generic · CBSE Class 12th English Medium · MATHEMATICS PART-2 · Page 7question
VECTOR ALGEBRA
Chapter 10: VECTOR ALGEBRA · MATHEMATICS PART-2 · EN medium
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| A Note From Fig . , using the triangle law, one may note that or (since which is parallelogram law. Thus, we may say that the two laws of vector addition are equivalent to each other. Properties of vector addition Property For any two vectors (Commutative property) Fig . Proof Consider the parallelogram ABCD (Fig . ). Let then using the triangle law, from triangle ABC, we have Now, since the opposite sides of a parallelogram are equal and parallel, from Fig . , we have, . Again using triangle law, from triangle ADC, we have Hence Property For any three vectors c (Associative property) Proof Let the vectors be represented by , respectively, as shown in Fig . (i) and (ii). Fig . Then So Fig . – Hence Remark The associative property of vector addition enables us to write the sum of three vectors without using brackets. Note that for any vector a , we have Here, the zero vector is called the additive identity for the vector addition. . Multiplication of a Vector by a Scalar Let be a given vector and λ a scalar. Then the product of the vector by the scalar λ , denoted as λ , is called the multiplication of vector by the scalar λ . Note that, λ is also a vector, collinear to the vector . The vector λ has the direction same (or opposite) to that of vector according as the value of λ is positive (or negative). Also, the magnitude of vector λ is | λ | times the magnitude of the vector , i.e., | λ | = | λ | A geometric visualisation of multiplication of a vector by a scalar is given in Fig . . Fig . When λ = – , then λ = – , which is a vector having magnitude equal to the magnitude of and direction opposite to that of the direction of . The vector – is called the negative (or additive inverse ) of vector and we always have + (– ) = (– ) + = Also, if | ||||
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| = | λ , provided ≠ i.e. is not a null vector, then | λ | = | λ | = So, λ represents the unit vector in the direction of . We write it as ˆ a = | r a A Note For any scalar k , = . . . Components of a vector Let us take the points A( , , ), B( , , ) and C( , , ) on the x -axis, y -axis and z -axis, respectively. Then, clearly | = , | = and | = The vectors , each having magnitude , are called unit vectors along the axes OX, OY and OZ, respectively, and denoted by ˆ ˆ , and i j k , respectively (Fig . ). Now, consider the position vector of a point P( x , y , z ) as in Fig . . Let P be the foot of the perpendicular from P on the plane XOY. We, thus, see that P P is parallel to z -axis. As ˆ ˆ , and i j k are the unit vectors along the x , y and z -axes, respectively, and by the definition of the coordinates of P, we have . Similarly, Fig . Fig . Therefore, it follows that Hence, the position vector of P with reference to O is given by This form of any vector is called its component form. Here, x , y and z are called as the scalar components of , and xi yj zk are called the vector components of along the respective axes. Sometimes x , y and z are also termed as rectangular components . The length of any vector = xi yj zk , is readily determined by applying the Pythagoras theorem twice. We note that in the right angle triangle OQP (Fig . ) and in the right angle triangle OP P, we have Hence, the length of any vector = ˆ+ xi yj zk is given by | = If are any two vectors given in |
| OP P , we have Using the properties of vector addition, the above equation becomes i.e. x i y j z k x i y j z k ˆ x x i y y z z k The magnitude of vector is given by | = x x y y z z Fig . Example Find the vector joining the points P( , , ) and Q(– , – , – ) directed from P to Q. Solution Since the vector is to be directed from P to Q, clearly P is the initial point and Q is the terminal point. So, the required vector joining P and Q is the vector given by ( ) ( ) ( ) −− + −− + −− i.e. . . . Section formula Let P and Q be two points represented by the position vectors , respectively, with respect to the origin O. Then the line segment joining the points P and Q may be divided by a third point, say R, in two ways – internally (Fig . ) and externally (Fig . ). Here, we intend to find the position vector for the point R with respect to the origin O. We take the two cases one by one. Case I When R divides PQ internally (Fig . ). If R divides such that where m and n are positive scalars, we say that the point R divides internally in the ratio of m : n . Now from triangles ORQ and OPR, we have Therefore, we have (Why?) or (on simplification) Hence, the position vector of the point R which divides P and Q internally in the ratio of m : n is given by Fig . Case II When R divides PQ externally (Fig . ). We leave it to the reader as an exercise to verify that the position vector of the point R which divides the line segment PQ externally in the ratio m : n PR i.e. QR m n is given by Remark If R is the midpoint of | |||||||||
