{"id":9412,"date":"2026-06-01T21:33:48","date_gmt":"2026-06-01T21:33:48","guid":{"rendered":"https:\/\/kapdec.com\/help\/?p=9412"},"modified":"2026-06-01T21:33:48","modified_gmt":"2026-06-01T21:33:48","slug":"vectors-for-motion-of-object","status":"publish","type":"post","link":"https:\/\/kapdec.com\/help\/vectors-for-motion-of-object\/","title":{"rendered":"Vectors For Motion Of Object"},"content":{"rendered":"

Unit: Functions involves Parameters, Vectors & Matrices<\/strong><\/h2>\n

Chapter: Vectors for Motion of Object<\/strong><\/h3>\n

Reference: – Basic vectors, Scalar, Vector, Vector representation, Vector addition, Vector subtraction, Position, Displacement & Velocity, Velocity vectors, Acceleration, Projectile motion, Effect of gravity, Tangent vectors, Normal vectors, Arc length, path Integral.<\/p>\n

After studying this chapter, you should be able to:
\n    Introduction to Vectors for motion & Their types.
\n    Position, Displacement & velocity.
\n    Effect of Gravity, Tangent vectors & Normal vectors.
\n    Arc length, Path Integral, Applications & Properties.<\/p>\n

Introduction to Vectors for motion
\n 
\nThe vectors are defined as an object containing both magnitude and direction. Vector describes the movement of an object from one point to another.  Vector math can be geometrically picturized by the directed line segment. The length of the segment of the directed line is called the magnitude of a vector and the angle at which the vector is inclined shows the direction of the vector. The beginning point of a vector is called "Tail," and the end side (having arrow) is called "Head." 
\n 
\nDefinition: A quantity that has magnitude, as well as direction, is called a vector.
\nNotice that a directed line segment is a vector (above figure), denoted as ¯AB or simply as a \u20d7, and read as ‘vector ¯AB’ or ‘vector a \u20d7’.
\nThe point A from where the vector ¯AB starts is called its initial point, and the point B where it ends is called its terminal point. The arrow indicates the direction of the vector.
\nPosition Vector
\nThe position vector is used to specify the position of a certain body. The position vector of an object is measured from the origin, in general.
\nPosition vector (r \u20d7) = xi \u0302 + yj \u0302 + zk \u0302 
\nwhere,
\ni \u0302 = unit vector along x direction
\nj \u0302 = unit vector along y direction
\nk \u0302 = unit vector along z direction
\nConsider a point P in space, having coordinates (x, y, z) with respect to the origin O(0, 0, 0). Then, the vector ¯OP having O and P as its initial and terminal points, respectively, is called the position vector of the point P with respect to O. Using the distance formula, the magnitude of ¯OP (or r \u20d7) is given by
\n|¯OP|= √(x^2  + y^2  + z^2 )
\nIn practice, the position vectors of points A, B, C, etc., with respect to the origin O are denoted by a \u20d7,b \u20d7,c \u20d7 etc., respectively.
\n 
\n <\/p>\n

\nDirection Cosines:
\n 
\nLet us consider a point P lying in space, and if its position vector makes positive angles (anticlockwise direction) of α, β and γ with the positive x, y, and z-axis, respectively, then these angles are known as direction angles and on taking the cosine of these angles, we get direction cosines. Taking direction cosines makes it easy to represent the direction of a vector in terms of angles with respect to the reference.
\nThe coordinates of the point P may also be expressed as the product of the magnitude of the given vector and the cosines of direction on the three axes, i.e.
\nx = l|r \u20d7 |
\ny = m|r \u20d7|
\nz = n|r \u20d7|
\nWhere l, m, n represent the direction cosines of the given vector on the axes x, y, z, respectively.<\/p>\n

