a vector is expressed in Cartesian form, then it’s easy tocalculate any scalar multiple of that vector in Cartesian form.The general rule is that given any vector v = xi + yj + zk andany scalar a, then
22. 22. For example, if v = 2i + j - k, then -4v = -8i - 4j + 4k.The same formula applies in two dimensions. The vector v =xi + yj can be thought of as v = xi + yj + 0k. Then if a is anyscalar,Thus if u = 2i + j then 2u = 4i + 2j, as illustrated in thefollowing picture.Example 1Find the magnitude of the vector x = 4i - 2j + 3k,the unit vector in the direction of x and the vectorof magnitude 5 in the direction opposite to x.SolutionWe have |x| = = = .Therefore the unit vector in the direction of x is thevector
23. 23. and the vector of magnitude 5 in the directionopposite to x isExample 2Given the vectors u = 3i - j + 2k and v = 2i + 3j - k,find Cartesian forms of the vectors u + v, u - v, 2u+ 3v.SolutionFirst,Next,and finally,Example 3Given that P1(-1, 2, 3) and P2(3, 3, 4) are twopoints in space, find the Cartesian form of thevector .Solution
24. 24. Example 4Show that the vectorsandhave the same magnitude.SolutionBoth vectors are in Cartesian form and theirlengths can be calculated using the formulaWe have and
25. 25. Therefore two given vectors have the same length.Vector algebra in geometric formWe discuss properties of the two operations, addition of vectors andmultiplication of a vector by a scalar. We do this first for free vectors. Equality of Vectors Two vectors u and v are equal if they have the same magnitude (length) and direction. The Negative of a Vector The negative of the vector u is written -u, and has the same magnitude but opposite direction to u. If u = , then -u = . Commutative Law of Addition for all vectors v and u. Associative Laws for all vectors u, v and w and for all scalars s and t. As an illustration of the first of these associative laws, we translate the three vectors u, v and w so that they are drawn head to tail, and then draw (u + v) + w in the first figure below and u + (v + w) underneath it, demonstrating that both equal .
26. 26. We may then simply write u + v + w, without using brackets. This associative law extends to sums of any number of vectors taken in any order, so that the expression u1 + u2 + u3 + ...... + un is well defined.Distributive Laws for all vectors v and u and for all scalars s and t. The first of these distributive laws is illustrated below in the case s = 2.
27. 27. Laws Involving the Zero Vector for all vectors u. Vector algebra in Cartesian formWe discuss properties of the two operations, addition ofvectors and multiplication of a vector by a scalar, but now forvectors in Cartesian form.Suppose vectors v1 and v2 are written in Cartesian form, Equality of Vectors Vectors v1 and v2 are equal if and only if their components are equal, that is, if and only if x1 = y1, x2 = y2 and x3 = y3. The Negative of a Vector The negative of the vector v1 = x1i + x2j + x3k is the vector Multiplication by a Scalar If s is any scalar, then
28. 28. Addition of Vectors The Zero Vector The zero vector is the vector 0 = 0i + 0j + 0k.Example 1Draw the vectors a + b and a - b in the diagramsbelow.SolutionExample 2Express the unknown vector v in terms of a and b.
29. 29. Solution v = b - a. v = a - b.Example 3Express the vector v in terms of a, b and c.Solutionv = a + b - 2c.Example 4Simplify the following expressions. a + 2b + 3( b - 5a) a - 7b + 7(b - a) 2a - 3b + c - (a - 3b)
30. 30. Solution -15a + b -6a a - 2b + c **Proof of the formula If one (or both) vectors are zero vectors then the above formula is obvious. Hence assume that u and v are non- zero. The vectors u, v and u - v then form a triangle. The edges of the above triangles have length |u|, |v| and |u - v|. By the law of cosines (see below) By definition u · v = |u||v| cos , so (1)
31. 31. By the formula for the magnitude of a vectorMultiplying the right hand side out we getIf we substitute the above into (1 ) we getso u · v = u1v1 + u2v2 + u3v3 as claimed.Appendix: The law of cosinesThe law of cosines asserts that in an arbitrary triangle a2= b2 + c2 - 2bc cos , where a, b, c and are as shownbelow.The formula follows by applying the Theorem ofPythagoras to the triangles ADC and DBC. Note that thered line AD has length |b cos | and the green line DBhas length c - b cos . Hence by the Theorem ofPythagorasMultiplying the right hand side out we get
32. 32. Adding b2 cos 2 and rearranging we get a2 = b2 + c2 - Vectoraddition and subtraction is simple in that we just add or subtractcorresponding terms. Vector multiplication is more complicated,there is not a general mutiplication that multiplies two vectors andproduces another vector, however there are two types ofmultiplication that combine two vectors and produce anotherquantity. A dot product takes two vectors and produces a scalar. A cross product takes two vectors and produces a vector. Vector Add Adding two vectors is equivalent to putting the tail of one vector against the head of the other. The order is not important, i.e. A + B = B + A. This can be easily calculated by separately adding the x,y and z components. Vector Subtract Subtracting two vectors is equivalent to putting the head of the vectors together, the result is then the vector between the tail of the vectors. An alternative way to think of it is to reverse the direction of the vector to be subtracted, then add the vectors. The resulting vector can be easily calculated by separately subtracting the x,y and z components.