| the angle between (Fig . ). If either then θ is not defined, and in this case, we define Observations . is a real number. . Let be two nonzero vectors, then if and only if are perpendicular to each other. i.e. . If θ = , then In particular, as θ in this case is . . If θ = π , then | = − In particular, , as θ in this case is π . . In view of the Observations and , for mutually perpendicular unit vectors ˆ ˆ k we have ˆ ˆ ˆ ˆ i i j j ⋅= = ˆ ˆ , k k ˆ ˆ j k = ˆ ˆ k i ⋅= . The angle between two nonzero vectors is given by or . The scalar product is commutative. i.e. (Why?) Two important properties of scalar product Property (Distributivity of scalar product over addition) Let be any three vectors, then Fig . (i) B C A l B l A C (ii) A B C l (iv) l C B A (iii) θ θ θ θ p p p p ( < < ) θ ( < < ) θ ( < < ) θ ( < < ) θ Property Let be any two vectors, and l be any scalar. Then If two vectors are given in component form as a i a j a k b i b j b k , then their scalar product is given as ) ( a i a j a k b i b j b k a i b i b j b k a j b i b j b k a k b i b j b k ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ a b i i a b i a b i k j i j j j k | |||||||||
| ˆ ˆ a b k i a b k a b k k (Using the above Properties and ) = a b + a b + a b (Using Observation ) Thus . . Projection of a vector on a line Suppose a vector makes an angle θ with a given directed line l (say), in the anticlockwise direction (Fig . ). Then the projection of on l is a vector (say) with magnitude , and the direction of being the same (or opposite) to that of the line l , depending upon whether cos θ is positive or negative. The vector Fig . is called the projection vector , and its magnitude | is simply called as the projection of the vector on the directed line l . For example, in each of the following figures (Fig . (i) to (iv)), projection vector of along the line l is vector Observations . If ˆ p is the unit vector along a line l , then the projection of a vector on the line l is given by ˆ p . Projection of a vector on other vector r b , is given by ˆ, b or . If θ = , then the projection vector of will be itself and if θ = π , then the projection vector of will be . If = π θ or = π θ , then the projection vector of will be zero vector. Remark If α , β and γ are the direction angles of vector a i a j a k , then its direction cosines may be given as Also, note that are respectively the projections of along OX, OY and OZ. i.e., the scalar components a , a and a of the vector , are precisely the projections of along x -axis, y -axis and z -axis, respectively. Further, if is a unit vector, then it may be expressed in terms of its direction cosines as cos cos cos α + | |||||||||
| β + γ Example Find the angle between two vectors with magnitudes and respectively and when = . Solution Given . We have Example Find angle ‘ θ ’ between the vectors Solution The angle θ between two vectors is given by cos θ = Now ) ( = −−= − . Therefore, we have cos θ = hence the required angle is θ = Example If , then show that the vectors are perpendicular. Solution We know that two nonzero vectors are perpendicular if their scalar product is zero. Here ( ) ) ( ) ) So ) . ( ( ) ( ) . Hence are perpendicular vectors. Example Find the projection of the vector on the vector Solution The projection of vector on the vector b is given by ( ) ( ) ( ) ( ) × + × + × Example Find , if two vectors are such that Solution We have B C A ( ) ( ) ( ) Therefore Example If is a unit vector and , then find Solution Since is a unit vector, . Also, or = or Therefore = (as magnitude of a vector is non negative). Example For any two vectors , we always have (Cauchy- Schwartz inequality). Solution The inequality holds trivially when either or r a . Actually, in such a situation we have . So, let us assume that Then, we have = | cos | θ ≤ Therefore Example For any two vectors , we always have (triangle inequality). Solution The inequality holds trivially in case either (How?). So, let . Then, (scalar product is commutative) ≤ (since | x x x ≤ ∀∈ R ) ≤ (from Example ) Fig . Hence Remark If the equality holds in triangle inequality (in the above Example ), i.e. then showing that the points A, B and C are collinear. Example Show that the points A( |
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