Vectors and Vector Operations in 2D and 3D Spaces, Summaries of Engineering

Download Vectors and Vector Operations in 2D and 3D Spaces and more Summaries Engineering in PDF only on Docsity! Unit I Vectors; Lines and Planes Unit I Vectors; Lines and Planes 1.1 Vectors 1.1.1 Scalar and Vector Quantities A scalar is a quantity that is determine by its magnitude (its number of units measured in a suitable scale). Examples 1 Mass, length, temperature, voltage are examples of scalar quantities. Quantities that have both magnitude and direction are called vectors. A vector is usually represented by an arrow, the length of the arrow represents the magnitude of the vector and the arrow head indicates the direction of the vector. Examples 2 Velocity, acceleration, displacement and force are examples of vector quantities. When a vector is represented by an arrow, say , the point A is called the initial point (tail) and B is called the terminal point (head) of the vector. Vectors can also be represented by a single letter (usually small letter) with a bar over it such as , etc. Example 3 Notations or - vector A or vector a. - a vector with initial point P and terminal point Q. - magnitude (or norm) of vector A. 1.1.2 Equality of two Vectors Definition 1.1 Two non-zero vectors and are said to be equal, denoted = , if and only if they have the same direction and magnitude, regardless of the position of their initial points. Note that: Equality of vectors is transitive. i.e. For three vectors , and , if = and = , then = . Definition 1.2 A Vector is called a free vector, provided that its magnitude and direction are fixed, but its position is indeterminate. If the initial point of a free vector is fixed, then it is called a localized vector. Definition 1.3 Two ( free) Vector are equal if and only if they have the same magnitude and direction. Definition 1.4 A vector of magnitude ( modulus) unity ( one) is called a unit vector. a initial point 1 terminal point Unit I Vectors; Lines and Planes Definition 1.5 Any vector whose magnitude is zero and direction indeterminate is called a null ( zero) vector. A null vector is denoted by . Note that: For any non-zero vector A , is a unit vector in the direction of that of vector A. 1.1.3 Vectors in 2 and 3 Definition 1.6 Position Vectors A non-zero vector in 2 (or 3) is called a position vector if and only if its initial point is at the origin and its terminal point is anywhere other than the origin. From this definition, the initial point (tail) of a vector can be anywhere with out changing the direction and the magnitude of the vector. Vector Addition Definition 1.7 Let and be two vectors in a plane. Then the sum + is the vector represented by . Triangle Law of Vector Addition Let and be any two vectors. To find + , join and head to tail. + is the vector whose initial point is that of and terminal point that of . This law of vector addition is called triangle law of vector addition. For any vector , + = . Theorem 1.1 a) For any two vectors and + = + . b) For any three vectors , and + ( + ) = ( + )+ . For any vector there exists a vector – such that + (– ) = . – , called the opposite of vector , has the same magnitude and opposite in direction to that of . Subtraction of Vectors For any two vectors and , – is the vector defined by adding and – pictorially illustrated as follows: Scalar Multiplication b a b a )( ba  2 Unit I Vectors; Lines and Planes Theorem. If = and = , then = Corollary. If = , = are non-zero vectors and  is the angle between and , then Cos  = The Two Important Inequalities. Let A and B be two vectors. 1.  (Cauchy-Schwarz Inequality) 2.  + (Triangle Inequality) Example. Compute the scalar product of i) = 3 + 4 and = 4  3 i) =  4 + 3 and = 8  6 Solutions. Using the above definition we get: i) = (3, 4) (4,  3) = 12  12 = 0. and ii) = ( 4, 3) (8,  6) =  32  18 =  50. Example Given: The angle between two unit vectors and is 60. Then find i) ii) the angle between and + Solutions i) = ( + ) ( + ) = + 2 cos 60 + = 2 (1 + cos 60) = 3. Therefore = . ii) Let  be the angle between and + . Then = + cos 60 = 1 + cos 60 = 1.5. i) On the other hand = cos  = cos  ii) From i) and ii) we get: cos  = 1.5  cos  =   = 30. Therefore the angle between and + 30. Example. Given: = and = + . Find the value of k such that i) + k is orthogonal to ii) + k is orthogonal to Solutions. = 1 and = . i) + k is orthogonal to if and only if ( + k ) = 0. Now ( + k ) = 0  + k = 0. 5 Unit I Vectors; Lines and Planes  1 + k = 0  k =  1. Therefore k =  1. ii) + k is orthogonal to if and only if ( + k ) = 0. Now ( + k ) = 0  k + = 0.  1 + 2k = 0  k =  0.5. Therefore k =  0.5. Example. Find the angle  between ( , 1,  1) and the positive x axis. Solution. Let A = ( , 1,  1) and B = (1, 0, 0) be two position vectors. Then = cos  Hence cos  =  cos  =   = . Therefore the angle between ( , 1,  1) and the positive x axis is . Direction Angles and Direction Cosines. Defn Let A = (a1, a2, a3) be a non-zero vector. The angles ,  and  (between 0 and  inclusively) that A makes with the positive x, y and z axes respectively are called the direction angles of A. Now take the unit vectors , and . From this definition we get: cos  = , cos  = and cos  = Furthermore; = cos , = cos  = and = cos  Therefore A = ( cos  + cos  + cos ) cos , cos  and cos  are called the direction cosines of A. Example. Let A = ( 2, 0, 3). Find the direction cosine of A. Solution. = = . Hence cos  = , cos  = 0 and cos  = .   = ,  = and  = . Therefore , and are the direction cosines of A. Projection and Resolution of Vectors. 6 Unit I Vectors; Lines and Planes Defn Let A be a non-zero vector. The projection of a vector B onto A, denoted by is defined as: Note that:- is a vector parallel to A. Example. Let A = ( 2,  3,  1) and B = (0, 1,  1). Find and . Solution. = , = and =  2. Therefore = A and =  B. Theorem Let A be a non-zero vector. Then for any vector B,  Proof.  . Therefore  . Now let A and B be orthogonal vectors and let C be a vector in the same plane as A and B. Then we can express C as a linear combination of vectors parallel to A and B as follows: C = + In this case, we say that vector C is resolved into vectors parallel to A and B. Example. Let A = (0, 1,  2), B = (0, 2, 1) and C = (0, 5,  4). Resolve C into vectors parallel to A and B. Solution. = 0 and these three vectors lie on the yz plane. = and = . Now = 13, = 6, = = . Hence = and = . Therefore C = + . Example. Let A = (1, 0, 3), B = ( 3, 0, 1) and C = (2, 0, 5). Resolve C into vectors parallel to A and B. Solution. = 0 and these three vectors lie on the xz plane. = and = . Now = 18, =  1, = = . Hence = and = . 7 Unit I Vectors; Lines and Planes Corollary. Two non-zero vectors A and B are parallel if and only if = . Proof. =  = 0  sin   sin  = 0   = 0 or  = .  A∥ B. Therefore Two non-zero vectors A and B are parallel if and only if = . Example. Let A (3, 2,  2) and B (0, 3, 7). i) Determine whether A and B are parallel or orthogonal or neither. ii) Find a vector orthogonal to both A and B. Solution. i) = (3  0) + (2  3) + ( 2  7) =  2, = and = . Hence neither  0 nor  . Therefore A and B are neither parallel nor orthogonal. Remark: is the area of a parallelogram with adjacent sides A and B. Triple Product. There are two types of triple products. i) Scalar triple product. For any three vectors A, B and C, is called the triple (box or mixed triple) product of A, B and C. Example. Show that for any three vectors A, B and C = Solution. = cos , where  is the angle between A and . = cos  sin , where  is the angle between B and and = cos  sin , where  is the angle between C and and  is the angle between A and B. Now cos  = sin  and sin  = cos , because co-functions of complementary angles are equal. Therefore = . ii) Vector Triple Product. For any three vectors A, B and C, is called the Vector triple product of A, B and C. Example. Show that for any three vectors A, B and C  Solution. = sin , where  is the angle between A and . = sin  sin , where  is the angle between B and and = sin  sin , where  is the angle between C and and  is the angle between A and B. Now sin   sin  and sin   sin . Therefore  . Remark: For any three vectors A, B and C , and are undefined operations. Some Properties of Triple Products. 