CHAPTER 3 : VECTORS. Definition 3.1 A vector is a quantity that has both magnitude and direction.

Transcription

1 EQT 101-Engineering Mathematics I Teaching Module CHAPTER 3 : VECTORS 3.1 Introduction Definition 3.1 A ector is a quantity that has both magnitude and direction. A ector is often represented by an arrow or a directed line segment. The length of the arrow represents the magnitude of the ector and the arrow points in the direction of the ector. For example, force, elocity, acceleration, etc are quantity that has magnitude and direction. Definition 3. A scalar is a quantity that has magnitud only. Length, area, olume, mass, time, etc are types of scalar that has magnitude only. 3. Vector Representation Suppose a particle moes along a line segment from point P to point Q. The corresponding displacement ector, shown in figure below, has initial point P (the tail) and the point Q is referred as its head and we indicate this by writing PQ. Notice that the ector QP has the same length but it is in the opposite direction. Q Q PQ QP P P

2 A ector could be written as a, b, u, and w. The magnitude of the ector, is denoted by. Note that would represent a ector of the same magnitude but with opposite direction. Q Q P P 3.3 Two Equal Vectors Definition 3.3 Two ectors u and are said to be equal if and only if they hae the same magnitude and the same direction Example 1 : w u u = and w = x 3.4 Addition of Vectors x Definition 3.4 If u and are ectors, then the addition of u and can be written as u +. i) The Triangle Law If u and are ectors, we connect the initial point of with the terminal point of u, then the sum u + is the ector from the initial point of u to the terminal point of. 41

3 Example : u u u + ii) The Parallelogram Law In figure below, we start with the same ectors u and as in figure aboe and we draw another copy of with the same initial point as u. Completing the parallelogram, we could see that u + = + u. This also gie another way to construct the sum : if we place u and so they start at the same point, then u + lies along the diagonal of the parallelogram with u and as sides. Example 3 : u u + iii) The Sum of a Number of Vectors u The sum of all ectors is gien by the single ector joining the start of the first to the end of last. Example 4: a x b c 1. a + b = x. a + b + c = w u w Note: Addition of ector is satisfied u + = + u (commutatie law) 4

4 3.5 Subtraction of Vectors Definition 3.5 If u and are ectors, then u - can be written as u + ( ) Example 5: u u Note: Subtraction of ectors is not satisfied the commutatie law, but their magnitudes are still the same een though the direction is opposite. 3.6 Scalar Multiplication of Vectors Definition 3.6 If is a ector and α is a scalar, then the scalar multiplication α is a ector whose length is α times the length of and whose direction is the same as if α > 0 and is opposite to if α < 0. Example 6 : Parallel Vectors Definition 3.7 Two ectors u and are said to be parallel if they are in the same direction or opposite direction and is equialent to αu. 43

5 Example 7: u -u αu = αu = 3.8 Components of Vectors in Terms of Unit Vectors If we place the initial point of a ector r at the origin of a rectangular coordinate system, then the terminal point of r has coordinate of the form x, y, z. These coordinates are called the components of r and we write r = xi + y j + zk Where, i, j and k are unit ectors in the x, y and z-axes, respectiely. Unit ectors i, j, k. i = 1,0, 0 to be a unit ector in the ox direction (x-axis) j = 0,1, 0 to be a unit ector in the oy direction (y-axis) k = 0,0, 1 to be a unit ector in the oz direction (z-axis) Vectors in Two Dimension (R ) Two unit ectors along ox and oy direction are denoted by the symbol i and j respectiely. Example 8 : u = 3 i + 4 j 44

6 3.8. Vectors in Three Dimension (R 3 ) Three unit ectors along ox, oy and oz direction are denoted by the symbol i, j and k respectiely. Example 9 : u = i + 3 j + 5k 3.9 Position Vectors Definition 3.8 If r represents the ector from the origin to the point P(a, b, c) in R 3, then the position ector of P can be defined by r = OP = (a, b, c) (0, 0, 0) = a, b, c Example 10 Two forces D and E are acts from the origin point O. If D = 46N, θ D = 0 o and E = 17N, θ E = 90 o. Find the resultant force. Solution 45

7 3.10 Components of Vectors Definition 3.9 If P and Q are two points in R 3 where P(a, b, c) and Q(x, y, z), then the components ector P and Q can be defined by PQ = OQ OP = ( x, y, z) ( a, b, c) = x a, y b, z c Example 11 Gien P(5,, 1) and Q(6, 1, 7) are two points in R 3. Find the components ector of PQ. Solution 3.11 Magnitude (Length) of Vectors Definition 3.10 If u is a ector with the components a, b, c, then the magnitude of ector u can be written as u = a + b + c Example 1 Find the magnitude of the following ectors: i) u = 5, =5i+j ii) u = 5,, 1 =5i+j+k Solutions 46