33. 33. Dot Product The dot product operation combines two vectors and produces a scalar output.Dot ProductIt is also given by A•B = |A| |B| cos(θ)where: |A| = magnitude of vector A |B| = magnitude of vector B θ = angle between vector A and vector B (0≤Ѳ≤п )It is also equal toA * projection of B on AorB * projection of A on B 2bc cos as required Algebraic rulesSuppose u, v and w are vectors (in the plane or in space)with magnitudes |u|, |v| and |w|. Then
34. 34. The rules are obtained by using the Cartesian representationof the scalar product. We let u = u1i + u2j + u3k, v = v1i + v2j +v3k and w = w1i + w2j + w3k. Then 1. By definition of the scalar product 2. If s is a scalar then (su) · v = su1v1 + su2v2 + su3v3 = u1sv1 + u2sv2 + u3sv3 = s(u1v1 + u2v2 + u3v3) = s(u · v) 3. For the distributive law note that u·v +u·w = u1(v1 + w1) + u2(v2 + w2) + u3(v3 + w3) = u1v1 + u1w1 + u2v2 + u2w2 + u3v3 + u3w3 = (u1v1 + u2v1 + u3v3) + (u1w1 + u2w2 + u3w3) = u · (v + w) 4. For the last rule simply note that
35. 35. Alternatively one could prove the algebraic rules using theCartesian representation of the scalar productConditions for perpendicularityBy definition of the scalar product, u·v = |u||v| cos , whereis the angle between u and v. If = 90° then cos = 0. Giventwo non-zero vectors u and vIf u = u1i + u2j + u3k and v = v1i + v2j + v3k are given inCartesian form then the above readsFor vectors in the plane drop the third terms. The scalar product of the standard basis vectorsBy assumption the standard basis vectors i, j and k aremutually perpendicular, and of magnitude one. Hence bydefinition of the scalar product 1. i · i = j · j = k · k = 1; 2. i · j = i · k = j · k = 0. The component of a vector parallel to a given vector 3. By definition of the scalar product 4.
36. 36. 5. is the projection of v into the direction of u as shown below. We call 6. 7. the component of v in the direction of u. The geometric picture is shown below. 8.Resolution of a vector into components parallel andperpendicular to a given vector 9. Given two vectors u and v we want to find the projections of v parallel and perpendicular to the vector v. We already know that the projection of v into the direction of u is 10. 11. Looking at the picture below it is evident that the projection of v into the direction perpendicular to u is n = v - p, that is, 12.
37. 37. 13. The vector product in terms of magnitudes, angles and the “right-hand rule”The vector product is a product of two vectors in space. Theresult of the multiplication is a vector, hence the name vectorproduct. Other names for the vector product are crossproduct or outer product.If u and v are vectors, their vector product is denoted by u ×v. We must assign that product a direction and a magnitude.We start by specifying the direction. Suppose that u and vare two non-zero and non-parallel vectors in space. Whenplaced tail-to-tail they define a plane. There are twopossibilities to choose a unit vector perpendicular to thatplane:
38. 38. We choose n such that the triple u, v, n is right-handed. Bythis we mean the following. Align a screw-driver along theline perpendicular to the plane defined by u and v. Then turnthe screw-driver such that u moves to v through the smallerangle. The direction of n is then given by the direction thescrew m oves.