Note: 
\n(i) The coordinates of the point P may also be expressed as (lr, mr, nr). The numbers lr, mr and nr, proportional to the direction cosines, are called as direction ratios of vector r \u20d7, and denoted as a, b and c, respectively.
\n(ii) l2 + m2 + n2 = 1 but a2 + b2 + c2 \uf0b9 1.
\nTypes of Vectors:
\nZero Vector: A vector whose initial and terminal points coincide is called a zero vector or (null vector). It is denoted by 0.
\nThe zero vector cannot be assigned a definite direction as it has zero magnitudes.
\nThe vectors ¯AA,¯BB represents the zero vector. 
\nUnit Vector: A vector whose magnitude is unity (i.e., 1 unit) is called a unit vector.
\nThe unit vector in the direction of a given vector a \u20d7 is denoted by a \u0302. 
\nCoinitial Vectors: Two or more vectors having the same initial point are called coinitial vectors.
\nCollinear Vectors: Two or more vectors are said to be collinear if they are parallel to the same line, irrespective of their magnitudes and directions.
\nFor example: Consider 3 vectors as shown in the figure, they all are parallel to each other but their magnitudes are different as well as the directions. But they are said to be collinear vectors because they are parallel to each other.
\n 
\nEqual Vectors: Two vectors a \u20d7 and b \u20d7 are said to be equal if they have the same magnitude and direction regardless of the positions of their initial points, and written as a \u20d7 and b \u20d7.
\nFor example: Consider 2 vectors whose magnitudes and their directions are the same irrespective of origin, then they are known as equal vectors. 
\n 
\nNegative of a Vector: A vector whose magnitude is the same as that of a given vector but direction is opposite to that of it is called negative of the given vector.
\n 
\nExample:
\nRepresent the following graphically:
\ni. a displacement of 40 km, 30° east of north
\nii. a displacement of 50 km south – east
\niii. a displacement of 70 km, 40° north of west.
\nSolution:
\ni. a displacement of 40 km, 30° east of north
\nStep 1: Draw north, south, east and west as shown below:<\/p>\n

\n <\/p>\n

\nStep 2: Plot a line  (OP) \u0305 30° east of north as shown below:<\/p>\n

\n <\/p>\n

\nStep 3: Define scale and mark 40km on line (OP) \u0305<\/p>\n

\nLet the scale be 10km = 1cm<\/p>\n

\n <\/p>\n

\n∴ (OP) \u0305 represents the displacement of 40 km, 30o East of North<\/p>\n

\nii. a displacement of 50 km south – east<\/p>\n

\nStep 1: Draw north, south, east and west as shown below:<\/p>\n

\n <\/p>\n

\nStep 2: As the displacement should be south – east, the angle between the displacement and east (or south) will be 45°. Now, plot a line (OP) \u0305 45° east of south as shown below:<\/p>\n

\n <\/p>\n

\nStep 3: Define scale and mark point R such that OR = 50km on line (OP) \u0305. Let the scale be 10km = 1cm<\/p>\n

\n <\/p>\n

\n∴ (OR) \u0305  represents the displacement of 50 km south – east<\/p>\n

\niii. A displacement of 70 km, 40o north of west.<\/p>\n

\nStep 1: Draw north, south, east and west as shown below:<\/p>\n

\n <\/p>\n

\nStep 2: Plot a line (OP) \u0305  40° north of west as shown below:<\/p>\n

\n <\/p>\n

\nStep 3: Define scale and mark point R such that OR = 70km on line(OP) \u0305 .Let the scale be 10km = 1cm<\/p>\n

\n <\/p>\n

\n∴  (OP) \u0305 represents the displacement of 70 km, 40o north of west<\/p>\n

Example: Represent graphically a displacement of 40 km, 30° west of south.
\nSolution: The vector ¯OP represents the required displacement.
\n 
\nExample: Classify the following measures as scalars and vectors.
\n(i)     200 g\/cm3                (ii)     20 km\/hr
\n(iii)     10 Newton                 (iv)     1000 cm3
\n(v)     500 minute                     (vi)     50 m\/s towards north
\nSolution
\n(i)     Density-scalar            (ii)     Speed-scalar 
\n(iii)     Force-vector            (iv)     Volume-scalar 
\n(v)     Time-scalar                     (vi)     Velocity-vector
\nExample: In the given figure, which of the vectors are:
\n(i)     Collinear         (ii)     Equal         (iii)     Coinitial
\n 
\nSolution
\n(i)     Collinear vectors: a \u20d7,c \u20d7  and d \u20d7.
\n(ii)     Equal vectors: a \u20d7  and c \u20d7.
\n(iii)     Coinitial vectors: b \u20d7,c \u20d7  and d \u20d7.
\nVector Addition
\nA variety of mathematical operations can be performed with and upon vectors. One such operation is the addition of vectors. Two vectors can be added together to determine the result (or resultant). This process of adding two or more vectors has already been discussed in an earlier unit.
\nThe two vectors a and b can be added, giving the sum to be a + b. This requires joining them head to tail.
\n 
\nCharacteristics of Vector Math Addition
\n    Commutative Law: a + b = b + a
\n    Associative law: (a + b) + c = a + (b + c)<\/p>\n