8 Definition 3.11 If P and Q are two points with coordinate (a, b, c) and (x, y, z), respectiely, then the length from P to Q can be written as PQ = ( x a) + ( y b) + ( z c ) Example 13 If P(5,, 1) and Q(6, 1, 7) in R 3. Find the length from P to Q. Solution 3.1 Unit Vectors Definition 3.1 If u is a ector, then the unit ector u is defined by u = u = 1. A unit ector is a ector whose length is 1. Normally, the symbol ^ is used to represent the unit ector. In general, if u 0, then the unit ector that has the same direction as u is, u u, where In order to erify this, let 1 u u = u = u u 1 c =. Then u = cu u and c is a positie scalar, so u has the same direction as u. Also 1 u = cu = c u = u = 1 u 47

9 Example 14 Find the unit ector in the direction of the ector u = 3i + 4j + k. Solutions 3.13 Direction Angles and Direction Cosines z P O y x The direction angles of a ector OP = xi + y j + zk are the angles α, β and γ that OP makes with the positie x-, y- and z-axes. The cosines of these direction angles, cos α, cos β and cos γ are called the direction cosines of the ector OP and defined by cos α = x OP, cos β = y OP, cos γ = z OP. [0 α, β, γ ] Example 15 Find the direction angles of the ector u = 6i -5j + 8k Solution 48

10 3.14 Operations of Vectors by Components Definition 3.13 If u = u 1, u, u 3, = 1,, 3 and k is a scalar, then i) u + = u 1 + 1, u +, u ii) u - = u 1-1, u -, u 3-3 iii) ku = ku 1, ku, ku 3 Example 16 Gien u =, -1, 1 and = 1, 3, -. Find i) u + ii) u iii) u i) 3u ) 4 Solutions i) ii) iii) i) Properties of A Vectors If a, b, and c are ectors, k and t are scalar, then i) a + b = b + a ii) a + (b + c) = (a + b) + c iii) a + 0 = 0 + a = a i) a + (-a) = (-a) + a = 0 ) k(a + b) = ka + kb i) (k + t)a = ka + ta ii) k(tb) = (kt)b iii) 0c = 0 ix) 1a = a x) -1b = -b 49

11 3.15 Product of Two Vectors Definition 3.14 (The Dot/Scalar Product) If u = u 1 i + u j + u 3 k and = 1 i + j + 3 k are ectors in R 3, then the dot product of u and can be written as u = u u + u 3 3 Teorem 1 If θ is the angle between the ectors u and, then cos θ = u u ; 0 < θ < π Example 17 Find the angle between the ectors u = 4i 5j k and = i + j + 3k. Solution 50

12 Properties of the Dot Product i) u. =. u ii) u. ( + w ) = u. + u. w iii) ( u + ). ( w + z ) = u. w +. w + u. z +. z i) If k is a scalar, then k(u. ) = (ku). = u. (k) ) i.i = j.j = k.k = 1 i) If u and are orthogonal, then u. = 0 ii) i. j = j. k = k. i = 0 iii) u. u = u Definition 3.15 (The Cross Product) If u = u 1, u, u 3 and = 1,, 3 are ectors, then the cross product of u and is the ector u = i j k u1 u u3 = ( u3 u3 ) i ( u13 u31) j + ( u1 u1) k 1 3 Example 18 If u = 3i + 6j + k and = 4i + 5j k. Find i) u ii) u Solutions 51

13 Theorem If u = u 1, u, u 3 and = 1,, 3 are ectors in R 3, then u x = u sin θ ^ n where θ is the angle between the ectors u and and normal unit ector that perpendicular to both u and. ^ n is a u ^ n θ ^ n θ u u x x u Note that: u x = u sin θ ^ n u x = u sin θ ^ n u x = u sin θ = Area of the parallelogram determined by u and. Example 19 Find the angle between the ectors u = 4i + 5j 3k and = i + j. Solution 5