39. 39. Note that it is irrelevant which way we align the screw-driver,the direction in which the screw moves will be the sameeither way.Having established the direction it remains to define themagnitude of the vector product. We let the magnitude be|u||v| sin , where is the smaller angle between u and v.Ѳ is the angle b/w vectors u,v, 0≤Ѳ≤п and is unit vectorper. To both u & v.Then we defineIf u and v are parallel or one (or both) of them are zerovectors then we set u×v = 0. Geometrically, the magnitudeequals the area of the parallelogram spanned by the vectorsu and v when placed tail-to-tail.Important Note: Unlike in the definition of the scalar product
40. 40. it is essential that the angle between u to v is the smallerone, that is, the one between 0° and 180°. The vector product in terms of componentsIf u = u1i + u2j + u3k and u = v1i + v2j + v3k are two vectors thentheir vector product isIf you are familiar with determinants of a 3 × 3 matrix, there isan easy way to remember the above formula. Formally, Vector Cross Product Both the two operands and the result of the cross product are vectors. The vector cross product has some useful properties, it produces a vector which is mutually perpendicular to the two vectors being multiplied.The vector product obeys the following algebraic rules
41. 41. Geometrically speaking, the vector that results from the crossproduct of vectors and has a magnitude given by the productof and times the sine of the angle from to ,or .The direction is given by the “right hand rule.” If you lay your right handpalm-up along , then curl your fingers toward , the resulting vector isin the direction of your thumb. Take a look:Properties of the cross product: [ lagrange’s identity]
42. 42. Proof: | sinѲ ∴ | |2 =( | |)2 (1-cos2Ѳ) [=| cosѲ ] ∴ | 2 =Scalar triple product identities The triple product can be evaluated using the relationThe triple product of vectors a, b, and c is given by =[ ] The value of the triple product is equal to the volume of the parallelepipedconstructed from the vectors. This can be seen from the figure since
43. 43. Magnitude of this projection = [ repts. The vector area ofthe base of the parallelopiped and the height of the parallelopiped isthe projection of along the normal to the plane containing vectors . i.e., along ]∴ volume of parallelepiped = [ ] {if repts. Three co-terminus edges of a parallelogram }[ ] = 0, if are coplanar vectors.The scalar triple product has the following properties ⇨ 1. [ ]=[ =[ If these vectors are cyclically permuted. The value remains unaltered. 2. the position of dot and cross can be interchanged, provided the cyclic order of vectors remains the same. 3. the value remains the same in mag. but changes the sign, if the cyclic order of vectors is changed.(as shown above) 4. if any two vectors are same and parallel( or collinear),then scalar triple product = 0 [ λ ] 5. if any three vectors and scalar λ , then [λ = λ[ ] Triple vector product:The triple vector product has the properties
44. 44. OrIf A,B and C are vectors and x is cross multiplication, then,A × (B × C) = B (A·C) - C (A·B)As explained above the dot product gives a scalar, so B and C arescaled by these scalars before they are subtracted.and also, reversing order of(A × B) × C = - C × (A × B) multiplication changes sign using first vector triple product = - A (B·C) + B (A·C) identityHere are some triple product identities for dot product of crossproduct:A · (B × C) = (B × C) · AA · (B × C) = C · (A × B) = B· (C × A)A · (B × C) = -A · (C × B)But ,The expression (u × v) × w makes sense, but
45. 45. in general. As a counterexample look at (i × i) × j. We have (i× i) × j = 0 × j = 0 but i × (i × j) = i × k = -j.This shows that writing u × v × w does not make sense as itis not clear which two vectors to multiply first!The easiest way to prove the algebraic rules of the vectorproduct is to make use of the Cartesian representation of thevector product. Hence assume that u = u1i + u2j + u3k, v = v1i+ v2j + v3k and w = w1i + w2j + w3k. 1. u×v = -(v×u) follows directly from the definition of the vector product; 2. For every scalar s 3. Using the Cartesian representations we getCondition of parallelismTwo vectors, u and v are parallel if and only if the anglebetween them is 0° or 180°. Hence, by definition of the vectorproduct we have the following characterisation of parallelvectors
