FACULTY OF MATHEMATICAL STUDIES MATHEMATICS FOR PART I ENGINEERING. Lectures

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1 FACULTY OF MATHEMATICAL STUDIES MATHEMATICS FOR PART I ENGINEERING Lectures MODULE 5 VECTORS II. Triple products 2. Differentiation and integration of vectors 3. Equation of a line 4. Equation of a plane. Triple products There are two triple products which make sense and are used. These will be introduced briefly and then considered in turn. Triple scalar product a b c (It should be clear from the definitions of scalar and vector products that the only possible interpretation of the above is (a b) c, when the answer is a scalar. The other possibility a (b c) does not make sense.) Triple vector product e.g. (a b) c (Here the brackets are essential since a (b c) also exists but gives a different answer. Both expressions lead to results which are vectors.) (i) Triple scalar product a b c Draw a parallelepiped with vectors a, b and c as sides. k c b a Now (a b) c = {(area base) ˆn} c= (area base) (ˆn c) = (area base)(± height) = ± volume. The alternative signs occur above because it is not known whether the unit vector ˆn points in the same direction as c, or in the opposite direction. It follows from the previous formula that volume of parallelpiped = (a b) c, where the modulus sign in this situation means taking the absolute value, since the triple product inside is a scalar. A number of results can now be proved, either geometrically by considering the equivalent parallelogram or algebraically from the properties of scalar and vector products. Properties: (a) If two, or more, of the vectors a, b and c are parallel then a b c =0 (e.g. a b a =0). (b) If a, b and c are coplanar (i.e. lie in the same plane) then a b c =0.

2 (c) a b c = a b c (can interchange and ) (d) a b c = b c a = c a b (can keep the operator signs fixed but rotate the three vectors in cyclic order a b, b c, c a) (e) The component answer for the triple scalar product is easily written as a determinant a a 2 a 3 a b c = b b 2 b 3 c c 2 c 3 (If you have not yet covered the matrix section then calculate the triple product (a b) c directly in two stages using the component forms of the vector and scalar products.) (ii) Triple vector product a (b c) It can be shown, from the component forms of the vectors for instance, that a (b c) =(a c)b (a b)c The above result, which is stated on the Formula Sheet and in the Data Book, can sometimes be useful in manipulations but it has no simple physical interpretation. The corresponding result for the triple vector product with brackets in different positions can easily be found using the above identity and known properties for vector products, as follows: (a b) c = c (a b) = {(c b)a (c a)b} =(c a)b (c b)a 2. Differentiation and integration of vectors In mechanics vectors are usually time-dependent. For instance the position of a particle normally depends on time so we write r(t). An obvious extension of the derivative of a scalar function of a single variable gives ( ) dr r(t + t) r(t) Def. = lim t 0 t r(t+ t) r(t) P Q Now r(t + t) r(t)= PQ,andas t 0thenPQ becomes closer to the tangential direction at P.The magnitude approaches the rate of change of distance with time and, therefore, the above definition for dr gives the velocity of a moving point with position vector r(t). In components, if position vector r(t) =(x(t),y(t),z(t)) then velocity acceleration ( dr dx = d2 r 2 = 2 ), dy, dz ( d 2 x 2, d 2 y 2, d 2 z 2 )

3 Properties: d (a) d (b) d (c) (d) dλ (λ(t) a(t)) = a + λ da da (a(t) +b(t)) = + db da (a(t) b(t)) = b + a db d da (a(t) b(t)) = b + a db Ex. If P has position r =sinti+costj find (a) the acceleration of P, (b) the speed of P. (a) Since i and j are constant velocity = dr acceleration = d2 r 2 = d =costi sin tj ( ) dr = sin ti cos tj (b) Using the above speed = dr = ( (cos t) 2 +( sin t) 2) /2 ( = cos 2 t +sin 2 t ) /2 = /2 =. Ex 2. The acceleration of a particle is ti+costb. Find the position of the particle at time t given that r =0 and dr =0 at time t =0. It is given that d 2 r 2 = ti+costj, and so we must integrate. Now integration is the inverse of differentiation, and since the vectors i and j are constant it is only necessary to integrate the components of the vector. Hence, integrating with respect to t implies dr = t2 2 i +sintj+c. A constant of integration must be introduced, as usual, but since each term in the vector equation is a vector the integration constant must also be a vector. Applying the initial condition that dr =0 when t =0 implies 0 =0i+0j+C, hence C =0. Integrating the simplified equation for the velocity then yields r(t) = t3 6 i cos tj+c 2, with C 2 denoting the new constant of integration, again a vector. At time t = 0 it follows that The final answer for the position, therefore, is 0 =0i j+c 2 C 2 =j. r(t) = t3 6 i+( cos t) j. 3

