1 Multivariable functions and vector geometry Due date and time: Wednesday, June 7th at 9:30 AM in lecture Please ensure that you have read and understood the document \"Homework Guidelines" on the Blackboard website before beginning this assignment. 1 Computational Problems Problem 1 (Functions). 1. Suppose f is a function defined by the formula f ( x, y, z, w) = x + yz w. Evaulate f whenever ( x, y, z, w) = (4, 2, 8, 3). 2. Suppose g is a function a value of 1 when \u0011 all its parameters\u0010are integers and a value of 0 \u0011 \u0010 that returns ln(8)+ln(2) tan(60 ) ln(8)+ln(2) tan(30 ) , 3 tan(30 ), ln(e)+ln(e/4) ? What about g , 3 tan(60 ), ln(e)+ln(4/e) ? otherwise. What is g 3 3 Note that there could not possibly be a formula for this function g! 3. Suppose P is a function that only takes integers as parameters and returns a value of 1 if all parameters are even, 1 if all parameters are odd and 0 otherwise. What is the value of P( P(1, 0, 2) + 1, 2 P(1, 1, 1) P(2, 2, 0), P(0, 2, 0) 1)? Problem 2 (Contours). Draw three or four contours in the plane for each of the following functions. Don't forget to label each contour with its corresponding function value. 1. 2. 3. 4. f ( x, y) f ( x, y) f ( x, y) f ( x, y) = x2 + 2y2 = sin( x + y) = e xy = xey ye x 5. 6. 7. 8. f ( x, y) f ( x, y) f ( x, y) f ( x, y) = x 2 + y3 = 2x y = 3y3 = cos( x ) + cos(y) Problem 3 (Level Surfaces). Match the following equations with the plotted level surfaces (if this takes you a lot of time, you may wish to get some ideas from http://wolframalpha.com/ but just make sure you cite it in your bibliography if you do use it - see \"Homework Guidelines\"): 1. 3x2 + 2y2 + z2 = 5 2. y2 = z 3x2 3. x2 + y2 + 3z2 = 5 4. x = z2 y2 5. z2 = 2(1 + x2 y2 ) 6. x + y2 + z2 = 0 7. 2x2 2y2 + 2z2 = 1 8. x2 = 10 3z2 9. x = y2 z2 10. z2 + 3y2 = 8 11. x2 = y2 + z2 12. x2 y2 = 5z2 1 1.1 Projections in general As we saw in class, if ~v is an arbitrary vector and ~i is the first standard basis vector, then we can interpret the dot product ~v ~i as the length of the \"shadow\" of ~v when projected down to the x-axis (think of projection as giving you the shadow cast by light shining vertically downwards): Let's denote the shadow vector by proj~i (~v). As we said in class, the length of this shadow vector is |~v| |~i | cos = |~v| cos (since |~i | = 1). Hence proj~i (~v) = (|~v| cos )~i. In fact, we can make a definition of proj~u (~v) for ~u more general than just a standard basis vector. We define this as follows: proj~u (~v) = (|~v| cos )u where is the angle between ~u and ~v = (|~v| cos ) = = \u0012 So proj~u (~v) = |~v| cos = ~v~u |~u| ~v~u |~u|2 ~u |~u| |~v| |~u| cos |~u|2 ! ~v ~u ~u |~u|2 ! ~u \u0013 ~u is the shadow/projection vector of ~v onto the line containing ~u. We call the number the scalar component of ~v in the direction of ~u. 2 Problem 4 (Dot product and projection). With the definitions from the previous page in mind, find (a) ~v ~u, |~v| , |~u| (b) the cosine of the angle between ~v and ~u (c) the scalar component of ~v in the direction of ~u (d) the vector proj~u (~v). for the following pairs of vectors: ~u = 3~i + 4~j + 6~k 1. ~v = 2~i + 4~j 6~k 2. ~v = (2/3)~j (1/2)~k ~u = ~i + 6~j ~u = h1, 2i 3. ~v = h2, 21 i Problem 5 (Angles between vectors). 1. Find the angles between the following pairs of vectors: ~v = 2~i + ~j (a) ~u = ~i + 2~j ~k ~i + 4~j ~i 2~j +~k ~v = 2 (b) ~u = 3 (c) ~u = 7~i + 2~j ~k ~v = 7~i ~j 2. Using what you have learnt in class, compute the angles of the triangle in the plane whose vertices are P = (2, 1), Q = (1, 2) and R = (1, 0). 3. Similarly, compute the angles between the diagonals of the rectangle in the plane whose vertices are A = (1, 0), B = (0, 3), C = (3, 4) and D = (4, 1). 1.2 More on cross products ~ are two vectors, At the end of last class, we started talking about cross-products of vectors. If ~u and w ~ is a number, whereas the cross product ~u w ~ is actually a vector. The definition then the dot product ~u w ~ was: we gave for which vector ~u w ~u w ~ = (|~u| |~ w| sin )n ~ , and n is the unit vector orthogonal to both ~u and w ~ pointing in the where is the angle between ~u and w direction determined by the right-hand rule: As was the case with the dot product, however, the definiton above for the cross product does not make ~ = h2, 1, 1i. To compute it very easy to compute the cross product of two vectors, say ~u = h1, 3, 7i and w ~ = hd, e, f i then cross products, we observe the following algebraic fact: if ~u = h a, b, ci and w ~u w ~ = hb f ce, ( a f cd), ae bdi This is quite a difficult calculation to remember, so here is another popular way to remember it: recall that ~ in terms of the standard basis vectors: ~u = a~i + b~j + c~k and w ~ = d~i + e~j + f~k. Then we can write ~u and w 3 we can compute the cross product as the following determinant (by expanding along the top row): ~i ~j ~k ~u w ~ = a b c d e f b = e c ~ a i d f c ~ a j + d f b ~ k e = (b f ce)~i ( a f cd)~j + ( ae bd)~k (If you don't know about determinants, or just need a reminder about them, have a read about them here: https://en.wikipedia.org/wiki/Determinant) One useful fact concerning cross products that we didn't get to in class was how the length of the cross product vector is related to the area of a parallelogram determined by the original vectors in a very straightforward way: ~ | = |~u| |~ because remember that |~u w w| sin . Problem 6 (Cross product). Compute the cross products of the following pairs of vectors: ~v = ~i + ~j 2~k 1. ~u = 3~i 2~j +~k ~v = h3, 2, 2i 2. ~u = h 31 , 1, 3i ~v = (7/4)~i + (1/2)~j ~k 3. ~u = h1, 5, 47 i 4. ~u = ~i (1/ 3)~j + (1/ 2)~k ~v = h1, 3, 1i 1.3 Triple Scalar Products In fact, not only do we find that the length of the cross product measures the area of a parallelogram, but ~ in 3-space - a triple scalar product (~u ~v) w ~ which is actually we also have - for three vectors ~u, ~v and w ~ ) equal to the volume of a a number (notice that it is ultimately a dot product of the vectors ~u ~v and w parallelepiped determined by the original vectors: 4 (a parallelepiped is a prism all of whose faces are parallelograms). ~ for the following triples and verify Problem 7 (Triple Scalar Products). Have a go at computing (~u ~v) w ~ ) ~u or (~ that in fact you get the same answer if you compute (~v w w ~u) ~v: ~u 1. 