46. 46. The vector product of the standard basis vectorsUsing the definition of the vector product it follows that 1. i × i = j × j = k × k = 0 2. i × j = -(j × i) = k; 3. i × k = -(k × i) = -j; 4. j × k = -(k × j) = i;Recall that, by convention, the triple i, j, k is always right-handed, so i, k, j is left handed and j, k, i is right handed Vector products and the area of a triangleThe magnitude of the product u × v is by definition the areaof the parallelogram spanned by u and v when placed tail-to-tail. Hence we can use the vector product to compute thearea of a triangle formed by three points A, B and C in space.It follows that the area of the triangle is
47. 47. . area of parallelogram.Identities anti-commutative, vector cross multiplication is notA×B + B×A = 0 commutative, changing the order reverses the direction of the resulting vector: A×B = - B×A.A×(B×C) +B ×(C×A) + Jacobi identity : anti-associative : anti-symmetryC ×(A×B) = 0 So cross multipication is not associative.A×A=0 A squared is a zero length vector because sin(0) = 0A × (B + C) =(A × B) + (A × cross multiplication for vectors is distributive over +C)0×A=0Where: A,B and C = any vectors 0 = zero length vector (all elements are zero)
48. 48. × = vector cross productFor example, the following expression:A×A×BWe might expect that this is zero, because A × A = 0, but this is notnecessarily so because cross multiplication is not associative. Weneed to specify the order, if it is:A × (A × B)then the result is not necessarily zero.Example19: For any two veceors , , we always have | . |≤| | | ( Cauchy-Schwartz inequality).Solution: It is hold trivially, when =0 or =0. If ≠0 or ≠0, thenwe get = |cosѲ| ≤ 1, hence proved.Example 20: For any two veceors , , we always have| + |≤| |+ |( triangle inequality).Solution: t is hold trivially, when =0 or =0. If ≠0 or ≠0, thenwe get , | + |² = ( + )² = ( + ). ( + ) = | |²+2 . |²≤| |²+2 . |² (∵ x≤|x| for all x∊R)≤ | |²+2 .| |² ( from ex. 19)= (| |+ |)²
49. 49. EXAMPLE 29: Three vectors , , are satisfy the condition + +=0. Evaluate the quantity μ = . + + . , if |=1, | =4and| =2.Solution: .( + + ) =0 ⇨ . + . = -| =-1………(i)Similarily , we get . + = -| =-16…………(ii) + . = -4 ………(iii), by adding, 2μ=-21 ⇨ μ =-21/2EXAMPLE 30: If with reference to the right handed system ofmutually per. Unit vectors I,j & k, = 3i-j, = 2i+j-3k, then expressIn the form = , where are // & per. To resp.Solution: let = λ , λ is scalar, i.e., = 3λi – λj , = - = (2-3λ)I + (1+λ)j - 3k. since is to be per. To , we should have . =0⇨ λ =1/2 ∴ = (1/2)i+(3/2)j-3k.NCERT QUESTIONS WITH HINTSEX.10.3 Question 4 Find the projection vector i+3j+7k on the vector 7i-j+8k. [ Ans. Projection of = ⇨ ]Question 8 Find the magnitude of two vectors , , having the samemag. and such that the angle b/w them is 600 and their scalar productis ½.Ans. = cosѲ ⇨ ½ = cos600 = (1/2) (∵ )⇨ =1
50. 50. Question 12 If . = 0 & . = 0, then what can be concludedabout the vector . [Ans. These eqns. For =0 ⇨ is any vector.]Question13 If , , are unit vectors such that + + =0, find thevalue of . + + .Ans. + =- ……..(i) take dot product with ⇨ ( + )= . (-⇨1+ . + = 0 ……(ii), similarily with & , we get 1 + . += 0 …..(iii) & 1 + . + = 0 …..(iv) [∵ |= |= 1], by adding(ii),(iii)&(iv) , we get . + + = -3/2Question 14 If either =0 or =0, then . = 0. But the converseneed not be true. Justify your answer with an example.Ans. Let = i-3j+4k, = 2i-2j+k are non- zero vectors but their dotproduct is zero.Question 16 Show that the points A(1,2,7), B(2,6,3),C(3,10,-1)arecollinear.ANS. The position vectors of A,B,C are i-2j+7k, 2i+6j+3k, 3i+10j-k = = i+4j-4k, | | = √33, | |=√33 & | |= 2√33| |+| |= | |= 2√33 ⇨ A,B, C are collinear.Question 17 show that the vectors 2i-j+k, i-3j-5k & 3i-4j-4k form thevertices of a right angled triangle.Ans. = = -i-2j-6k ,| |= √41, | |=√6 & | |= √35| |² = | |²+ | |² ⇨ ABC IS a right angled triangle.