4 3. Equation of a line The points A and B have position vectors a and b respectively. The vector equation of the straight line through A and B is now derived. a A r P b B O Consider a general point P which lies on the line, with the position vector of this point being r. Itisclear from the diagram that OP = OA + AP i.e. r = a + AP Now AB = b a and, since the point P always lies on the straight line passing through A and B,the vectors AP and AB are parallel. Thus AP = t AB = t (b a), where t is any scalar, i.e. <t<+. Combining the two results the vector equation of a line through two given points can be written r = a + t (b a), <t<+. In almost all cases you will be working with the vector equation of a line and not its Cartesian equivalent. However, the Cartesian version is easily calculated. If r =(x, y, z), a =(a,a 2,a 3 ) and b =(b,b 2,b 3 ), then substituting into the vector equation of the line and equating components gives x = a + t(b a ), y = a 2 +t(b 2 a 2 ), z = a 3 +t(b 3 a 3 ). Rearranging each of these expressions to produce a formula for t leads to the equations x a = y a 2 = z a 3, b a b 2 a 2 b 3 a 3 which is the Cartesian equation of a line. (= t), Ex 3. Show that the line through A (0, 4, 4) in the direction of b =(,2,) intersects the line through the points C (, 4, ) and D (, 2, 3). A a b O r P For the first line, L, we are given a point on L and a vector parallel to L.IfP is any point on the line then it is clear that AP is always parallel to the line, hence AP = tb=t(,2,). It follows immediately that for line L OP = OA + AP or r =(0,4,4) + t (, 2, ). 4 C D L L 2

5 The second line, L 2, passes through the two points C and D, and is identical to the general case considered at the beginning of this section. Thus CD = OD OC = ((, 2 4, 3 ( )) = (0, 2, 4) so equation of L 2 is : r =(,4, ) + s (0, 2, 4). (Note in the above that L 2 is the equation of the line passing through the point C which is parallel to CD. We could have written down the line passing through the point D which is parallel to CD. This would also lead to a correct equation for the line the details would be different, but the final part of the question would lead to the same answer. It is also important to note that line L 2 involves a different scalar constant to that used for L. This is essential because a general point on the first line has no connection with a general point on the second line.) The two lines intersect if there is a point in common between the lines. This will occur when ALL THREE coordinates of the general points on the two lines are equal. Thus the lines intersect if which requires (0, 4, 4) + t (, 2, ) = (, 4, ) + s (0, 2, 4) i.e. ( t, 4+2t, 4+t)=(,4 2s, +4s), t = 4+2t=4 2s 4+t= +4s. The first of the above equations implies t =. Substituting this value into the second equation yields 4 + 2( ) = 4 2s, 4 2 = 4 2s, hence s =. Finally, substituting the calculated values for s and t into the third equation gives 4 = +4, i.e. 3 = 3. This equation is clearly satisfied identically so the lines do intersect. (It is essential to look at the third equation because if this was not satisfied then the lines would not intersect.) The point of intersection is easily calculated by substituting the calculated value for t into the equation for L (or the value for s into the equation for L 2 ) leading to point of intersection = (0, 4, 4) + ( )(, 2, ) = (0 +, 4 2, 4 ) = (, 2, 3) Shortest distance between two skew lines The calculation of this distance is illustrated by the following example. Ex 4. Find the shortest between the two skew lines r =(0,9,2) + t (3,, ) and r =( 6, 5,0) + s ( 3, 2, 4). From the earlier discussions of equations of lines it is known that the first line above passes through the point (0, 9, 2) and is parallel to the vector (3,, ), whereas the second line passes through the point ( 6, 5, 0) and is parallel to the vector ( 3, 2, 4). It should be obvious after a little thought that the shortest line which joins the two given lines will be perpendicular to both lines. Thus, if the ends of this shortest line are P and P 2 with P and P 2 lying on the first and second lines respectively, then the shortest line will be parallel to the vector product of the vectors parallel to the two given straight lines. Define n =(3,,) ( 3, 2, 4) then from the components written it follows that n =( (4) (2), ( 3) 3(4), 3(2) ( )( 3)) = ( 6, 5, 3). The line with shortest length will be parallel to n and it is helpful to calculate the unit vector ˆn. Since n =(( 6) 2 +( 5) )) /2 = ( ) /2 = (270) /2 =3 30, 5