10 Unit I Vectors; Lines and Planes For any three vectors A, B and C i) = = ii) = “ bac – cab” rule. Remark: For any three non-zero vectors A, B and C; is the volume of a parallelepiped with sides A, B and C. Lines in . A line in space is determined by a point p0 (x0, y0, z0) on ℓ and a non-zero vector L parallel to it. Now let ℓ be a line parallel to a non-zero vector L and let p0 (x0, y0, z0) be a fixed point on ℓ. Let p (x, y, z) be an arbitrary point on ℓ. We need to express p in terms of p0 and L. ℓ ∥ L  ∥ L  (x  x0, y  y0, z  z0) = t L ; for some t   and t  0.  (x, y, z) = (x0, y0, z0) +  = + t L; for some t   and t  0; where = (x, y, z) and = (x0, y0, z0). Therefore = + t L; for some t   and t  0 is the vector form of the equation of a line. Example. Find a vector equation of the line that contains ( 1, 3, 5) and is parallel to Solution. Now = and L = . = ( ) + t ( ) Therefore = is the required vector form of the equation of the line. Now let L = be a given non-zero vector and let (x0, y0, z0) be a point on ℓ. Then for any point (x, y, z) on ℓ that is parallel to L, the vector equation form of ℓ is given by: = + t L. Hence (x, y, z) =  x = , y = and z = . (i) These equations are called the parametric equations of ℓ and t is called the parameter. Example. Find the parametric equation of the line that contain ( 2, 1, 3) and is parallel to . Solution. Now (x0, y0, z0) = ( 2, 1, 3) and L = . Then x = , y = and z = . Therefore x = , y = and z = is the required solution. In the above parametric equations of a line ℓ if a, b, and c are non-zero real numbers then We can express (i) as follows: ; where t  . This form of the equations of a line is called the Symmetric form of the equation of a line. Example. Find the symmetric equations of the line containing the points P1 (2, 3,  1) and P2 (5, 0, 4). Solution. Now take L = = and (x0, y0, z0) = (2, 3,  1). Therefore ; where t   is the required equation. 11 Unit I Vectors; Lines and Planes Example. Find the vector, parametric and symmetric equations of the line containing the point P ( 3, 4, 5) which is parallel to . Solution. (x0, y0, z0) = ( 3, 4, 5). i) Vector equation Hence = ( 3, 4, 5) + t (4, 0,  3). = ( 3 + 4 t, 4, 5  3 t) = ( 3 + 4 t) + + (5  3 t) Therefore ; where t   is the required equation. ii) Parametric equations. (x, y, z) = ( 3 + 4 t, 4, 5  3 t)  x =  3 + 4 t, y = 4 and z = 5  3t ; where t   is the required equation. iii) Symmetric equations. x =  3 + 4 t, y = 4 and z = 5  3t ; where t    ; where t  . Therefore ; where t   is the required equation. Example. Show that the line containing the points (0, 0, 5) and (1,  1, 4) is perpendicular to the line with equation . Solution. Let P (0, 0, 5) and Q (1,  1, 4). We need to show that and = (7, 4, 3) are perpendicular. Now = 0. Therefore the two lines are perpendicular. Distance from a Point to a Line. Given a line ℓ and a non-zero vector L parallel to ℓ, we wish to determine the distance D between ℓ and a point P1 (not on ℓ). To do so choose a point P0 on ℓand let  be the angle between ℓ and , where 0 ≤  ≤ π. Now D = sin , but = sin . Therefore is the distance of P1 from the line ℓ. Example. Find the distance D from the point (2, 1, 0) to the line with equation x =  2, y + 1 = z = t. Solution. Take any point P0 on the line. Say P0 ( 2,  1, 0). P1 (2, 1, 0) and L = (0, 1, 1). Hence = (4, 2, 0). Now =  2 + 4  4 , = 6 and = . Therefore D = 3 units. Planes in 3. Given a point P0 and a non-zero vector , there exists one and only one plane J containing P0 and perpendicular to . Let P0 (x0, y0, z0) be a given point, = (a, b, c) be a non-zero vector and let P (x, y, z) be an arbitrary point on the plane J containing P0 and which is perpendicular to . is perpendicular to the plane J. 12