Triangle law of vector addition
\nThe Triangle law of vector addition is appropriate to deal with such a situation. If two vectors are represented by two sides of a triangle in sequence, then the third closing side of the triangle, in the opposite direction of the sequence, represents the sum (or resultant) of the two vectors in both magnitude and direction.
\nA vector ¯AB simply means the displacement from a point A to the point B. Now consider a situation that a girl moves from A to B and then from B to C.
\n 
\nThe net displacement made by the girl from point A to the point C, is given by the vector ¯AC and expressed as
\n        ¯AC=¯AB+¯BC
\nThis is known as the triangle law of vector addition.
\nNote:
\n(i)  The associative property of vector addition enables us to write the sum of three vectors a \u20d7,b \u20d7,c \u20d7 as a \u20d7+b \u20d7+c \u20d7 without using brackets.
\n(ii) a \u20d7+0 \u20d7=0 \u20d7+a \u20d7=a \u20d7
\nHere, the zero vector a \u20d7 is called the additive identity for the vector addition.
\nVector Subtraction
\nTo subtract two vectors, you put their feet (or tails, the non-pointy parts) together, then draw the resultant vector, which is the difference of the two vectors, from the head of the vector you're subtracting to the head of the vector you're subtracting it from.
\nA reverse vector (-a), which is opposite of (a) has a similar magnitude as (a) but pointed in the opposite direction.
\n 
\nMultiplication of a Vector by a Scalar
\nMultiplication of a vector by a scalar quantity is called “Scaling.” In this type of multiplication, only the magnitude of a vector is changed not the direction.
\n    S(a + b) = Sa + Sb
\n    (S + T)a = Sa + Ta
\n    a.1 = a
\n    a.0 = 0
\n    a.(-1) = -a
\nGeometric visualization of multiplication of a vector by a scalar
\nA scalar, however, cannot be multiplied by a vector. To multiply a vector by a scalar, simply multiply the similar components, that is, the vector's magnitude by the scalar's magnitude. This will result in a new vector with the same direction but the product of the two magnitudes.
\n 
\nWhen \uf06c = –1, then \uf06ca \u20d7=-a \u20d7, which is a vector having a magnitude equal to the magnitude a \u20d7 of and direction opposite to that of the direction of a \u20d7. The vector -a \u20d7 is called the negative (or additive inverse) of vector a \u20d7 and we always have 
\na \u20d7+(-a \u20d7 )=(-a \u20d7 )+a \u20d7=0 \u20d7 
\nAlso, if \uf06c=1\/(|a \u20d7|), provided a \u20d7≠0 i.e. a \u20d7 is not a null vector, then
\n|\uf06ca \u20d7| =|\uf06c||a \u20d7| = 1\/(|a \u20d7|) |a \u20d7 |=1
\nSo, \uf06ca \u20d7 represents the unit vector in the direction of a \u20d7. We write it as
\na \u0302=1\/(|a \u20d7|) a \u20d7 
\nNote: For any scalar k,k0 \u20d7=0 \u20d7.<\/p>\n