2~i 2. ~i ~j +~k 3. h2, 1, 0i 2 ~v ~ w 2~j 2~i + ~j 2~k h2, 1, 1i 2~k ~i + 2~j ~k h1, 2, 0i Conceptual Problems Problem 8 (Functions). Are f ( x, y, z) = x2 + y2 and f ( x, y) = x2 + y2 the same function? Explain your reasoning. Problem 9 (Cauchy-Schwarz Inequality). We already saw the triangle inequality in class. Here you will prove another inequality involving dot products of vectors, known as the Cauchy-Schwarz Inequality. ~ = |~v| |~ Recall the definition of the dot product from lecture: ~v w w| cos where is the angle between ~v ~ . Show that |~v w ~ | |~v| |~ ~ (in the plane or in 3-space). Under what and w w| for any pair of vectors ~v and w circumstances is the inequality actually an equality? Problem 10 (Cancellation for dot products). When we take products of real numbers u, v and w, if we know that u 6= 0 and uv = uw then we can always cancel the u to conclude that v = w. Is the same true ~ ? i.e. if ~u ~v = ~u w ~ , can we \"cancel\" ~u to conclude that when we take dot products of vectors ~u, ~v and w ~v = w ~ ? If so, prove it. If not, give a counter-example. Problem 11 (Cross products). Which of the following equalities are always true and which of them are not always true? Give reasons for your answers, even if it was mentioned in class. 1. 2. 3. ~u ~v = (~v ~u) (~u) ~v = (~u ~v) (~v ~v) ~v = 0 ~ lies Problem 12 (Triple cross products). Show that, apart from in some special cases, the vector (~u ~v) w ~ ) lies in the plane of ~v and w ~ . What are the special cases? This in the plane of ~u and ~v, whereas ~u (~v w means, in particular, that cross products are not associative (i.e. it matters where you put the parentheses!). 5 Homework Assignment 1 Multivariable functions and vector geometry 1 Computational Problems Problem 1 (Functions). f 1. Suppose f is a function defined by the formula f ( x , y , z , w )=x + yzw . Evaluate ( x , y , z , w )=( 4, 2 , 8 , 3 ) . whenever x=4, y= 2 , z= 8 , w=3 f ( 4, 2 , 8 ,3 ) =4 + 2 83=4+ 2 83 f ( 4, 2 , 8 , 3 ) =4 + 163=4 +43 f ( 4, 2 , 8 , 3 ) =5 g 2. Suppose is a function that returns a value 1 when all its parameters are integers and a value of 0 otherwise. What is about g ( g ( tan ( 30 ) ln ( 8 ) + ln (2 ) , 3 tan ( 30 ) , ln ( e ) + ln ( e/4 ) 3 tan ( 60 ) ln ( 8 )+ ln ( 2 ) , 3 tan ( 60 ) , ln ( e )+ ln ( 4 /e ) 3 ) ? Note that there could not possibly a formula for this function ( ( )) tan ( 30 ) ln ( 8 ) + ln ( 2 ) g , 3 tan ( 30 ) , =g 3 e ln ( e ) + ln 4 ( g ( 2 ) ( g ! ) 3 3 3 3 ln ( 2 )+ ln ( 2 ) , 3 , = 3 1+ln ( e )ln ( 4 ) 3 3 1 , ( 3 ) , 3 ln ( 2 ) + ln ( 2 ) =g 1 , 3 , 4 ln (2 ) =g 1 ,1, 4 ln ( 2 ) = 3 3 3 3 3 22 ln ( 2 ) 3 2 [ 1ln ( 2 ) ] 1+1ln ( 22 ) ) ( ) ( 4 ( 0.693147181 ) 2.772588722 =g 0.333333333,1, = 2 ( 10.693147181 ) 2 ( 0.306852819 ) ( 2.772588722 =g ( 0.333333333,1, 4.517782707 ) 0.613705639 g 0.333333333,1, g 0.333333333,1, ) ( ) ) ) ? What g g ( tan ( 30 ) ln ( 8 ) + ln ( 2 ) , 3 tan ( 30 ) , =0 e 3 ln ( e ) + ln 4 ( tan ( 60 ) ln ( 8 )+ ln (2 ) , 3 tan ( 60 ) , =g 3 4 ln ( e )+ ln e () ( )) ( 3 ln ( 2 )+ ln ( 2 ) 2 g 1, ( 3 ) , g ( ) 2 1+ln ( 2 )1 ) ( =g 1,3, ( 3 3 , 3 3 , ln ( 2 ) +ln ( 2 ) = 1+ln ( 4 )ln ( e ) 3 ) 4 ln ( 2 ) =g ( 1, 3, 2 ) 2 ln ( 2 ) ) tan ( 60 ) ln ( 8 )+ ln (2 ) , 3 tan ( 60 ) , =g ( 1, 3,2 ) =1 4 3 ln ( e )+ ln e 3. Suppose () P ) is a function that only takes integers as parameters and returns a value of 1 if all parameters are even, -1 if all parameters are odd and 0 otherwise. What is the value of P ( P ( 1,0,2 ) +1,2 P ( 1,1,1 )P ( 2,2,0 ) , P ( 0,2,0 )1 ) ? 0 is an even number. P (1,0,2 ) =0 ; P ( 1,1,1 )=1; P ( 2,2,0 )=1 ; P ( 0,2,0 )=1 P ( P ( 1,0,2 ) +1,2 P ( 1,1,1 )P ( 2,2,0 ) , P ( 0,2,0 )1 )=P ( 0+1, 2 (1 )1, 21 ) = P (1,21, 1 )=P ( 1,3, 1 )=1 P ( P ( 1,0,2 ) +1,2 P ( 1,1,1 )P ( 2,2,0 ) , P ( 0,2,0 )1 )=1 Problem 2 (Contours). Draw three or four contours in the plane for each of the following functions. Don't forget to label each contour with its corresponding function value. 1. 2 f ( x , y ) =x +2 y 2 2. f ( x , y ) =sin ( x+ y ) 3. f ( x , y ) =e xy 4. f ( x , y ) =x e y y e x 5. f ( x , y ) =x2 + y 3 6. f ( x , y ) =2 x y 7. f ( x , y )= y3 8. f ( x , y ) =cos ( x )+ cos ( y ) Problem 3 (Level Surfaces) Match the following equations with the plotted surfaces (if this takes you a lot of time, you may wish to get some ideas from http://wolframalpha.com/ but just make sure you cite it in your bibliography if you use it - see \"Homework Guidelines\"): 1. 2. 3 x2 +2 y 2+ z2 =5 2 2 2 2 2 2 3x 2 y z 5 x y z + + = + + =1 5 5 5 5 5 5 5 3 2 y 2=z3 x 2 y 2+ z=3 x 2 z=3 x 2 y 2 z = 2 x 2 y 2 x2 y2 x2 y2 (1 ) z= z= + + =z ; z 0 1 1 1 1 1 1 3 3 3 ( 3. 4. ) x 2+ y 2 +3 z 2=5 2 x=z y 2 2 2 2 2 2 2 2 x y 1 1 3 x y 3z 5 x y z + + = + + =1 5 5 5 5 5 5 5 3 2 2 2 2 5. z z z =2 ( 1+ x 2 y 2) =1+ x 2 y 2 + y 2x 2=1 6. x+ y 2+ z 2=0 x= y 2z 2 (1 ) ( x= y 2z 2 ) x= y 2 + z 2 2 2 2 y + z =x , x 0 7. 2 x 22 y 2 +2 z2=1 2 x 2 +2 z 22 y 2=1 x2 z2 y2 + =1 1 1 1 2 2 2 8. x 2 3 z 2 10 x 2 z2 x =103 z x + 3 z =10 + = + =1 10 10 10 10 10 3 9. x= y 2z 2 2 2 2 2 z2 3 y 2 8 z 2 y 2 z +3 y =8 + = + =1 8 8 8 8 8 10. 3 2 2 2 2 2 11. x = y + z 2 2 2 2 2 2 2 2 2 2 2 2 12. x y =5 z x =5 z + y x +5 z = y y =x + 5 z Problem 4 (Dot product and projection). Find for the following pair of vectors: 1. v =2 i +4 j 6 k u =3 i + 4 j+ 6 k 2 v u =( 2 ) (3 )+ ( 4 )( 4 ) + ( 6 ) ( 6 )=6+16( 6 ) =6 +166 v u=4 (a) |v|= ( 2 )2+ ( 4 )2+ ( 6 ) = 4+16 +6 |v|= 26 2 |u|= (3 )2 + ( 4 )2 + ( 6 ) = 9+16+6 |u|= 31 2 (b) the cosine of the angle between v and u v u 4 4 4 = = cos = |v||u| ( 26 ) ( 31 ) 26 31 806 cos = v (c) the scalar component of v u u = 2 u 4 ( | | ) ( ( ) ) ( 31 2 ( 23 ) j ( 12 ) k u=i +6 j v = v u 4 = |u| 31 4 (3 i +4 j+ 6 k ) 3 i +4 j+ 6 k )= 31 ( 1231 ) i+( 1631 ) j +( 431 6 )k proju ( v )= 2. u= proju ( v ) . (d) the vector proju ( v )= in the direction of ( ) (a) ( 23 ) ( 6 )+( 12 ) ( 0) =0+ 123 0 v u=4 v u =( 0 )( 1 ) + |v|= 4 ( 4 )+ 9 (1 ) 2 2 1 2 4 1 16+9 25 25 + = 0+ + = = = = 3 2 9 4 36 36 36 36 ()( ) |v|= ( 0 )2 + 5 6 |u|= ( 1 ) + ( 6 ) + ( 0 ) = 1+ 36+0 |u|= 37 2 2 2 (b) the cosine of the angle between v and u v u 4 4 6 24 = = =4 cos = |v||u| 5 5 37 5 37 ( 37 ) 5 37 6 6 cos = () (c) the scalar component of in the direction of u = v u 4 = |u| 37 proju ( v ) . (d) the vector 3. v 4 4 v u ( i + 6 j )= ( i +6 j ) u = 2 2 37 |u| ( 37 ) proju ( v )= ( ) ( proju ( v )= ( 374 ) i+( 2437 ) j ) ( ) v = 2, (a) 1 u= 1,2 2 v u =( 2 ) ( 1 )+ |v|= ( 2 )2+ ( 12 ) ( 2)=2+ 22=2+ 1 v u =3 4 ( 4 )+ 1 1 2 1 16+1 17 17 17 = 4+ = = = = |v|= 2 4 4 4 4 2 4 () |u|= ( 1 ) + ( 2 ) = 1+ 4 |u|= 5 2 2 (b) the cosine of the angle between v u = |v||u| cos = and u 3 3 3 2 6 = = =3 cos = 85 85 17 ( 5 ) 17 5 85 2 2 2 ( ) (c) the scalar component of (d) the vector v v in the direction of u= v u 3 = |u| 5 proju ( v ) . ( ) ( ) v u 3 u = 1,2 = 3 1,2 = 3 ( 1 ) , 3 ( 2 ) 2 2 5 5 5 |u| ( 5) proju ( v )= ( ) ( ) proju ( v )= () 3 6 , 5 5 Problem 5 (Angle between vectors). 