51. 51. EX. 10.4 Question 3 If a unit vector makes angles п/3 with I, п/4with j & an acute angle Ѳ with k, then find Ѳ and hence, thecomponent of .Ans. Let = a1i+a2j+a3k, | =1 , according to ques.(a1i+a2j+a3k).i= п/3 ⇨ a1= ½, similarily a2=1/√2 & a3 = ½[∵a1²+a2²+a3²=1+ , (1/2i+1/√2j+1/2k).k=|1/4+1/2+1/4|cosѲ⇨Ѳ=п/3Question 6 given that . = 0 and = 0. What can you concludeabout the vectors and ?Ans. given that . = 0 and =0⇨ =0 or =0 or per. ToAnd =0 or =0 or // ⇨ =0 =0 [∵ per. To & // canNever hold at a time]Question 8 if either =0 or =0, then = 0. is converse true?Justify your answer with an example.Ans.| |=| |sinѲ = 0 [∵ =0 ] similarily for =0Converse: let ≠0=i+j+k ≠0=2i+2j+2k, = 0 i.e., // soѲ=0 but | |=| |sinѲ = 0Question 9 find the area of △ with vertices A(1,1,2), B(2,3,5), C(1,5,5).Ans. Area of △ABC = ½| |=½ = ½ √61 * positionvectors of and are i+j+2k & 4j+3k]Question 10 find the area of //gm. Whose adjacent sides aredetermined by the vectors i-j+3k and 2i-7j-k.Find unit vector // to its
52. 52. diagonal [Ans. Area of //gm whose adjacent sides are given | |= 15 √2, where = ,unit vector=( + )/ + |MISC. Question 3 A girl walks 4 km. towards west, then she walks 3km in a direction 300 east of north and stops. Determine the girl’sdisplacement from her initial point of departure.Ans. N A 300 3 km displacement W B (4 km) OLet the girl starts from O and the girl walks 4 km towards W. again shewalks 3 km along WA.(let repts. and repts. j) = +=-4i+(3cos600i+3sin600j) ⇨ = -4i+3j/2+3√3j/2 = -5i/2+3√3j/2[ = + = 3cos600i+3sin600j ] , so displacement from O to A = )² = √13 km along .Question 4 If = + , then is it true = +| ? justify youranswer.Ans. ²= + = + + ) = =| |²+ | |²+2| || |cosѲWhen Ѳ=00 ² =( + , if Ѳ≠00 , ≠ +|Question 9 Find the position vector of a point R which divides the linejoining two points P and Q whose position vectors are (2 + ) & ( -
53. 53. 3 ) externally in the ratio 1:2. Also, show that P is the mid point ofthe line segment RQ.Ans. R divides PQ externally in the ratio 1:2 ∴ position vector of apoint R is =(3 + ) mid point of RQ is (2 + )=position vector of P ⇨ P is mid point of RQ and the point R is (3 + )Question 12 let = i+4j+2k, = 3i-2j+7k and = 2i-j+4k. find a vector which is per. To both & , and . = 15.Ans. Let =d1i+d2j+d3k, since is per. To = i+4j+2k, = 3i-2j+7k,then d1+4d2+2d3=0 & d1-2d2+7d3=0 , by solving d1=32λ, d2=-λ, d3=-14λ (by cross multi.) , put in . = 15 ⇨ (2i-j+4k).(32λi -λj -14λk)⇨λ=5/3, then = 160i/3-5j/3-70k/3Question 13 the scalar product of the vectors i+j+k with a unit vectorAlong the sum of vectors 2i+4j-5k and λi+2j+3k is equal to one. Findthe value of λ.Ans. + = = (2+λ)i+6j-2k , unit vector of = [(2+λ)i+6j-2k]/ |(i+j+k). *(2+λ)i+6j-2k]/ |= 1 ⇨ (λ+6)/ |= 1, where |=⇨λ =1Question 14 If are mutually per. Vectors of equal mag., showthat the vectors is equally inclined to .Ans. )=λ |cosѲ [ ∵ |= λ,=0 . ] ⇨ cosѲ= , similarily dot product with