\nComponents of a vector
\nWhen you break a vector into its parts, those parts are called its components. For example, in the vector (4, 1), the x-axis (horizontal) component is 4, and the y-axis (vertical) component is 1.
\nLet us take the points A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1) on the x-axis, y-axis and z-axis, respectively. 
\n 
\nThen, clearly
\n|¯OA| = 1, |¯OB|= 1 and |¯OC| = 1
\nThe vectors ¯OA,¯OB and ¯OC, each having magnitude 1, are called unit vectors along the axes OX, OY and OZ, respectively, and denoted by i \u0302,j \u0302 and k \u0302, respectively (Fig. 10.13).
\nNow, consider the position vector ¯OP of a point P(x, y, z). Let P1 be the foot of the perpendicular from P on the plane XOY.
\n 
\nWe, thus, see that P1 P is parallel to z-axis. As i \u0302,j \u0302 and k \u0302 are the unit vectors along the x, y and z-axes, respectively, and by the definition of the coordinates of P, we have ¯(P_1 P)=¯QR=zk \u0302 similarly, ¯(OP_1 )=¯OS=yj \u0302 and ¯OQ=xi \u0302.
\nTherefore, it follows that ¯(OP_1 )=¯OQ+¯(\u3016QP\u3017_1 )=xi \u0302+yj \u0302
\nand ¯OP=¯(OP_1 )+¯(P_1 P)=xi \u0302+yj \u0302+zk \u0302
\nHence, the position vector of P with reference to O is given by
\n¯OP  (or r \u20d7 )=xi \u0302+yj \u0302+zk \u0302 
\nThis form of any vector is called its component form. Here, x, y and z are called as the scalar components of r \u20d7, and xi \u0302+yj \u0302  and zk \u0302 are called the vector components of r \u20d7 along the respective axes. Sometimes x, y and z are also termed as rectangular components.
\nThe length of any vector r \u20d7=xi \u0302+yj \u0302+zk \u0302, is readily determined by applying the Pythagoras theorem twice. We note that in the right angle triangle OQP1
\n|¯(\u3016OP\u3017_1 )|=√(\u3016|¯OQ|\u3017^2+\u3016|¯(QP_1 )|\u3017^2 )=√(x^2+y^2 ) 
\nand in the right angle triangle OP1P, we have
\nHence, the length of any vector r \u20d7=xi \u0302+yj \u0302+zk \u0302 is given by
\n|r \u20d7 |=|xi \u0302+yj \u0302+zk \u0302 |=√(x^2+y^2+z^2 ) 
\nIf a \u20d7 and b \u20d7 are any two vectors given in the component form a_1 i \u0302+a_2 j \u0302+a_3 k \u0302 and b_1 i \u0302+b_2 j \u0302+b_3 k \u0302, respectively, then.
\n(i)     the sum (or resultant) of the vectors a \u20d7 and b \u20d7 is given by
\n    a \u20d7+b \u20d7=(a_1+b_1 ) i \u0302+(a_2+b_2 ) j \u0302+(a_3+b_3 ) k \u0302
\n(ii)     the difference of the vector a \u20d7 and b \u20d7 is given by
\n    a \u20d7-b \u20d7=(a_1-b_1 ) i \u0302+(a_2-b_2 ) j \u0302+(a_3-b_3 ) k \u0302
\n(iii)     the vectors a \u20d7  and b \u20d7 are equal if and only if 
\n    a_1=b_1,a_2=a_2  and a_3=a_3
\n(iv)     the multiplication of vector a \u20d7 by any scalar \uf06c is given by
\n    \uf06ca \u20d7=(\uf06ca_1 ) i \u0302+(\u3016\uf06ca\u3017_2 ) j \u0302+(\u3016\uf06ca\u3017_3 ) k \u0302
\nThe addition of vectors and the multiplication of a vector by a scalar together give the following distributive laws:
\nLet a \u20d7 and b \u20d7 be any two vectors, and k and m be any scalars. Then
\n    ka \u20d7+ma \u20d7=(k+ma)a \u20d7
\n    k(ma \u20d7 )=(km)a \u20d7
\n    k(a \u20d7+b \u20d7 )=ka \u20d7+kb \u20d7
\nNote:
\n(i)     One may observe that whatever be the value of \uf06c, the vector \uf06ca \u20d7 is always collinear to the vector a \u20d7. In fact, two vectors a \u20d7 and b \u20d7 are collinear if and only if there exists a nonzero scalar \uf06c such that b \u20d7=\uf06ca \u20d7. If the vectors a \u20d7 and b \u20d7 are given in the component form, i.e. a \u20d7=a_1 i \u0302+a_2 j \u0302+a_3 k \u0302 and b \u20d7=b_1 i \u0302+b_2 j \u0302+b_3 k \u0302 then the two vectors are collinear if and only if
\n        b_1 i \u0302+b_2 j \u0302+b_3 k \u0302=\uf06c\u3016(a\u3017_1 i \u0302+a_2 j \u0302+a_3 k \u0302)
\n    b_1 i \u0302+b_2 j \u0302+b_3 k \u0302=\u3016(\uf06ca\u3017_1)i \u0302+(\uf06ca_2)j \u0302+\u3016(\uf06ca\u3017_3)k \u0302
\n    b_1 i \u0302=\u3016\uf06ca\u3017_1,+b_2=\uf06ca_2,b_3=\u3016\uf06ca\u3017_3
\n    b_1\/a_1 =b_2\/a_2 =b_3\/a_3 =\uf06c
\n(ii)     If a \u20d7=a_1 i \u0302+a_2 j \u0302+a_3 k \u0302, then a1, a2, a3 are also called direction ratios 
\nof a \u20d7.
\n(iii)     In case if it is given that l, m, n are direction cosines of a vector, then li \u0302+mj \u0302+nk \u0302=(cos\u2061α ) i \u0302+(cosβ) j \u0302+(cosγ)k \u0302 is the unit vector in the direction of that vector, where \uf061, \uf062 and \uf067 are the angles which the vector makes with x, y and z axes respectively.
\nExample: If a=xi+2j−zk and b=3i−yj+k are two equal vectors, then write the value of x + y + z
\nSolution: 
\nGiven: a=xi+2j−zk and b=3i−yj+k are two equal vectors.
\nTo find: value of x + y + z …(i)<\/p>\n