1. Find the angles between the following pairs of vectors: (a) u=i +2 j k v =2 i + j u v =( 1 ) ( 2 )+ ( 2 )( 1 ) + (1 )( 0 )=2+20 u v =4 |u|= ( 1 ) + ( 2 ) + (1 ) = 1+ 4+ 1|u|= 6 2 2 2 |v|= ( 2 )2+ ( 1 )2+ ( 0 )2= 4+ 1+ 0 |v|= 5 u v 4 4 4 4 = = = = cos =0.730296743 5.477225575 |u||v| ( 6 ) ( 5 ) 6 5 30 cos = =cos1 ( 0.730296743 ) =43.08872313 43.09 (b) u=3 i +4 j v =2 i 2 j + k u v =( 3 ) ( 2 ) + ( 4 )(2 ) + ( 0 ) ( 1 )=68+ 0 u v =2 |u|= ( 3 ) + ( 4 ) + ( 0 ) = 9+16+0= 25 |u|=5 2 2 2 |v|= ( 2 ) + (2 ) + ( 1 ) = 4 +4 +1= 9 |v|=3 2 2 2 u v 2 2 2 = = = cos =0.133333333 |u||v| ( 5 ) ( 3 ) 15 15 cos = 1 =cos (0.133333333 ) =97.66225566 97.66 (c) u= 7 i +2 j k v = 7 i j 2 u v =( 7 ) ( 7 ) + ( 2 ) (1 ) + (1 ) ( 0 )=( 7 ) 20=72 u v =5 |u|= ( 7 ) + ( 2 )2+ (1 )2= 7+ 4+1=12= 4 3= 4 3|u|=2 3 2 |v|= ( 7 ) + (1 )2 + ( 0 )2= 7+1+0= 8= 4 2= 4 2|v|=2 2 2 u v 5 5 5 5 5 = = = = = |u||v| ( 2 3 )( 2 2 ) 4 3 2 4 6 4 ( 2.449489743 ) 9.797958971 cos = 1 cos =0.510310363 =cos ( 0.510310363 ) =59.31549491 59.32 2. Using what you have learnt in class, compute the angles of the triangle in the plane whose vertices are P=( 2,1 ) , Q=( 1,2 ) and R=(1,0 ) . Graph: The angle in vertex R RP RQ cos = |RP||RQ| ( ) is the angle between the vectors RP and RQ , then: RP=PR=( 2,1 )(1,0 )= 2(1 ) , 10 = 2+1,1 RP= 3,1 | RP|= ( 3 ) + ( 1 ) = 9+1 | RP|= 10 2 2 RQ=QR=( 1,2 ) (1,0 ) = 1(1 ) ,20 = 1+1,2 RQ= 2,2 | RQ|= ( 2 ) + (2 ) = 4+ 4 | RQ|= 8 2 2 RP RQ= 3,1 2,2 =( 3 ) ( 2 )+ (1 )(2 )=62 RP RQ=4 cos = RP RQ 4 4 4 4 4 4 = = = = = = | RP|| RQ| 10 8 10 8 5 2 8 5 16 5 16 5 ( 4 ) cos = 1 1 = cos =0.447213596 =cos1 ( 0.447213596 ) 2.236067978 5 =63.43494882 63.435 The angle in vertex P ( ) is the angle between the vectors P R P R PQ cos = |P R||P Q| and PQ , then: PR= RP= 3,1 PR= 3,1 | PR|=| RP|| PR|= 10 PQ=QP=( 1,2 ) (2 , 1 )= 12 ,21 P Q= 1,3 | P Q|= (1 ) + (3 ) = 1+9 | P Q|= 10 2 2 P R PQ=3,1 1 ,3 =(3 )(1 )+ (1 ) (3 )=3+3 P R P Q=6 cos = P R PQ 6 6 6 3 = = = cos = =0.6 =cos1 ( 0. 6 ) 2 5 | P R|| P Q| 10 10 ( 10 ) 10 =53.13010235 5 3.130 The angle in vertex Q QR QP cos = | Q R|| Q P| QR ( ) is the angle between the vectors and Q R= RQ= 2,2 = 2,(2 ) Q R=2 , 2 | QR|=| RQ|| Q R|= 8 Q P= PQ=1 ,3 = (1 ) ,(3 ) Q P= 1 , 3 | Q P|=| P Q|| QP|= 10 Q R Q P=2 , 2 1 ,3 =(2 ) ( 1 )+ (2 )( 3 ) =2+6 Q R Q P=4 Q R QP 4 4 4 4 4 4 = = = = = = |Q R||Q P| 8 10 8 10 8 2 5 16 5 16 5 4 5 cos = Q P , then: cos = 1 1 = cos =0.447213596 =cos1 ( 0.447213596 ) 2.236067978 5 =63.43494882 63.435 To check, the sum of the internal angles of any triangle must be equal to 180: + +=63.435 + 53.130 +63.435 + +=180 Ok 3. Similarly, compute the angles between the diagonals of the rectangle in the plane whose vertices are A= (1,0 ) , B=( 0,3 ) , C=( 3,4 ) and D=( 4,1 ) . Graph: and . The angle is the angle between the vectors DB AC cos = | DB|| AC | DB The angles between the diagonals are DB=BD=( 0 ,3 )( 4 ,1 )= 04 , 31 DB=4 , 2 and AC , then: | DB|= (4 ) + ( 2 ) = 16 +4= 20= 4 5= 4 5 | DB|=2 5 2 2 AC=C A=( 3 , 4 )( 1,0 )= 31, 40 AC= 2, 4 | AC|= ( 2 ) + ( 4 ) = 4+16= 20= 4 5= 4 5 | AC|=2 5 2 2 DB AC= 4 , 2 2, 4 =(4 ) ( 2 ) + ( 2 )( 4 )=8+8 DB AC=0 cos = DB AC 0 0 0 0 = = 2 = = cos =0 2 | DB|| AC | ( 2 5 ) ( 2 5 ) ( 2 ) ( 5 ) ( 4 ) ( 5 ) 20 =cos1 ( 0 ) =90 The angle is the angle between the vectors B D AC cos = | B D|| AC| BD and AC , then: B D= DB=4,2 =(4 ) ,2 BD= 4 ,2 | BD|=| DB|| BD|=2 5 AC = 2,4 ;| AC|=2 5 B D AC = 4 ,2 2,4 = ( 4 )( 2 ) + (2 )( 4 ) =88 B D AC=0 cos = B D AC 0 0 0 0 = = 2 = = cos =0 2 | B D|| AC| ( 2 5 ) ( 2 5 ) ( 2 ) ( 5 ) ( 4 ) ( 5 ) 20 =cos1 ( 0 ) =90 To check, and are adjacent angle, then their sum must be equal to 180: + =90 +90 + =180 Ok Homework Assignment 1 Multivariable functions and vector geometry 1 Computational Problems Problem 1 (Functions). f 1. Suppose f is a function defined by the formula f ( x , y , z , w )=x + yzw . Evaluate ( x , y , z , w )=( 4, 2 , 8 , 3 ) . whenever x=4, y= 2 , z= 8 , w=3 f ( 4, 2 , 8 ,3 ) =4 + 2 83=4+ 2 83 f ( 4, 2 , 8 , 3 ) =4 + 163=4 +43 f ( 4, 2 , 8 , 3 ) =5 g 2. Suppose is a function that returns a value 1 when all its parameters are integers and a value of 0 otherwise. What is about g ( g ( tan ( 30 ) ln ( 8 ) + ln (2 ) , 3 tan ( 30 ) , ln ( e ) + ln ( e/4 ) 3 tan ( 60 ) ln ( 8 )+ ln ( 2 ) , 3 tan ( 60 ) , ln ( e )+ ln ( 4 /e ) 3 ) ? Note that there could not possibly a formula for this function ( ( )) tan ( 30 ) ln ( 8 ) + ln ( 2 ) g , 3 tan ( 30 ) , =g 3 e ln ( e ) + ln 4 ( g ( 2 ) ( g ! ) 3 3 3 3 ln ( 2 )+ ln ( 2 ) , 3 , = 3 1+ln ( e )ln ( 4 ) 3 3 1 , ( 3 ) , 3 ln ( 2 ) + ln ( 2 ) =g 1 , 3 , 4 ln (2 ) =g 1 ,1, 4 ln ( 2 ) = 3 3 3 3 3 22 ln ( 2 ) 3 2 [ 1ln ( 2 ) ] 1+1ln ( 22 ) ) ( ) ( 4 ( 0.693147181 ) 2.772588722 =g 0.333333333,1, = 2 ( 10.693147181 ) 2 ( 0.306852819 ) ( 2.772588722 =g ( 0.333333333,1, 4.517782707 ) 0.613705639 g 0.333333333,1, g 0.333333333,1, ) ( ) ) ) ? What g g ( tan ( 30 ) ln ( 8 ) + ln ( 2 ) , 3 tan ( 30 ) , =0 e 3 ln ( e ) + ln 4 ( tan ( 60 ) ln ( 8 )+ ln (2 ) , 3 tan ( 60 ) , =g 3 4 ln ( e )+ ln e () ( )) ( 3 ln ( 2 )+ ln ( 2 ) 2 g 1, ( 3 ) , g ( ) 2 1+ln ( 2 )1 ) ( =g 1,3, ( 3 3 , 3 3 , ln ( 2 ) +ln ( 2 ) = 1+ln ( 4 )ln ( e ) 3 ) 4 ln ( 2 ) =g ( 1, 3, 2 ) 2 ln ( 2 ) ) tan ( 60 ) ln ( 8 )+ ln (2 ) , 3 tan ( 60 ) , =g ( 1, 3,2 ) =1 4 3 ln ( e )+ ln e 3. Suppose () P ) is a function that only takes integers as parameters and returns a value of 1 if all parameters are even, -1 if all parameters are odd and 0 otherwise. What is the value of P ( P ( 1,0,2 ) +1,2 P ( 1,1,1 )P ( 2,2,0 ) , P ( 0,2,0 )1 ) ? 0 is an even number. P (1,0,2 ) =0 ; P ( 1,1,1 )=1; P ( 2,2,0 )=1 ; P ( 0,2,0 )=1 P ( P ( 1,0,2 ) +1,2 P ( 1,1,1 )P ( 2,2,0 ) , P ( 0,2,0 )1 )=P ( 0+1, 2 (1 )1, 21 ) = P (1,21, 1 )=P ( 1,3, 1 )=1 P ( P ( 1,0,2 ) +1,2 P ( 1,1,1 )P ( 2,2,0 ) , P ( 0,2,0 )1 )=1 Problem 2 (Contours). Draw three or four contours in the plane for each of the following functions. Don't forget to label each contour with its corresponding function value. 