54. 54. Give same angle Ѳ = cos-1 )EXTRA QUESTIONS:Q. 1 If are unit vectors and Ѳ is the angle b/w them, thenshow that sin(Ѳ/2)= ½ | |.Ans. | |²= ( ). ( )= | |²= 2-2cosѲ[| |=1, A•B = |A| |B| cos(θ)] ⇨ | |²= 4sin2(Ѳ/2)Q.2 If = , = , show that - is // to -Ans. By subtracting above results , we get - )= -⇨ - )= - ⇨( - ) - )=0 ⇨ - is // to-Q.3 If the sum of two unit vectors is a unit vector, show that the mag.of their diff. is √3. [ Hint :use | + |² ⇨ 2 . = -1, put in | - |²]Q.4 If are three unit vectors such that = =0 andangle b/w is п/6, prove that = ∓ 2( ).Ans. = =0 ⇨ , ⇨ & ⇨ ⇨ ) …….(i) , λ is scalar ⇨| )|= |sinп/6 ⇨ |λ| = 2 ∴ |λ| = ∓2 , put in (i)Q. 5 If are resp. the position vectors of the vertices A,B,C of△ABC, Prove that area of the △ ABC is given by =1/2 | )+ )|
55. 55. Ans. Area of △ABC = ½ | )| = ½ |( ) ( )|∵ = - , =Q. 6 If | |=12 | |=10 & | | = 2, find | |.Ans. We know that | |² +| |² = | |² ⇨ 16 [ bylagrange’s identity, | |²sin²Ѳ=| |²(1-cos²Ѳ) ]Q.7 if = 0 , show that = = .Ans. )= ⇨ = , simily take crossproduct with vector b , we get = , by above results ]Q.8 ABCD is quad. Such that = , = , =m +p.Show that the area of the quad. ABCD is ½ |m+p|| |Ans. + = ⇨ = = ∴ area ofquad. ABCD = ½ | | = ½ |( m +p ) ( )| = ½ | m )- p( )| = ½ |m+p|| |∵ )= - ( )]Q. 9 Let A,B,C &D be any four points in space. Prove that| + + | = 4 ( area of △ABC)Ans. Taking A as orgin , let the position vectors of B,C &D beresp. then are resp. & = , = , = ∴| + + |=| )+ ) +( ) ( )|= 2| |Q.10 Three vectors are coplanar vectors ⇨ ). =0
56. 56. Questions on scalar triple product:1.If = 2i-3j+4k, =i+2j-k, =3i+4j-k then find .( )=. ? [ ans. 36=36 ]2. find the volume of parallelepiped whose edges are =2i-3j+4k,=i+2j-k and = 2i-j+2k [ ans. Use scalar triple product 2 cubic units]3. Show that three vectors -4i-6j-2k, -i+4j+3k and -8i-j+3k are co-planar. [ find scalar triple product is zero]4. Show that three vectors i-2j+3k, -2i+3j-4k and i-3j+λk are co-planar.If λ = 5 [ take scalar triple product is zero , put λ=5]5. Show that four points with position vectors 6i-7j, 16i-19j-4k, 3j-6kand 2i+5j+10k are not co-planar.[hint: = =10i-12j-4k , = =-6i+10j-6k, = =-4i+12j+10kAnd show scalar triple product =840≠ 0]6. For any three vectors prove that[ + ]= 2 [ ][hint: [ + X( )] =[ X + X )]7. If the vectors =ai+j+k , = i+bj+k, =i+j+ck are co-planar, then + + =1, where a,b,c ≠ 1.8. Show that if vectors are co-planar, then are alsoco-planar.9.Find the value of λ if the points A(-1,4,-3) , B(3,λ,-5) , C(-3,8,-5), D(-3,2,1) areco-planar. [ ans. Is 2]
57. 57. 10. Let = 5i-j+7k & = i-j+mk . find m, such that + & - areorthogonal. * ans. Is √(73) +*11. Let , be two given non-cillinear vectors. Then any vectorcoplanar with vector & can be uniquely expressed as = x + y,where x , y are scalars.Find the value of p which makes the vectors co-planar , where = 2i-j+k, = i+2j-3k, = 3i-pj+5k.[hint: by using above result , we get x+3y=2, 2x-py=-1 & 5y-3x =1⇨x=1/2, y=1/2 , p= 4]12. if & are unit vectors and Ѳ is the angle b/w them, then showthat sin = ½ | - |[hint: - |2 = ( - ). - )= 2 + | |2 -2 || |cosѲ =2(1-cosѲ)As | |= | |=1]13. If two vectors =i+j+k , = j-k , find a vector such that X =& . =3. [ ans. Is 5/3i+2/3j+2/3k ]14 . Find angle b/w two vectors with mag. 1 & 2 resp. and whenX |= √3. [ hint: use sinѲ= , Ѳ=п/3]15. If X = X , X = X , show that – is // to - .[ hint: by subtracting above given results X( - )=( - )X ⇨( - )X( - )=0 ⇨ sinѲ=0, where Ѳ is angle b/w two vectors, ≠ ; ≠ ]