\nAccording to question,<\/p>\n

\n <\/p>\n

\nComparing both the sides, we get<\/p>\n

\nx = 3, 2 = -y and -z = 1<\/p>\n

\n⇒ x = 3, y = -2 and z = -1<\/p>\n

\nPutting the value of x, y and z in eq. (i), we get<\/p>\n

\nx + y + z = 3 + (-2) + (-1)<\/p>\n

\n= 3 – 2 – 1<\/p>\n

\n= 0<\/p>\n

\nHence, the value of x + y + z is 0
\nExample: Find the values of x, y and z so that the vectors 
\na \u20d7=xi \u0302+2j \u0302+zk \u0302 and b \u20d7=2i \u0302+yj \u0302+k \u0302 are equal.
\nSolution: Given vectors are equal, therefore
\na \u20d7=b \u20d7 
\nxi \u0302+2j \u0302+zk \u0302=2i \u0302+yj \u0302+k \u0302 
\nx = 2, y = 2, z = 1.
\nExample: Let a \u20d7=(5i) \u0302+10j \u0302 and b \u20d7=10i \u0302+(5j) \u0302. Is |a \u20d7|=|b \u20d7|? 
\nAre the vectors a \u20d7 and b \u20d7  equal?
\nSolution: a \u20d7=(5i) \u0302+10j \u0302 and b \u20d7=10i \u0302+(5j) \u0302
\n|a \u20d7 |=√(5^2+10^2 )=√125 and |b \u20d7 |=√(10^2+25^2 )=√125
\nSo, |a \u20d7|=|b \u20d7|. 
\nBut, the two vectors are not equal since their corresponding components are distinct.
\nExample: Find unit vector in the direction of vector a \u20d7=5i \u0302+3j \u0302+2k \u0302.
\nSolution: a \u20d7=5i \u0302+3j \u0302+2k \u0302
\nThe unit vector in the direction of a vector a \u20d7 is given by a \u20d7=1\/(|a \u20d7|) a \u20d7
\n|a \u20d7 |=√(5^2+3^2+2^2 )=√(25+9+4)=√38 
\nTherefore        a \u20d7=1\/√38 (5i \u0302+3j \u0302+2k \u0302 )=5\/√38 i \u0302+3\/√38 j \u0302+2\/√38 k \u0302
\nExample: Find a vector in the direction of vector a \u20d7=5i \u0302-(5j) \u0302 that has magnitude 11 units.
\nSolution: a \u20d7=5i \u0302-(5j) \u0302
\nThe unit vector in the direction of the given vector a \u20d7 is
\n|a \u20d7 |=√(5^2+5^2 )=√50=5√2
\na \u0302=1\/(|a \u20d7|) a \u20d7=1\/(5√2) ((5i) \u0302-5j \u0302 )=1\/√2 i \u0302-1\/√2 j \u0302
\nTherefore, 
\nRequired vector = 11a \u0302=11(1\/√2 i \u0302-1\/√2 j \u0302 )=11\/√2 i \u0302-11\/√2 j \u0302
\nExample: Write the direction ratio’s of the vector a \u20d7=(2i) \u0302+4j \u0302-2k \u0302 and hence calculate its direction cosines.
\nSolution: a \u20d7=(2i) \u0302+4j \u0302-2k \u0302
\nWe know, direction ratio’s a, b, c of a vector r \u20d7=xi \u0302+yj \u0302+zk \u0302 are just the respective components x, y and z of the vector.
\nSo, a = 2, b = 4 and c = –2
\n|r \u20d7 |=√(4+16+4)=√24 
\nLet l, m and n are the direction cosines of the given vector.
\n    l=a\/(|r \u20d7|)=2\/√24  ,   m=b\/(|r \u20d7|)=4\/√24  ,n=c\/(|r \u20d7|)=(-2)\/√24  
\nThus, the direction cosines are (2\/√24  ,4\/√24  ,-2\/√24).
\nVector joining two points
\nIf P1(x1, y1, z1) and P2(x2, y2, z2) are any two points, then the vector joining P1 and P2 is the vector ¯(P_1 P_2 ).
\n 
\nJoining the points P1 and P2 with the origin O, and applying triangle law, from the triangle OP1P2, we have
\n¯(\u3016OP\u3017_1 )+¯(P_1 P_2 )=¯(OP_2 ) 
\nUsing the properties of vector addition, the above equation becomes
\n¯(P_1 P_2 )=¯(\u3016OP\u3017_2 )-¯(OP_1 ) 
\n¯(P_1 P_2 )=(x_2 i \u0302+y_2 j \u0302+z_2 k \u0302  )–(x_1 i \u0302+y_1 j \u0302+z_1 k \u0302  ) 
\n      =(x_2-x_1 ) i \u0302+(y_2-y_1 ) j \u0302+(z_2-z_1 ) k \u0302
\nThe magnitude of vector ¯(P_1 P_2 ) is given by
\n        |¯(P_1 P_2 )|=√(\u3016(x_2-x_1)\u3017^2+\u3016(y_2-y_1)\u3017^2+\u3016(z_2-z_1)\u3017^2 )
\nExample: Find the position vector of the mid-point of the vector joining the points 
\n  
\nand 
\n 
\nSolution:
\n <\/p>\n