14 Properties of the Cross Product If u,, w are ectors and k is a scalar, then i) u x = - x u ii) u x + w = u x + u x w iii) u x k = (ku) x = k u x i) u x 0 = 0 x u = 0 ) u x = 0 if and only if u // i) u x = u sin θ ii) i x i = j x j = k x k = 0 iii) i x j = k ; j x k = i ; k x i = j ix) j x i = -k ; ; k x j = -i ; i x k = -j 3.16 Applications of the Dot Product and Cross Product i) Projections Figure below shows representations PQ and PR of two ectors a and b with the same initial point P. If S is the foot of the perpendicular from R to the lines containing PQ, then the ector with representation PS is called the ector projection of b onto a and is denoted by proj a b. b R P proj a b S a Q The scalar projection of b onto a (also called the component of b along a) is defined to be magnitude of the ector projection, which is the number b cos θ, where θ is the angle between a and b. This is denoted by comp a b. The equation a b = a b cosθ 53

15 shows that the dot product of a and b can be interpreted as the length of a times the scalar projection of b onto a. Since b cosθ = a b a = b a a the component of b along a can be computed by taking the dot product of b with the unit ector in the direction of a. Theorem 3 Scalar projection of b onto a: Vector projection of b onto a: a b comp a b = a a b a a b proj a b = = a a a a Example 0 Gien b = i + j + k and a = -i + 3j + k. Find the scalar projection and ector projection of b onto a. Solution ii) The Area of Triangle and Parallelogram A C a θ O b B 54

16 Area of triangle OBC = ½ a b sin θ = ½ a x b Area of parallelogram OBCA = a b sin θ = a x b Example 1 Find the area of triangle and parallelogram with A(1,-1, 0), B(, 1, -1) and C(-1, 1, ). Solution i) The Volume of Parallelepiped and Tetrahedron Parallelepiped determined by three ectors a, b and c. 55

17 The olume of the parallelepiped determined by the ectors a, b and c is the magnitude of their scalar triple product Notice that: V = a. ( b x c ) i) a. (b x c) = (a x b). c a. (b x c) if a = a1, a, a3, b = b1, b, b3 and c = c1, c, c3 so a. (b x c)= a b c a b c a b c ii) If a. (b x c) = 0, then the all ectors on a same plane. A a b B O c C The olume of the tetrahedron by the ectors a, b and c is 6 1 of the olume of parallelepiped, that is V = 6 1 a. (b x c). i) Equations of Planes n = a, b, c P 1 = (x 1, y 1, z 1 ) S P = (x, y, z) 56

18 Let P(x, y, z) a point on the plane S and n is normal ector to the plane S. If P 1 (x 1, y 1, z 1 ) is an arbitrary point in the plane S, then P 1 P. n = 0 x x 1, y y 1, z z 1. a, b, c = 0 a(x x 1 ) + b(y y 1 ) + c(z z 1 ) = 0 So that, the equation of the plane is gien by ax + by + cz + k = 0; where k = ax 1 by 1 cz 1 Example Find an equation of the plane through the point P(3,-1,7) with normal ector n = 4,, 5. Solution ) Parametric Equations of A Line in R 3 l B(x, y, z) A(x 1, y 1, z 1 ) a,b,c 57

19 A line l in three-dimensional space is determined when we know a point A(x 1, y 1, z 1 ) on l and the direction of l. The direction of line is coneniently described by a ector, so we let be a ector parallel to l. If B(x, y, z) be an arbitrary point on l, then AB = t. ; t is a scalar become x x 1, y y 1, z z 1 = t. a, b, c = ta, tb, tc So that, the parametric equations of a line can be written as or x = x 1 + ta y = y 1 + tb z = z 1 + tc x x1 a y = y + 1 t b z z1 c These equations can be described in terms of Cartesian equations when we eliminate the parameter of t from parametric equations. Therefore, we obtain x x a y y b z z c t = = =, a, b, c 0. Example 3 a) Find a parametric equations of the line that passes through the point P(1,, 3) and is parallel to the ector = 1, -1,. b) From the result of (a), state the parametric equations to the Cartesian equations. Solutions 58

20 i) Distance from a Point to the Plane P 1 b n D P 0 Let P 0 (x 0, y 0, z 0 ) be any point in the gien plane and let b be the ector corresponding to P 0P1, then b = x x0, y y0, z z0 From figure aboe, we can se that the distance D from P 1 to the plane is equal to the absolute alue of the ector projection of b onto the normal ector n = a, b, c. Thus D = W 1 = = n b n n n b n 59

21 = = a( x x 0 ) + b( y y0 ) + c( z z0 ) a + b + c ( ax + by + cz) ( ax + by a + b cz0 ) + c Since P 0 (x 0, y 0, z 0 ) lies in the plane, its coordinates satisfy the equation of the plane and so we hae ax0 + by0 + cz0 = d. Thus, the formula for D can be written as D = ax + by + cz + d a + b + c Example 4 Find the distance between the parallel planes 10 x + y z = 5 and 5 x + y z = 1. Solution 60