1. 2 f ( x , y ) =x +2 y 2 2. f ( x , y ) =sin ( x+ y ) 3. f ( x , y ) =e xy 4. f ( x , y ) =x e y y e x 5. f ( x , y ) =x2 + y 3 6. f ( x , y ) =2 x y 7. f ( x , y )= y3 8. f ( x , y ) =cos ( x )+ cos ( y ) Problem 3 (Level Surfaces) Match the following equations with the plotted surfaces (if this takes you a lot of time, you may wish to get some ideas from http://wolframalpha.com/ but just make sure you cite it in your bibliography if you use it - see \"Homework Guidelines\"): 1. 2. 3 x2 +2 y 2+ z2 =5 2 2 2 2 2 2 3x 2 y z 5 x y z + + = + + =1 5 5 5 5 5 5 5 3 2 y 2=z3 x 2 y 2+ z=3 x 2 z=3 x 2 y 2 z = 2 x 2 y 2 x2 y2 x2 y2 (1 ) z= z= + + =z ; z 0 1 1 1 1 1 1 3 3 3 ( 3. 4. ) x 2+ y 2 +3 z 2=5 2 x=z y 2 2 2 2 2 2 2 2 x y 1 1 3 x y 3z 5 x y z + + = + + =1 5 5 5 5 5 5 5 3 2 2 2 2 5. z z z =2 ( 1+ x 2 y 2) =1+ x 2 y 2 + y 2x 2=1 6. x+ y 2+ z 2=0 x= y 2z 2 (1 ) ( x= y 2z 2 ) x= y 2 + z 2 2 2 2 y + z =x , x 0 7. 2 x 22 y 2 +2 z2=1 2 x 2 +2 z 22 y 2=1 x2 z2 y2 + =1 1 1 1 2 2 2 8. x 2 3 z 2 10 x 2 z2 x =103 z x + 3 z =10 + = + =1 10 10 10 10 10 3 9. x= y 2z 2 2 2 2 2 z2 3 y 2 8 z 2 y 2 z +3 y =8 + = + =1 8 8 8 8 8 10. 3 2 2 2 2 2 11. x = y + z 2 2 2 2 2 2 2 2 2 2 2 2 12. x y =5 z x =5 z + y x +5 z = y y =x + 5 z Problem 4 (Dot product and projection). Find for the following pair of vectors: 1. v =2 i +4 j 6 k u =3 i + 4 j+ 6 k 2 v u =( 2 ) (3 )+ ( 4 )( 4 ) + ( 6 ) ( 6 )=6+16( 6 ) =6 +166 v u=4 (a) |v|= ( 2 )2+ ( 4 )2+ ( 6 ) = 4+16 +6 |v|= 26 2 |u|= (3 )2 + ( 4 )2 + ( 6 ) = 9+16+6 |u|= 31 2 (b) the cosine of the angle between v and u v u 4 4 4 = = cos = |v||u| ( 26 ) ( 31 ) 26 31 806 cos = v (c) the scalar component of v u u = 2 u 4 ( | | ) ( ( ) ) ( 31 2 ( 23 ) j ( 12 ) k u=i +6 j v = v u 4 = |u| 31 4 (3 i +4 j+ 6 k ) 3 i +4 j+ 6 k )= 31 ( 1231 ) i+( 1631 ) j +( 431 6 )k proju ( v )= 2. u= proju ( v ) . (d) the vector proju ( v )= in the direction of ( ) (a) ( 23 ) ( 6 )+( 12 ) ( 0) =0+ 123 0 v u=4 v u =( 0 )( 1 ) + |v|= 4 ( 4 )+ 9 (1 ) 2 2 1 2 4 1 16+9 25 25 + = 0+ + = = = = 3 2 9 4 36 36 36 36 ()( ) |v|= ( 0 )2 + 5 6 |u|= ( 1 ) + ( 6 ) + ( 0 ) = 1+ 36+0 |u|= 37 2 2 2 (b) the cosine of the angle between v and u v u 4 4 6 24 = = =4 cos = |v||u| 5 5 37 5 37 ( 37 ) 5 37 6 6 cos = () (c) the scalar component of in the direction of u = v u 4 = |u| 37 proju ( v ) . (d) the vector 3. v 4 4 v u ( i + 6 j )= ( i +6 j ) u = 2 2 37 |u| ( 37 ) proju ( v )= ( ) ( proju ( v )= ( 374 ) i+( 2437 ) j ) ( ) v = 2, (a) 1 u= 1,2 2 v u =( 2 ) ( 1 )+ |v|= ( 2 )2+ ( 12 ) ( 2)=2+ 22=2+ 1 v u =3 4 ( 4 )+ 1 1 2 1 16+1 17 17 17 = 4+ = = = = |v|= 2 4 4 4 4 2 4 () |u|= ( 1 ) + ( 2 ) = 1+ 4 |u|= 5 2 2 (b) the cosine of the angle between v u = |v||u| cos = and u 3 3 3 2 6 = = =3 cos = 85 85 17 ( 5 ) 17 5 85 2 2 2 ( ) (c) the scalar component of (d) the vector v v in the direction of u= v u 3 = |u| 5 proju ( v ) . ( ) ( ) v u 3 u = 1,2 = 3 1,2 = 3 ( 1 ) , 3 ( 2 ) 2 2 5 5 5 |u| ( 5) proju ( v )= ( ) ( ) proju ( v )= () 3 6 , 5 5 Problem 5 (Angle between vectors). 1. Find the angles between the following pairs of vectors: (a) u=i +2 j k v =2 i + j u v =( 1 ) ( 2 )+ ( 2 )( 1 ) + (1 )( 0 )=2+20 u v =4 |u|= ( 1 ) + ( 2 ) + (1 ) = 1+ 4+ 1|u|= 6 2 2 2 |v|= ( 2 )2+ ( 1 )2+ ( 0 )2= 4+ 1+ 0 |v|= 5 u v 4 4 4 4 = = = = cos =0.730296743 5.477225575 |u||v| ( 6 ) ( 5 ) 6 5 30 cos = =cos1 ( 0.730296743 ) =43.08872313 43.09 (b) u=3 i +4 j v =2 i 2 j + k u v =( 3 ) ( 2 ) + ( 4 )(2 ) + ( 0 ) ( 1 )=68+ 0 u v =2 |u|= ( 3 ) + ( 4 ) + ( 0 ) = 9+16+0= 25 |u|=5 2 2 2 |v|= ( 2 ) + (2 ) + ( 1 ) = 4 +4 +1= 9 |v|=3 2 2 2 u v 2 2 2 = = = cos =0.133333333 |u||v| ( 5 ) ( 3 ) 15 15 cos = 1 =cos (0.133333333 ) =97.66225566 97.66 (c) u= 7 i +2 j k v = 7 i j 2 u v =( 7 ) ( 7 ) + ( 2 ) (1 ) + (1 ) ( 0 )=( 7 ) 20=72 u v =5 |u|= ( 7 ) + ( 2 )2+ (1 )2= 7+ 4+1=12= 4 3= 4 3|u|=2 3 2 |v|= ( 7 ) + (1 )2 + ( 0 )2= 7+1+0= 8= 4 2= 4 2|v|=2 2 2 u v 5 5 5 5 5 = = = = = |u||v| ( 2 3 )( 2 2 ) 4 3 2 4 6 4 ( 2.449489743 ) 9.797958971 cos = 1 cos =0.510310363 =cos ( 0.510310363 ) =59.31549491 59.32 2. Using what you have learnt in class, compute the angles of the triangle in the plane whose vertices are P=( 2,1 ) , Q=( 1,2 ) and R=(1,0 ) . Graph: The angle in vertex R RP RQ cos = |RP||RQ| ( ) is the angle between the vectors RP and RQ , then: RP=PR=( 2,1 )(1,0 )= 2(1 ) , 10 = 2+1,1 RP= 3,1 | RP|= ( 3 ) + ( 1 ) = 9+1 | RP|= 10 2 2 RQ=QR=( 1,2 ) (1,0 ) = 1(1 ) ,20 = 1+1,2 RQ= 2,2 | RQ|= ( 2 ) + (2 ) = 4+ 4 | RQ|= 8 2 2 RP RQ= 3,1 2,2 =( 3 ) ( 2 )+ (1 )(2 )=62 RP RQ=4 cos = RP RQ 4 4 4 4 4 4 = = = = = = | RP|| RQ| 10 8 10 8 5 2 8 5 16 5 16 5 ( 4 ) cos = 1 1 = cos =0.447213596 =cos1 ( 0.447213596 ) 2.236067978 5 =63.43494882 63.435 The angle in vertex P ( ) is the angle between the vectors P R P R PQ cos = |P R||P Q| and PQ , then: PR= RP= 3,1 PR= 3,1 | PR|=| RP|| PR|= 10 PQ=QP=( 1,2 ) (2 , 1 )= 12 ,21 P Q= 1,3 | P Q|= (1 ) + (3 ) = 1+9 | P Q|= 10 2 2 P R PQ=3,1 1 ,3 =(3 )(1 )+ (1 ) (3 )=3+3 P R P Q=6 cos = P R PQ 6 6 6 3 = = = cos = =0.6 =cos1 ( 0. 6 ) 2 5 | P R|| P Q| 10 10 ( 10 ) 10 =53.13010235 5 3.130 The angle in vertex Q QR QP cos = | Q R|| Q P| QR ( ) is the angle between the vectors and Q R= RQ= 2,2 = 2,(2 ) Q R=2 , 2 | QR|=| RQ|| Q R|= 8 Q P= PQ=1 ,3 = (1 ) ,(3 ) Q P= 1 , 3 | Q P|=| P Q|| QP|= 10 Q R Q P=2 , 2 1 ,3 =(2 ) ( 1 )+ (2 )( 3 ) =2+6 Q R Q P=4 Q R QP 4 4 4 4 4 4 = = = = = = |Q R||Q P| 8 10 8 10 8 2 5 16 5 16 5 4 5 cos = Q P , then: cos = 1 1 = cos =0.447213596 =cos1 ( 0.447213596 ) 2.236067978 5 =63.43494882 63.435 To check, the sum of the internal angles of any triangle must be equal to 180: + +=63.435 + 53.130 +63.435 + + =180 Ok 3. Similarly, compute the angles between the diagonals of the rectangle in the plane whose vertices are A= (1,0 ) , B=( 0,3 ) , C=( 3,4 ) and D=( 4,1 ) . Graph: and . The angle is the angle between the vectors DB AC cos = | DB|| AC | DB The angles between the diagonals are DB=BD=( 0,3 )( 4,1 )= 04,31 DB= 4,2 and AC , then: | DB|= (4 ) + ( 2 ) = 16 +4= 20= 4 5= 4 5 | DB|=2 5 2 2 AC=C A=( 3,4 ) ( 1,0 )= 31,40 AC= 2,4 | AC|= ( 2 ) + ( 4 ) = 4+16= 20= 4 5= 4 5 | AC|=2 5 2 2 DB AC= 4,2 2,4 =(4 ) ( 2 )+ ( 2 )( 4 )=8+ 8 DB AC=0 cos = DB AC 0 0 0 0 = = 2 = = cos =0 2 | DB|| AC | ( 2 5 ) ( 2 5 ) ( 2 ) ( 5 ) ( 4 ) ( 5 ) 20 =cos1 ( 0 ) =90 The angle is the angle between the vectors