\n <\/p>\n

\nFormula to be used – The midpoint of a line joining points a and b is given by .(a+b)\/2<\/p>\n

\nThe position vector of the midpoint<\/p>\n

\n <\/p>\n

\n <\/p>\n

\nExample: Find the vector joining the points A(1, 2, 3) and 
\nB(–2, –3, –4) directed from A to B.
\nSolution: Since the vector is to be directed from A to B, so
\nA is the initial point and B is the terminal point. 
\nTherefore, 
\nRequired vector = ¯AB=(-2-1) i \u0302+(-3-2) j \u0302+(-4-3)k \u0302
\n¯AB=-3i \u0302-5j \u0302-7k \u0302 
\nSection formula
\nIf O is taken as reference origin and A is an arbitrary point in space then the vector (OA) \u20d7 is called as the position vector of the point. Let us consider two points P and Q denoted by position vectors (OP) \u20d7 and (OQ) \u20d7 with respect to origin O.
\n 
\nLet us consider that the line segment connecting P and Q is divided by a point R lying on PQ. The point R can divide the line segment PQ in two ways: internally and externally. Let us consider both these cases individually.
\nCase 1: Line segment PQ is divided by R internally
\nLet us consider that the point R divides the line segment PQ in the ratio m: n; given that m and n are positive scalar quantities we can say that,
\nm¯RQ=n¯PR 
\nConsider the triangles, \u2206ORQ and \u2206OPR.
\n¯RQ=¯OQ-¯OR=b \u20d7-r \u20d7 
\nand ¯PR=¯OR-¯OP=r \u20d7-a \u20d7,
\nTherefore,
\nm(b \u20d7-r \u20d7 )=n(r \u20d7-a \u20d7) 
\nr \u20d7=(mb \u20d7+na \u20d7)\/(m+n) 
\nTherefore the position vector of point R dividing P and Q internally in the ratio m:n is given by:
\n(OR) \u20d7=(mb \u20d7+na \u20d7)\/(m+n) 
\nCase 2: Line segment PQ is divided by R externally
\n 
\n(OR) \u20d7=(mb \u20d7-na \u20d7)\/(m-n) 
\nNote: If R is the midpoint of PQ, then m = n. 
\nTherefore,     (OR) \u20d7=(a \u20d7+b \u20d7)\/2     
\nExample:
\nShow that the points A, B and C having position vectors.
\n   
\nAnd
\n   
\nRespectively, form the vertices of a right-angled triangle.<\/p>\n

Solution:
\n <\/p>\n

\n <\/p>\n

\n <\/p>\n

\n <\/p>\n

\n <\/p>\n

\n <\/p>\n

\n <\/p>\n

\n <\/p>\n

\n <\/p>\n

 <\/p>\n

 <\/p>\n

 <\/p>\n

Tip – For any 2 perpendicular vectors
\na \u20d7 & b \u20d7, a \u20d7 . b \u20d7= 0<\/p>\n