6 ˆn = n n = 3 ( 6, 5, 3) = ( 2, 5, ). Since P and P 2 have been defined to lie on the first and second lines respectively, then P P 2 = {( 6, 5, 0) + s ( 3, 2, 4)} {(0, 9, 2) + t (3,, )}. When the joining line has shortest length then P P 2 = d ˆn,where d denotes the required distance. Equating the two expressions gives {( 6, 5, 0) + s ( 3, 2, 4)} {(0, 9, 2) + t (3,, )} = d ˆn. () The unknown d can be found by equating the three components of the above equation, and then solving the three resulting equations. However, it is much quicker to use the fact that, by construction, the vector ˆn is perpendicular to the vectors parallel to the two original lines. Therefore, distance d is found more quickly by taking the scalar product of the above equation with ˆn, leading to {( 6, 5, 0) + s ( 3, 2, 4)} {(0, 9, 2) + t (3,, )} ˆn=dˆn ˆn. Using the calculated value for ˆn and expanding leads to ( 6, 5, 0) ( 2, 5, ) + s( 3, 2, 4) ( 2, 5, ) (0, 9, 2) 30 ( 2, 5, ) t(3,, ) Evaluating the scalar products gives 30 ( 2, 5, ) = d ˆn ˆn = d. ( ) + s(0) ( ) + t(0) = d, i.e. d = 47 ( 43) = 90 = = [To find the coordinates of the end-points P and P 2 it is necessary to equate the three components in equation () and solve: 6 3s 3t = d 4 + 2s + t = d 8+4s t=d ( ) 2 =(3 ( ) 2 30) = 6, 3s +3t=0, ( ) 5 =(3 ( ) 5 30) = 5, 2s + t =, ( ) =(3 ( ) 30) =3, 4s t= 5. The solution of the three equations in s and t is s = and t = leading to the points P 2 =( 3, 7,6) and P =(3,8,3).] 6

7 4. Equation of a plane The key feature about a plane is that at any point a line perpendicular to the plane is always pointing in the same direction. n A P a r O Consider a plane which passes through the point A with position a. If the vector n is perpendicular to the plane and P, with position vector r, is any point on the plane, then the vector AP must always lie in the plane and hence always be perpendicular to n. Now AP = r a, so the vectors being perpendicular implies (r a) n =0 r n a n=0 i.e. r n = a n The right-hand side of the above contains two constant vectors and so the general vector equation of a plane is usually written r n = C, where n is the normal to the plane and C is a constant determined by a point on the plane. Ex 5. Find the equation of the plane which passes through the point A (2, 3, ) and is perpendicular to the vector 2 i 2 j + k. The equation of a plane with given normal is r n = C, or r (2, 2, ) = C. The constant C is determined by recalling that A (2, 3, ) lies on the plane so must satisfy the above equation when the position of the general point is replaced by the position of A. It follows that C =(2,3, ) (2, 2, ) = 2(2) + 3( 2) + ( )() = 4 6 = 3, and so the required equation of the plane is r (2, 2, ) = 3. [The Cartesian equation for the plane is easily found from the above, if required: (x, y, z) (2, 2, ) = 3 or2x 2y+z= 3. Note that if you were given the latter Cartesian equation of a plane then the coefficients of the unknowns give you the components of the vector which is perpendicular to the plane, namely (2, 2, ) in the above example.] Perpendicular distance from a point to a plane Consider the plane introduced at the beginning of this section, that is passing through the point A with normal n. Suppose we need to calculate the distance from point Q to the plane. n A N Q 7

8 The easiest way to find the answer is to say that i.e. QN = component of QA in the direction perpendicular to plane QN = QA ˆn, where ˆn denotes the usual unit vector and the modulus sign is necessary because we do not know the relative directions of ˆn and QA. Ex 6. Find the distance from the point Q (2, 3, 4) to the plane x +2y+2z= 3. As mentioned at the end of Ex 5 the given equation of the plane shows that a vector perpendicular to the plane is (, 2, 2), = n, say. It is necessary to find the coordinates of a point that lies on the plane. Clearly, any point is sufficient so choose y = z = 0 in which case the equation x +2y+2z =3 is satisfied provided x = 3. Hence the point (3, 0, 0) is on the given plane. Call this point A, then QA =(3 2,0 ( 3), 0 4) = (, 3, 4). Now hence ˆn = n n = (, 2, 2) (, 2, 2) = = ( ) /2 3 distance from Q to the plane = QA ˆn = ( 3, 2 3, 2 ), 3 ( (, 3, 4) 3, 2 3, 2 = 3) =3. 8

Created by T. Madas VECTOR PRACTICE Part B Created by T. Madas

Created by T. Madas VECTOR PRACTICE Part B Created by T. Madas VECTOR PRACTICE Part B THE CROSS PRODUCT Question 1 Find in each of the following cases a) a = 2i + 5j + k and b = 3i j b) a = i + 2j + k and b = 3i j k c) a = 3i j 2k and b = i + 3j + k d) a = 7i + j