B D AC cos = | B D|| AC| BD and AC , then: B D= DB=4,2 =(4 ) ,2 BD= 4,2 | BD|=| DB|| BD|=2 5 AC = 2,4 ;| AC|=2 5 B D AC = 4 ,2 2,4 = ( 4 )( 2 ) + (2 )( 4 ) =88 B D AC=0 cos = B D AC 0 0 0 0 = = 2 = = cos =0 2 | B D|| AC| ( 2 5 ) ( 2 5 ) ( 2 ) ( 5 ) ( 4 ) ( 5 ) 20 =cos1 ( 0 ) =90 To check, and are adjacent angle, then their sum must be equal to 180: + =90 +90 + =180 Ok Problem 6 (Cross Product). Compute the cross products of the following pair of vectors: 1. u=3 i 2 j+ k v =i + j 2 k | || j i k 1 3 2 2 1 i 3 u v = 3 2 j+ k 1 = 1 2 1 2 1 1 1 1 2 || || | u v =[ (2 ) ( 2 )( 1 ) ( 1 ) ] i [ ( 3 ) ( 2 )( 1 ) ( 1 ) ] j+ [ ( 3 )( 1 ) ( 2 ) ( 1 ) ] k 2 (3 21 ) j+ ( 3+ 2 ) k u v =[ ( 2 ) 1 ] i u v =( 21 ) i + ( 3 2+1 ) j + ( 3+ 2 ) k u v =i + ( 3 2+1 ) j + ( 3+ 2 ) k 2. u= 1 , 1,3 v = 3,2,2 3 | || j k i 1 1 1 3 u v = = i 1 3 3 2 2 3 3 3 2 2 | | || | [( ) [( ) 1 3 1 j+ 3 k 2 3 2 1 1 ( 2 ) ( 3 )( 3 ) j+ (2 )( 1 ) ( 3 ) k u v =[ ( 1 )( 2 ) ( 3 )(2 ) ] i 3 3 23 ( 9 ) 2+3 ( 3 ) 2 2 3 k =8 i j k u v =( 2+6 ) i 9 j + 3 3 3 3 ( ) ( ) ( ) ( j 2+9 k =8 i 25 j 11 k ( 227 ( 3 ) (3) 3 ) ( 3 ) u v =8 i 25 11 u v =8 i + j k 3 3 ( ) ( ) 3. u= 1 ,5 , () () 4 7 1 v = i+ jk 7 4 2 ) | || | | | | i u v = 1 7 4 j k 4 5 1 4 5 7 i 7 = 1 7 1 1 1 2 4 2 [ u v = (5 ) (1 ) | ( 47 )( 12 )] i[( 1) (1) ( 47 )( 74 )] j +[ ( 1)( 12 )(5) ( 74 )] k 2 ( 1 ) +35 2 1 35 7 ( 5 ) 2 i (11 ) j+ + k= i (2 ) j + k 7 2 4 7 4 ( ) u v = 5 4 1 5 7 j+ 7 1 k 1 4 2 ) ( ( ) ( ) 2+35 33 37 i +2 j+ ( k u v =( )i +2 j+( ) k ( 352 ) ) 7 4 7 4 u v = 4. 1 1 j+ k v = 1,3,1 u=i 3 2 ( ) ( ) | i j 1 u v = 1 3 1 3 || k 1 1 = 3 2 3 1 || || | 1 1 1 1 1 2 i 2 j+ 3 k 1 1 1 1 3 [( ) ( ) ] [ u v = ( ) ( 13 32 )i(1 12 ) j+(3+ 13 ) k u v = 3 i 21 j+ 3 3+1 k ( 23 3 2 ) ( 2 ) ( 3 ) u v = ( u v = Problem 7 (Triple [ 1 1 1 1 (1 ) (3 ) i (1 )(1 ) ( 1 ) j+ (1 )( 3 ) (1 ) k 3 2 2 3 23 3 i 21 j 3 31 k 6 2 3 ) ( ) ( ) ( ) Homework Assignment 1 Multivariable functions and vector geometry 1 Computational Problems Problem 1 (Functions). f 1. Suppose f is a function defined by the formula f ( x , y , z , w )=x + yzw . Evaluate ( x , y , z , w )=( 4, 2 , 8 , 3 ) . whenever x=4, y= 2 , z= 8 , w=3 f ( 4, 2 , 8 ,3 ) =4 + 2 83=4+ 2 83 f ( 4, 2 , 8 , 3 ) =4 + 163=4 +43 f ( 4, 2 , 8 , 3 ) =5 g 2. Suppose is a function that returns a value 1 when all its parameters are integers and a value of 0 otherwise. What is about g ( g ( tan ( 30 ) ln ( 8 ) + ln (2 ) , 3 tan ( 30 ) , ln ( e ) + ln ( e/4 ) 3 tan ( 60 ) ln ( 8 )+ ln ( 2 ) , 3 tan ( 60 ) , ln ( e )+ ln ( 4 /e ) 3 ) ? Note that there could not possibly a formula for this function ( ( )) tan ( 30 ) ln ( 8 ) + ln ( 2 ) g , 3 tan ( 30 ) , =g 3 e ln ( e ) + ln 4 ( g ( 2 ) ( g ! ) 3 3 3 3 ln ( 2 )+ ln ( 2 ) , 3 , = 3 1+ln ( e )ln ( 4 ) 3 3 1 , ( 3 ) , 3 ln ( 2 ) + ln ( 2 ) =g 1 , 3 , 4 ln (2 ) =g 1 ,1, 4 ln ( 2 ) = 3 3 3 3 3 22 ln ( 2 ) 3 2 [ 1ln ( 2 ) ] 1+1ln ( 22 ) ) ( ) ( 4 ( 0.693147181 ) 2.772588722 =g 0.333333333,1, = 2 ( 10.693147181 ) 2 ( 0.306852819 ) ( 2.772588722 =g ( 0.333333333,1, 4.517782707 ) 0.613705639 g 0.333333333,1, g 0.333333333,1, ) ( ) ) ) ? What g g ( tan ( 30 ) ln ( 8 ) + ln ( 2 ) , 3 tan ( 30 ) , =0 e 3 ln ( e ) + ln 4 ( tan ( 60 ) ln ( 8 )+ ln (2 ) , 3 tan ( 60 ) , =g 3 4 ln ( e )+ ln e () ( )) ( 3 ln ( 2 )+ ln ( 2 ) 2 g 1, ( 3 ) , g ( ) 2 1+ln ( 2 )1 ) ( =g 1,3, ( 3 3 , 3 3 , ln ( 2 ) +ln ( 2 ) = 1+ln ( 4 )ln ( e ) 3 ) 4 ln ( 2 ) =g ( 1, 3, 2 ) 2 ln ( 2 ) ) tan ( 60 ) ln ( 8 )+ ln (2 ) , 3 tan ( 60 ) , =g ( 1, 3,2 ) =1 4 3 ln ( e )+ ln e 3. Suppose () P ) is a function that only takes integers as parameters and returns a value of 1 if all parameters are even, -1 if all parameters are odd and 0 otherwise. What is the value of P ( P ( 1,0,2 ) +1,2 P ( 1,1,1 )P ( 2,2,0 ) , P ( 0,2,0 )1 ) ? 0 is an even number. P (1,0,2 ) =0 ; P ( 1,1,1 )=1; P ( 2,2,0 )=1 ; P ( 0,2,0 )=1 P ( P ( 1,0,2 ) +1,2 P ( 1,1,1 )P ( 2,2,0 ) , P ( 0,2,0 )1 )=P ( 0+1, 2 (1 )1, 21 ) = P (1,21, 1 )=P ( 1,3, 1 )=1 P ( P ( 1,0,2 ) +1,2 P ( 1,1,1 )P ( 2,2,0 ) , P ( 0,2,0 )1 )=1 Problem 2 (Contours). Draw three or four contours in the plane for each of the following functions. Don't forget to label each contour with its corresponding function value. 1. 2 f ( x , y ) =x +2 y 2 2. f ( x , y ) =sin ( x+ y ) 3. f ( x , y ) =e xy 4. f ( x , y ) =x e y y e x 5. f ( x , y ) =x2 + y 3 6. f ( x , y ) =2 x y 7. f ( x , y )= y3 8. f ( x , y ) =cos ( x )+ cos ( y ) Problem 3 (Level Surfaces) Match the following equations with the plotted surfaces (if this takes you a lot of time, you may wish to get some ideas from http://wolframalpha.com/ but just make sure you cite it in your bibliography if you use it - see \"Homework Guidelines\"): 1. 2. 3 x2 +2 y 2+ z2 =5 2 2 2 2 2 2 3x 2 y z 5 x y z + + = + + =1 5 5 5 5 5 5 5 3 2 y 2=z3 x 2 y 2+ z=3 x 2 z=3 x 2 y 2 z = 2 x 2 y 2 x2 y2 x2 y2 (1 ) z= z= + + =z ; z 0 1 1 1 1 1 1 3 3 3 ( 3. 4. ) x 2+ y 2 +3 z 2=5 2 x=z y 2 2 2 2 2 2 2 2 x y 1 1 3 x y 3z 5 x y z + + = + + =1 5 5 5 5 5 5 5 3 2 2 2 2 5. z z z =2 ( 1+ x 2 y 2) =1+ x 2 y 2 + y 2x 2=1 6. x+ y 2+ z 2=0 x= y 2z 2 (1 ) ( x= y 2z 2 ) x= y 2 + z 2 2 2 2 y + z =x , x 0 7. 2 x 22 y 2 +2 z2=1 2 x 2 +2 z 22 y 2=1 x2 z2 y2 + =1 1 1 1 2 2 2 8. x 2 3 z 2 10 x 2 z2 x =103 z x + 3 z =10 + = + =1 10 10 10 10 10 3 9. x= y 2z 2 2 2 2 2 z2 3 y 2 8 z 2 y 2 z +3 y =8 + = + =1 8 8 8 8 8 10. 3 2 2 2 2 2 11. x = y + z 2 2 2 2 2 2 2 2 2 2 2 2 12. x y =5 z x =5 z + y x +5 z = y y =x + 5 z Problem 4 (Dot product and projection). Find for the following pair of vectors: 1. v =2 i +4 j 6 k u =3 i + 4 j+ 6 k 2 v u =( 2 ) (3 )+ ( 4 )( 4 ) + ( 6 ) ( 6 )=6+16( 6 ) =6 +166 v u=4 (a) |v|= ( 2 )2+ ( 4 )2+ ( 6 ) = 4+16 +6 |v|= 26 2 |u|= (3 )2 + ( 4 )2 + ( 6 ) = 9+16+6 |u|= 31 2 (b) the cosine of the angle between v and u v u 4 4 4 = = cos = |v||u| ( 26 ) ( 31 ) 26 31 806 cos = v (c) the scalar component of v u u = 2 u 4 ( | | ) ( ( ) ) ( 31 2 ( 23 ) j ( 12 ) k u=i +6 j v = v u 4 = |u| 31 4 (3 i +4 j+ 6 k ) 3 i +4 j+ 6 k )= 31 ( 1231 ) i+( 1631 ) j +( 431 6 )k proju ( v )= 2. u= proju ( v ) . (d) the vector proju ( v )= in the direction of ( ) (a) ( 23 ) ( 6 )+( 12 ) ( 0) =0+ 123 0 v u=4 v u =( 0 )( 1 ) + |v|= 4 ( 4 )+ 9 (1 ) 2 2 1 2 4 1 16+9 25 25 + = 0+ + = = = = 3 2 9 4 36 36 36 36 ()( ) |v|= ( 0 )2 + 5 6 |u|= ( 1 ) + ( 6 ) + ( 0 ) = 1+ 36+0 |u|= 37 2 2 2 (b) the cosine of the angle between v and u v u 4 4 6 24 = = =4 cos = |v||u| 5 5 37 5 37 ( 37 ) 5 37 6 6 cos = () (c) the scalar component of in the direction of u = v u 4 = |u| 37 proju ( v ) . (d) the vector 3. v 4 4 v u ( i + 6 j )= ( i +6 j ) u = 2 2 37 |u| ( 37 ) proju ( v )= ( ) ( proju ( v )= ( 374 ) i+( 2437 ) j ) ( ) v = 2, (a) 1 u= 1,2 2 v u =( 2 ) ( 1 )+ |v|= ( 2 )2+ ( 12 ) ( 2)=2+ 22=2+ 1 v u =3 4 ( 4 )+ 1 1 2 1 16+1 17 17 17 = 4+ = = = = |v|= 2 4 4 4 4 2 4 () |u|= ( 1 ) + ( 2 ) = 1+ 4 |u|= 5 2 2 (b) the cosine of the angle between v u = |v||u| cos = and u 3 3 3 2 6 = = =3 cos = 85 85 17 ( 5 ) 17 5 85 2 2 2 ( ) (c) the scalar component of (d) the vector v v in the direction of u= v u 3 = |u| 5 proju ( v ) . ( ) ( ) v u 3 u = 1,2 = 3 1,2 = 3 ( 1 ) , 3 ( 2 ) 2 2 5 5 5 |u| ( 5) proju ( v )= ( ) ( ) proju ( v )= () 3 6 , 5 5 Problem 5 (Angle between vectors). 