 <\/p>\n

 <\/p>\n

 <\/p>\n

 <\/p>\n

The triangle is right-angled.<\/p>\n

Example: Consider two points P and Q with position vectors 
\n¯OP=5a \u20d7-3b \u20d7 and ¯OQ=a \u20d7+2b \u20d7. Find the position vector of a point R which divides the line joining P and Q in the ratio 3:1, (i) internally, and 
\n(ii) externally.
\nSolution
\n(i)     Required position vector is 
\n        ¯OR=(2(a \u20d7+2b \u20d7 )+(5a \u20d7-3b \u20d7))\/(3+1)=(7a \u20d7+b \u20d7)\/4
\n(ii)     Required position vector is
\n        ¯OR=(2(a \u20d7+2b \u20d7 )-(5a \u20d7-3b \u20d7))\/(3+1)=(-3a \u20d7+7b \u20d7)\/4
\nExample: Show that the points A=(2i \u0302-j \u0302+k \u0302 ),B=(i \u0302-3j \u0302-5k \u0302 ),  C=(3i \u0302-4j \u0302-4k \u0302 ), are the vertices of a right angled triangle.
\nSolution: A=(2i \u0302-j \u0302+k \u0302 ),B=(i \u0302-3j \u0302-5k \u0302 ),  C=(3i \u0302-4j \u0302-4k \u0302 )
\n¯AB=(1-2) i \u0302+(-3+1) j \u0302+(-5-1) k \u0302=-i \u0302-2j \u0302-6k \u0302 
\n¯BC=(3-1) i \u0302+(-4+3) j \u0302+(-4+5) k \u0302=2i \u0302-j \u0302+k \u0302 
\n¯CA=(2-3) i \u0302+(-1+4) j \u0302+(1+4) k \u0302=-i \u0302+3j \u0302+5k \u0302 
\nSo, |¯AB|^2=41=6+35=|¯BC|^2+|¯CA|^2
\nTherefore, A, B and C are the vertices of the right-angled triangle.<\/p>\n

Key Points
\n    Vectors represent quantities that have both magnitude and direction, such as displacement, velocity, and acceleration.
\n    Scalars, on the other hand, are quantities that only have magnitude, like distance and speed.
\n    Position vectors describe the location of an object in a coordinate system, often using components like x, y, and z.
\n    Displacement vectors represent the change in position of an object and are calculated by subtracting initial position from final position.
\n    Velocity vectors describe the rate at which an object's position changes with respect to time and are calculated by dividing displacement by time.
\n    Acceleration vectors represent the rate of change of velocity and are calculated by dividing the change in velocity by time.
\n    Average velocity is the total displacement divided by the total time taken.
\n    Instantaneous velocity is the velocity of an object at a specific instant in time and is found by taking the derivative of the position vector.
\n    Average acceleration is the change in velocity divided by the change in time.
\n    Instantaneous acceleration is the acceleration of an object at a specific instant in time and is found by taking the derivative of the velocity vector.
\n    Constant acceleration equations can be used to analyze motion under uniform acceleration, such as free fall or motion along an inclined plane.
\n    Projectile motion involves objects moving in two dimensions under the influence of gravity, with horizontal and vertical components of velocity.
\n    Parametric equations and vector functions are used to describe the motion of objects in space.
\n    The derivative of a vector function gives the velocity vector, while the second derivative gives the acceleration vector.
\n    Integrals of vector functions can be used to calculate arc length, work done, or other quantities related to motion.
\n <\/p>\n","protected":false},"excerpt":{"rendered":"

Unit: Functions involves Parameters, Vectors & Matrices Chapter: Vectors for Motion of Object Reference: – Basic vectors, Scalar, Vector, Vector representation, Vector addition, Vector subtraction, Position, Displacement & Velocity, Velocity vectors, Acceleration, Projectile motion, Effect of gravity, Tangent vectors, Normal vectors, Arc length, path Integral. After studying this chapter, you should be able to:    […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[628],"tags":[],"class_list":["post-9412","post","type-post","status-publish","format-standard","hentry","category-ap-precalculus"],"_links":{"self":[{"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/posts\/9412","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/comments?post=9412"}],"version-history":[{"count":0,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/posts\/9412\/revisions"}],"wp:attachment":[{"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/media?parent=9412"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/categories?post=9412"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/tags?post=9412"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}