1. Find the angles between the following pairs of vectors: (a) u=i +2 j k v =2 i + j u v =( 1 ) ( 2 )+ ( 2 )( 1 ) + (1 )( 0 )=2+20 u v =4 |u|= ( 1 ) + ( 2 ) + (1 ) = 1+ 4+ 1|u|= 6 2 2 2 |v|= ( 2 )2+ ( 1 )2+ ( 0 )2= 4+ 1+ 0 |v|= 5 u v 4 4 4 4 = = = = cos =0.730296743 5.477225575 |u||v| ( 6 ) ( 5 ) 6 5 30 cos = =cos1 ( 0.730296743 ) =43.08872313 43.09 (b) u=3 i +4 j v =2 i 2 j + k u v =( 3 ) ( 2 ) + ( 4 )(2 ) + ( 0 ) ( 1 )=68+ 0 u v =2 |u|= ( 3 ) + ( 4 ) + ( 0 ) = 9+16+0= 25 |u|=5 2 2 2 |v|= ( 2 ) + (2 ) + ( 1 ) = 4 +4 +1= 9 |v|=3 2 2 2 u v 2 2 2 = = = cos =0.133333333 |u||v| ( 5 ) ( 3 ) 15 15 cos = 1 =cos (0.133333333 ) =97.66225566 97.66 (c) u= 7 i +2 j k v = 7 i j 2 u v =( 7 ) ( 7 ) + ( 2 ) (1 ) + (1 ) ( 0 )=( 7 ) 20=72 u v =5 |u|= ( 7 ) + ( 2 )2+ (1 )2= 7+ 4+1=12= 4 3= 4 3|u|=2 3 2 |v|= ( 7 ) + (1 )2 + ( 0 )2= 7+1+0= 8= 4 2= 4 2|v|=2 2 2 u v 5 5 5 5 5 = = = = = |u||v| ( 2 3 )( 2 2 ) 4 3 2 4 6 4 ( 2.449489743 ) 9.797958971 cos = 1 cos =0.510310363 =cos ( 0.510310363 ) =59.31549491 59.32 2. Using what you have learnt in class, compute the angles of the triangle in the plane whose vertices are P=( 2,1 ) , Q=( 1,2 ) and R=(1,0 ) . Graph: The angle in vertex R RP RQ cos = |RP||RQ| ( ) is the angle between the vectors RP and RQ , then: RP=PR=( 2,1 )(1,0 )= 2(1 ) , 10 = 2+1,1 RP= 3,1 | RP|= ( 3 ) + ( 1 ) = 9+1 | RP|= 10 2 2 RQ=QR=( 1,2 ) (1,0 ) = 1(1 ) ,20 = 1+1,2 RQ= 2,2 | RQ|= ( 2 ) + (2 ) = 4+ 4 | RQ|= 8 2 2 RP RQ= 3,1 2,2 =( 3 ) ( 2 )+ (1 )(2 )=62 RP RQ=4 cos = RP RQ 4 4 4 4 4 4 = = = = = = | RP|| RQ| 10 8 10 8 5 2 8 5 16 5 16 5 ( 4 ) cos = 1 1 = cos =0.447213596 =cos1 ( 0.447213596 ) 2.236067978 5 =63.43494882 63.435 The angle in vertex P ( ) is the angle between the vectors P R P R PQ cos = |P R||P Q| and PQ , then: PR= RP= 3,1 PR= 3,1 | PR|=| RP|| PR|= 10 PQ=QP=( 1,2 ) (2 , 1 )= 12 ,21 P Q= 1,3 | P Q|= (1 ) + (3 ) = 1+9 | P Q|= 10 2 2 P R PQ=3,1 1 ,3 =(3 )(1 )+ (1 ) (3 )=3+3 P R P Q=6 cos = P R PQ 6 6 6 3 = = = cos = =0.6 =cos1 ( 0. 6 ) 2 5 | P R|| P Q| 10 10 ( 10 ) 10 =53.13010235 5 3.130 The angle in vertex Q QR QP cos = | Q R|| Q P| QR ( ) is the angle between the vectors and Q R= RQ= 2,2 = 2,(2 ) Q R=2 , 2 | QR|=| RQ|| Q R|= 8 Q P= PQ=1 ,3 = (1 ) ,(3 ) Q P= 1 , 3 | Q P|=| P Q|| QP|= 10 Q R Q P=2 , 2 1 ,3 =(2 ) ( 1 )+ (2 )( 3 ) =2+6 Q R Q P=4 Q R QP 4 4 4 4 4 4 = = = = = = |Q R||Q P| 8 10 8 10 8 2 5 16 5 16 5 4 5 cos = Q P , then: cos = 1 1 = cos =0.447213596 =cos1 ( 0.447213596 ) 2.236067978 5 =63.43494882 63.435 To check, the sum of the internal angles of any triangle must be equal to 180: + +=63.435 + 53.130 +63.435 + + =180 Ok 3. Similarly, compute the angles between the diagonals of the rectangle in the plane whose vertices are A= (1,0 ) , B=( 0,3 ) , C=( 3,4 ) and D=( 4,1 ) . Graph: and . The angle is the angle between the vectors DB AC cos = | DB|| AC | DB The angles between the diagonals are DB=BD=( 0,3 )( 4,1 )= 04,31 DB= 4,2 and AC , then: | DB|= (4 ) + ( 2 ) = 16 +4= 20= 4 5= 4 5 | DB|=2 5 2 2 AC=C A=( 3,4 ) ( 1,0 )= 31,40 AC= 2,4 | AC|= ( 2 ) + ( 4 ) = 4+16= 20= 4 5= 4 5 | AC|=2 5 2 2 DB AC= 4,2 2,4 =(4 ) ( 2 )+ ( 2 )( 4 )=8+ 8 DB AC=0 cos = DB AC 0 0 0 0 = = 2 = = cos =0 2 | DB|| AC | ( 2 5 ) ( 2 5 ) ( 2 ) ( 5 ) ( 4 ) ( 5 ) 20 =cos1 ( 0 ) =90 The angle is the angle between the vectors B D AC cos = | B D|| AC| BD and AC , then: B D= DB=4,2 =(4 ) ,2 BD= 4,2 | BD|=| DB|| BD|=2 5 AC = 2,4 ;| AC|=2 5 B D AC = 4 ,2 2,4 = ( 4 )( 2 ) + (2 )( 4 ) =88 B D AC=0 cos = B D AC 0 0 0 0 = = 2 = = cos =0 2 | B D|| AC| ( 2 5 ) ( 2 5 ) ( 2 ) ( 5 ) ( 4 ) ( 5 ) 20 =cos1 ( 0 ) =90 To check, and are adjacent angle, then their sum must be equal to 180: + =90 +90 + =180 Ok Problem 6 (Cross Product). Compute the cross products of the following pair of vectors: 1. u=3 i 2 j+ k v =i + j 2 k | || j i k 1 3 2 2 1 i 3 u v = 3 2 j+ k 1 = 1 2 1 2 1 1 1 1 2 || || | u v =[ (2 ) ( 2 )( 1 ) ( 1 ) ] i [ ( 3 ) ( 2 )( 1 ) ( 1 ) ] j+ [ ( 3 )( 1 ) ( 2 ) ( 1 ) ] k 2 (3 21 ) j+ ( 3+ 2 ) k u v =[ ( 2 ) 1 ] i u v =( 21 ) i + ( 3 2+1 ) j + ( 3+ 2 ) k u v =i + ( 3 2+1 ) j + ( 3+ 2 ) k 2. u= 1 , 1,3 v = 3,2,2 3 | || j k i 1 1 1 3 u v = = i 1 3 3 2 2 3 3 3 2 2 | | || | [( ) [( ) 1 3 1 j+ 3 k 2 3 2 1 1 ( 2 ) ( 3 )( 3 ) j+ (2 )( 1 ) ( 3 ) k u v =[ ( 1 )( 2 ) ( 3 )(2 ) ] i 3 3 23 ( 9 ) 2+3 ( 3 ) 2 2 3 k =8 i j k u v =( 2+6 ) i 9 j + 3 3 3 3 ( ) ( ) ( ) ( j 2+9 k =8 i 25 j 11 k ( 227 ( 3 ) (3) 3 ) ( 3 ) u v =8 i 25 11 u v =8 i + j k 3 3 ( ) ( ) 3. u= 1 ,5 , () () 4 7 1 v = i+ jk 7 4 2 ) | || | | | | i u v = 1 7 4 j k 4 5 1 4 5 7 i 7 = 1 7 1 1 1 2 4 2 [ u v = (5 ) (1 ) | ( 47 )( 12 )] i[( 1) (1) ( 47 )( 74 )] j +[ ( 1)( 12 )(5) ( 74 )] k 2 ( 1 ) +35 2 1 35 7 ( 5 ) 2 i (11 ) j+ + k= i (2 ) j + k 7 2 4 7 4 ( ) u v = 5 4 1 5 7 j+ 7 1 k 1 4 2 ) ( ( ) ( ) 2+35 33 37 i +2 j+ ( k u v =( )i +2 j+( ) k ( 352 ) ) 7 4 7 4 u v = 4. 1 1 j+ k v = 1,3,1 u=i 3 2 ( ) ( ) | i j 1 u v = 1 3 1 3 || k 1 1 = 3 2 3 1 || || | 1 1 1 1 1 2 i 2 j+ 3 k 1 1 1 1 3 [( ) ( ) ] [ u v = ( ) ( 13 32 )i(1 12 ) j+(3+ 13 ) k u v = 3 i 21 j+ 3 3+1 k ( 23 3 2 ) ( 2 ) ( 3 ) u v = ( u v = [ 1 1 1 1 (1 ) (3 ) i (1 )(1 ) ( 1 ) j+ (1 )( 3 ) (1 ) k 3 2 2 3 23 3 i 21 j 3 31 k 6 2 3 ) ( ) ( ) ( ) Problem 7 (Triple Scalar Products). Have a go at computing ( u v ) w for the following triples and verify that in fact you get the same answer if you compute ( w u ) v : 1. u=2 i v =2 j w =2 k ( v w ) u | || i j k v w= 0 2 0 = 2 0 i 0 0 j+ 0 2 k 0 2 0 2 0 0 0 0 2 || || | v w=[ ( 2 ) ( 2 )( 0 )( 0 ) ] i [ ( 0 ) ( 2 )( 0 )( 0 ) ] j+ [ ( 0 )( 0 )( 2 ) ( 0 ) ] k v w=( 40 ) i ( 00 ) j+ ( 00 ) k =4 i 0 j+ 0 k v w =4 i ( v w ) u =( 4 i ) ( 2 i )=( 4 ) ( 2 ) ( v w ) u =8 ( w u ) v | || i j k w u = 0 0 2 = 0 2 i 0 2 j+ 0 0 k 0 0 2 0 2 0 2 0 0 || | | | w u =[ ( 0 ) ( 0 ) ( 2 )( 0 ) ] i [ ( 0 )( 0 )( 2 )( 2 ) ] j + [ ( 0 )( 0 ) ( 0 ) ( 2 ) ] k w u =( 00 ) i ( 04 ) j+ ( 00 ) k =0 i (4 ) j+0 k w u =4 j ( w u ) v =( 4 j ) ( 2 j )=( 4 ) ( 2 ) ( w u ) v =8 Then: 2. ( v w ) u =8= ( w u ) v u=i j+ k v =2 i + j2 k w =i +2 j k ( v w ) u ( v w ) u or | || i j k 1 2 2 2 2 1 v w= 2 1 2 = i j+ k 2 1 1 1 1 2 1 2 1 | | | | | v w=[ ( 1 ) (1 ) (2 )( 2 ) ] i [ ( 2 )(1 )(2 )(1 ) ] j+ [ ( 2 )( 2 ) ( 1 )(1 ) ] k v w=(1+4 ) i (22 ) j+ ( 4 +1 ) k =3 i (4 ) j+5 k v w =3 i + 4 j +5 k ( v w ) u =( 3 i + 4 j +5 k ) ( i j+ k )=( 3 ) ( 1 )+ ( 4 )(1 ) + ( 5 ) (1 ) =34+5 ( v w ) u =4 ( w u ) v | || j k i w u = 1 2 1 = 2 1 i 1 1 j + 1 2 k 1 1 1 1 1 1 1 1 1 || || | w u =[ ( 2 ) ( 1 )(1 )(1 ) ] i [ (1 ) (1 ) (1 ) ( 1 ) ] j+ [ (1 ) (1 ) ( 2 )( 1 ) ] k w u =( 21 ) i (1+1 ) j+ ( 12 ) k =i 0 j+ (1 ) k w u =i k ( w u ) v =( i k ) (2 i + j2 k )=( 1 ) ( 2 )+ ( 0 )( 1 ) + (1 ) (2 )=2+ 0+2 ( w u ) v =4 Then: 3. ( v w ) u =4=( w u ) v u= 2,1,0 v = 2,1,1 w = 1,2,0 ( v w ) u | || i j k 1 1 2 1 2 1 v w= 2 1 1 = i j+ k 2 0 1 0 1 2 1 2 0 | | | | | v w=[ (1 ) ( 0 )( 1 ) ( 2 ) ] i [ ( 2 ) ( 0 ) (1 )( 1 ) ] j+ [ ( 2 )( 2 ) (1 ) ( 1 ) ] k v w=( 02 ) i ( 01 ) j+ ( 4+1 ) k =2 i (1 ) j+5 k =2 i + j+5 k v w= 2,1,5 ( v w ) u =2,1,5 2,1,0 =(2 ) ( 2 )+ (1 )( 1 ) + ( 5 ) ( 0 )=4 +1+0 ( v w ) u =3 ( w u ) v | || i j k w u = 1 2 0 = 2 0 i 1 0 j + 1 2 k 1 0 2 0 2 1 2 1 0 || || | w u =[ ( 2 ) ( 0 )( 0 )( 1 ) ] i [ ( 1 )( 0 )( 0 ) ( 2 ) ] j+ [ ( 1 )( 1 ) (2 )( 2 ) ] k w u =( 00 ) i ( 00 ) j+ ( 14 ) k =0 i 0 j + (3 ) k =3 k w u = 0,0,3 ( w u ) v = 0,0,3 2,1,1 =( 0 )( 2 ) + ( 0 )(1 ) + (3 )( 1 )=0+03 ( w u ) v =3 Then: ( v w ) u =3=( w u ) v Homework Assignment 1 Multivariable functions and vector geometry 1 Computational Problems Problem 1 (Functions). f 1. Suppose f is a function defined by the formula f ( x , y , z , w )=x + yzw . Evaluate ( x , y , z , w )=( 4, 2 , 8 , 3 ) . whenever x=4, y= 2 , z= 8 , w=3 f ( 4, 2 , 8 ,3 ) =4 + 2 83=4+ 2 83 f ( 4, 2 , 8 , 3 ) =4 + 163=4 +43 f ( 4, 2 , 8 , 3 ) =5 g 2. Suppose is a function that returns a value 1 when all its parameters are integers and a value of 0 otherwise. What is about g ( g ( tan ( 30 ) ln ( 8 ) + ln (2 ) , 3 tan ( 30 ) , ln ( e ) + ln ( e/4 ) 3 tan ( 60 ) ln ( 8 )+ ln ( 2 ) , 3 tan ( 60 ) , ln ( e )+ ln ( 4 /e ) 3 ) ? Note that there could not possibly a formula for this function ( ( )) tan ( 30 ) ln ( 8 ) + ln ( 2 ) g , 3 tan ( 30 ) , =g 3 e ln ( e ) + ln 4 ( g ( 2 ) ( g ! ) 3 3 3 3 ln ( 2 )+ ln ( 2 ) , 3 , = 3 1+ln ( e )ln ( 4 ) 3 3 1 , ( 3 ) , 3 ln ( 2 ) + ln ( 2 ) =g 1 , 3 , 4 ln (2 ) =g 1 ,1, 4 ln ( 2 ) = 3 3 3 3 3 22 ln ( 2 ) 3 2 [ 1ln ( 2 ) ] 1+1ln ( 22 ) ) ( ) ( 4 ( 0.693147181 ) 2.772588722 =g 0.333333333,1, = 2 ( 10.693147181 ) 2 ( 0.306852819 ) ( 2.772588722 =g ( 0.333333333,1, 4.517782707 ) 0.613705639 g 0.333333333,1, g 0.333333333,1, ) ( ) ) ) ? What g g ( tan ( 30 ) ln ( 8 ) + ln ( 2 ) , 3 tan ( 30 ) , =0 e 3 ln ( e ) + ln 4 ( tan ( 60 ) ln ( 8 )+ ln (2 ) , 3 tan ( 60 ) , =g 3 4 ln ( e )+ ln e () ( )) ( 3 ln ( 2 )+ ln ( 2 ) 2 g 1, ( 3 ) , g ( ) 2 1+ln ( 2 )1 ) ( =g 1,3, ( 3 3 , 3 3 , ln ( 2 ) +ln ( 2 ) = 1+ln ( 4 )ln ( e ) 3 ) 4 ln ( 2 ) =g ( 1, 3, 2 ) 2 ln ( 2 ) ) tan ( 60 ) ln ( 8 )+ ln (2 ) , 3 tan ( 60 ) , =g ( 1, 3,2 ) =1 4 3 ln ( e )+ ln e 3. Suppose () P ) is a function that only takes integers as parameters and returns a value of 1 if all parameters are even, -1 if all parameters are odd and 0 otherwise. What is the value of P ( P ( 1,0,2 ) +1,2 P ( 1,1,1 )P ( 2,2,0 ) , P ( 0,2,0 )1 ) ? 0 is an even number. P (1,0,2 ) =0 ; P ( 1,1,1 )=1; P ( 2,2,0 )=1 ; P ( 0,2,0 )=1 P ( P ( 1,0,2 ) +1,2 P ( 1,1,1 )P ( 2,2,0 ) , P ( 0,2,0 )1 )=P ( 0+1, 2 (1 )1, 21 ) = P (1,21, 1 )=P ( 1,3, 1 )=1 P ( P ( 1,0,2 ) +1,2 P ( 1,1,1 )P ( 2,2,0 ) , P ( 0,2,0 )1 )=1 Problem 2 (Contours). Draw three or four contours in the plane for each of the following functions. Don't forget to label each contour with its corresponding function value. 1. 2 f ( x , y ) =x +2 y 2 2. f ( x , y ) =sin ( x+ y ) 3. f ( x , y ) =e xy 4. f ( x , y ) =x e y y e x 5. f ( x , y ) =x2 + y 3 6. f ( x , y ) =2 x y 7. f ( x , y )= y3 8. f ( x , y ) =cos ( x )+ cos ( y ) Problem 3 (Level Surfaces) Match the following equations with the plotted surfaces (if this takes you a lot of time, you may wish to get some ideas from http://wolframalpha.com/ but just make sure you cite it in your bibliography if you use it - see \"Homework Guidelines\"): 1. 2. 3 x2 +2 y 2+ z2 =5 2 2 2 2 2 2 3x 2 y z 5 x y z + + = + + =1 5 5 5 5 5 5 5 3 2 y 2=z3 x 2 y 2+ z=3 x 2 z=3 x 2 y 2 z = 2 x 2 y 2 x2 y2 x2 y2 (1 ) z= z= + + =z ; z 0 1 1 1 1 1 1 3 3 3 ( 3. 4. ) x 2+ y 2 +3 z 2=5 2 x=z y 2 2 2 2 2 2 2 2 x y 1 1 3 x y 3z 5 x y z + + = + + =1 5 5 5 5 5 5 5 3 2 2 2 2 5. z z z =2 ( 1+ x 2 y 2) =1+ x 2 y 2 + y 2x 2=1 6. x+ y 2+ z 2=0 x= y 2z 2 (1 ) ( x= y 2z 2 ) x= y 2 + z 2 2 2 2 y + z =x , x 0 7. 2 x 22 y 2 +2 z2=1 2 x 2 +2 z 22 y 2=1 x2 z2 y2 + =1 1 1 1 2 2 2 8. x 2 3 z 2 10 x 2 z2 x =103 z x + 3 z =10 + = + =1 10 10 10 10 10 3 9. x= y 2z 2 2 2 2 2 z2 3 y 2 8 z 2 y 2 z +3 y =8 + = + =1 8 8 8 8 8 10. 3 2 2 2 2 2 11. x = y + z 2 2 2 2 2 2 2 2 2 2 2 2 12. x y =5 z x =5 z + y x +5 z = y y =x + 5 z Problem 4 (Dot product and projection). Find for the following pair of vectors: 1. v =2 i +4 j 6 k u =3 i + 4 j+ 6 k 2 v u =( 2 ) (3 )+ ( 4 )( 4 ) + ( 6 ) ( 6 )=6+16( 6 ) =6 +166 v u=4 (a) |v|= ( 2 )2+ ( 4 )2+ ( 6 ) = 4+16 +6 |v|= 26 2 |u|= (3 )2 + ( 4 )2 + ( 6 ) = 9+16+6 |u|= 31 2 (b) the cosine of the angle between v and u v u 4 4 4 = = cos = |v||u| ( 26 ) ( 31 ) 26 31 806 cos = v (c) the scalar component of v u u = 2 u 4 ( | | ) ( ( ) ) ( 31 2 ( 23 ) j ( 12 ) k u=i +6 j v = v u 4 = |u| 31 4 (3 i +4 j+ 6 k ) 3 i +4 j+ 6 k )= 31 ( 1231 ) i+( 1631 ) j +( 431 6 )k proju ( v )= 2. u= proju ( v ) . (d) the vector proju ( v )= in the direction of ( ) (a) ( 23 ) ( 6 )+( 12 ) ( 0) =0+ 123 0 v u=4 v u =( 0 )( 1 ) + |v|= 4 ( 4 )+ 9 (1 ) 2 2 1 2 4 1 16+9 25 25 + = 0+ + = = = = 3 2 9 4 36 36 36 36 ()( ) |v|= ( 0 )2 + 5 6 |u|= ( 1 ) + ( 6 ) + ( 0 ) = 1+ 36+0 |u|= 37 2 2 2 (b) the cosine of the angle between v and u v u 4 4 6 24 = = =4 cos = |v||u| 5 5 37 5 37 ( 37 ) 5 37 6 6 cos = () (c) the scalar component of in the direction of u = v u 4 = |u| 37 proju ( v ) . (d) the vector 3. v 4 4 v u ( i + 6 j )= ( i +6 j ) u = 2 2 37 |u| ( 37 ) proju ( v )= ( ) ( proju ( v )= ( 374 ) i+( 2437 ) j ) ( ) v = 2, (a) 1 u= 1,2 2 v u =( 2 ) ( 1 )+ |v|= ( 2 )2+ ( 12 ) ( 2)=2+ 22=2+ 1 v u =3 4 ( 4 )+ 1 1 2 1 16+1 17 17 17 = 4+ = = = = |v|= 2 4 4 4 4 2 4 () |u|= ( 1 ) + ( 2 ) = 1+ 4 |u|= 5 2 2 (b) the cosine of the angle between v u = |v||u| cos = and u 3 3 3 2 6 = = =3 cos = 85 85 17 ( 5 ) 17 5 85 2 2 2 ( ) (c) the scalar component of (d) the vector v v in the direction of u= v u 3 = |u| 5 proju ( v ) . ( ) ( ) v u 3 u = 1,2 = 3 1,2 = 3 ( 1 ) , 3 ( 2 ) 2 2 5 5 5 |u| ( 5) proju ( v )= ( ) ( ) proju ( v )= () 3 6 , 5 5 Problem 5 (Angle between vectors). 1. Find the angles between the following pairs of vectors: (a) u=i +2 j k v =2 i + j u v =( 1 ) ( 2 )+ ( 2 )( 1 ) + (1 )( 0 )=2+20 u v =4 |u|= ( 1 ) + ( 2 ) + (1 ) = 1+ 4+ 1|u|= 6 2 2 2 |v|= ( 2 )2+ ( 1 )2+ ( 0 )2= 4+ 1+ 0 |v|= 5 u v 4 4 4 4 = = = = cos =0.730296743 5.477225575 |u||v| ( 6 ) ( 5 ) 6 5 30 cos = =cos1 ( 0.730296743 ) =43.08872313 43.09 (b) u=3 i +4 j v =2 i 2 j + k u v =( 3 ) ( 2 ) + ( 4 )(2 ) + ( 0 ) ( 1 )=68+ 0 u v =2 |u|= ( 3 ) + ( 4 ) + ( 0 ) = 9+16+0= 25 |u|=5 2 2 2 |v|= ( 2 ) + (2 ) + ( 1 ) = 4 +4 +1= 9 |v|=3 2 2 2 u v 2 2 2 = = = cos =0.133333333 |u||v| ( 5 ) ( 3 ) 15 15 cos = 1 =cos (0.133333333 ) =97.66225566 97.66 (c) u= 7 i +2 j k v = 7 i j 2 u v =( 7 ) ( 7 ) + ( 2 ) (1 ) + (1 ) ( 0 )=( 7 ) 20=72 u v =5 |u|= ( 7 ) + ( 2 )2+ (1 )2= 7+ 4+1=12= 4 3= 4 3|u|=2 3 2 |v|= ( 7 ) + (1 )2 + ( 0 )2= 7+1+0= 8= 4 2= 4 2|v|=2 2 2 u v 5 5 5 5 5 = = = = = |u||v| ( 2 3 )( 2 2 ) 4 3 2 4 6 4 ( 2.449489743 ) 9.797958971 cos = 1 cos =0.510310363 =cos ( 0.510310363 ) =59.31549491 59.32 2. Using what you have learnt in class, compute the angles of the triangle in the plane whose vertices are P=( 2,1 ) , Q=( 1,2 ) and R=(1,0 ) . Graph: The angle in vertex R RP RQ cos = |RP||RQ| ( ) is the angle between the vectors RP and RQ , then: RP=PR=( 2,1 )(1,0 )= 2(1 ) , 10 = 2+1,1 RP= 3,1 | RP|= ( 3 ) + ( 1 ) = 9+1 | RP|= 10 2 2 RQ=QR=( 1,2 ) (1,0 ) = 1(1 ) ,20 = 1+1,2 RQ= 2,2 | RQ|= ( 2 ) + (2 ) = 4+ 4 | RQ|= 8 2 2 RP RQ= 3,1 2,2 =( 3 ) ( 2 )+ (1 )(2 )=62 RP RQ=4 cos = RP RQ 4 4 4 4 4 4 = = = = = = | RP|| RQ| 10 8 10 8 5 2 8 5 16 5 16 5 ( 4 ) cos = 1 1 = cos =0.447213596 =cos1 ( 0.447213596 ) 2.236067978 5 =63.43494882 63.435 The angle in vertex P ( ) is the angle between the vectors P R P R PQ cos = |P R||P Q| and PQ , then: PR= RP= 3,1 PR= 3,1 | PR|=| RP|| PR|= 10 PQ=QP=( 1,2 ) (2 , 1 )= 12 ,21 P Q= 1,3 | P Q|= (1 ) + (3 ) = 1+9 | P Q|= 10 2 2 P R PQ=3,1 1 ,3 =(3 )(1 )+ (1 ) (3 )=3+3 P R P Q=6 cos = P R PQ 6 6 6 3 = = = cos = =0.6 =cos1 ( 0. 6 ) 2 5 | P R|| P Q| 10 10 ( 10 ) 10 =53.13010235 5 3.130 The angle in vertex Q QR QP cos = | Q R|| Q P| QR ( ) is the angle between the vectors and Q R= RQ= 2,2 = 2,(2 ) Q R=2 , 2 | QR|=| RQ|| Q R|= 8 Q P= PQ=1 ,3 = (1 ) ,(3 ) Q P= 1 , 3 | Q P|=| P Q|| QP|= 10 Q R Q P=2 , 2 1 ,3 =(2 ) ( 1 )+ (2 )( 3 ) =2+6 Q R Q P=4 Q R QP 4 4 4 4 4 4 = = = = = = |Q R||Q P| 8 10 8 10 8 2 5 16 5 16 5 4 5 cos = Q P , then: cos = 1 1 = cos =0.447213596 =cos1 ( 0.447213596 ) 2.236067978 5 =63.43494882 63.435 To check, the sum of the internal angles of any triangle must be equal to 180: + +=63.435 + 53.130 +63.435 + + =180 Ok 3. Similarly, compute the angles between the diagonals of the rectangle in the plane whose vertices are A= (1,0 ) , B=( 0,3 ) , C=( 3,4 ) and D=( 4,1 ) . Graph: and . The angle is the angle between the vectors DB AC cos = | DB|| AC | DB The angles between the diagonals are DB=BD=( 0,3 )( 4,1 )= 04,31 DB= 4,2 and AC , then: | DB|= (4 ) + ( 2 ) = 16 +4= 20= 4 5= 4 5 | DB|=2 5 2 2 AC=C A=( 3,4 ) ( 1,0 )= 31,40 AC= 2,4 | AC|= ( 2 ) + ( 4 ) = 4+16= 20= 4 5= 4 5 | AC|=2 5 2 2 DB AC= 4,2 2,4 =(4 ) ( 2 )+ ( 2 )( 4 )=8+ 8 DB AC=0 cos = DB AC 0 0 0 0 = = 2 = = cos =0 2 | DB|| AC | ( 2 5 ) ( 2 5 ) ( 2 ) ( 5 ) ( 4 ) ( 5 ) 20 =cos1 ( 0 ) =90 The angle is the angle between the vectors B D AC cos = | B D|| AC| BD and AC , then: B D= DB=4,2 =(4 ) ,2 BD= 4,2 | BD|=| DB|| BD|=2 5 AC = 2,4 ;| AC|=2 5 B D AC = 4 ,2 2,4 = ( 4 )( 2 ) + (2 )( 4 ) =88 B D AC=0 cos = B D AC 0 0 0 0 = = 2 = = cos =0 2 | B D|| AC| ( 2 5 ) ( 2 5 ) ( 2 ) ( 5 ) ( 4 ) ( 5 ) 20 =cos1 ( 0 ) =90 To check, and are adjacent angle, then their sum must be equal to 180: + =90 +90 + =180 Ok Problem 6 (Cross Product). Compute the cross products of the following pair of vectors: 1. u=3 i 2 j+ k v =i + j 2 k | || j i k 1 3 2 2 1 i 3 u v = 3 2 j+ k 1 = 1 2 1 2 1 1 1 1 2 || || | u v =[ (2 ) ( 2 )( 1 ) ( 1 ) ] i [ ( 3 ) ( 2 )( 1 ) ( 1 ) ] j+ [ ( 3 )( 1 ) ( 2 ) ( 1 ) ] k 2 (3 21 ) j+ ( 3+ 2 ) k u v =[ ( 2 ) 1 ] i u v =( 21 ) i + ( 3 2+1 ) j + ( 3+ 2 ) k u v =i + ( 3 2+1 ) j + ( 3+ 2 ) k 2. u= 1 , 1,3 v = 3,2,2 3 | || j k i 1 1 1 3 u v = = i 1 3 3 2 2 3 3 3 2 2 | | || | [( ) [( ) 1 3 1 j+ 3 k 2 3 2 1 1 ( 2 ) ( 3 )( 3 ) j+ (2 )( 1 ) ( 3 ) k u v =[ ( 1 )( 2 ) ( 3 )(2 ) ] i 3 3 23 ( 9 ) 2+3 ( 3 ) 2 2 3 k =8 i j k u v =( 2+6 ) i 9 j + 3 3 3 3 ( ) ( ) ( ) ( j 2+9 k =8 i 25 j 11 k ( 227 ( 3 ) (3) 3 ) ( 3 ) u v =8 i 25 11 u v =8 i + j k 3 3 ( ) ( ) 3. u= 1 ,5 , () () 4 7 1 v = i+ jk 7 4 2 ) | || | | | | i u v = 1 7 4 j k 4 5 1 4 5 7 i 7 = 1 7 1 1 1 2 4 2 [ u v = (5 ) (1 ) | ( 47 )( 12 )] i[( 1) (1) ( 47 )( 74 )] j +[ ( 1)( 12 )(5) ( 74 )] k 2 ( 1 ) +35 2 1 35 7 ( 5 ) 2 i (11 ) j+ + k= i (2 ) j + k 7 2 4 7 4 ( ) u v = 5 4 1 5 7 j+ 7 1 k 1 4 2 ) ( ( ) ( ) 2+35 33 37 i +2 j+ ( k u v =( )i +2 j+( ) k ( 352 ) ) 7 4 7 4 u v = 4. 1 1 j+ k v = 1,3,1 u=i 3 2 ( ) ( ) | i j 1 u v = 1 3 1 3 || k 1 1 = 3 2 3 1 || || | 1 1 1 1 1 2 i 2 j+ 3 k 1 1 1 1 3 [( ) ( ) ] [ u v = ( ) ( 13 32 )i(1 12 ) j+(3+ 13 ) k u v = 3 i 21 j+ 3 3+1 k ( 23 3 2 ) ( 2 ) ( 3 ) u v = ( u v = [ 1 1 1 1 (1 ) (3 ) i (1 )(1 ) ( 1 ) j+ (1 )( 3 ) (1 ) k 3 2 2 3 23 3 i 21 j 3 31 k 6 2 3 ) ( ) ( ) ( ) Problem 7 (Triple Scalar Products). Have a go at computing ( u v ) w for the following triples and verify that in fact you get the same answer if you compute ( w u ) v : 1. u=2 i v =2 j w =2 k ( v w ) u | || i j k v w= 0 2 0 = 2 0 i 0 0 j+ 0 2 k 0 2 0 2 0 0 0 0 2 || || | v w=[ ( 2 ) ( 2 )( 0 )( 0 ) ] i [ ( 0 ) ( 2 )( 0 )( 0 ) ] j+ [ ( 0 )( 0 )( 2 ) ( 0 ) ] k v w=( 40 ) i ( 00 ) j+ ( 00 ) k =4 i 0 j+ 0 k v w =4 i ( v w ) u =( 4 i ) ( 2 i )=( 4 ) ( 2 ) ( v w ) u =8 ( w u ) v | || i j k w u = 0 0 2 = 0 2 i 0 2 j+ 0 0 k 0 0 2 0 2 0 2 0 0 || | | | w u =[ ( 0 ) ( 0 ) ( 2 )( 0 ) ] i [ ( 0 )( 0 )( 2 )( 2 ) ] j + [ ( 0 )( 0 ) ( 0 ) ( 2 ) ] k w u =( 00 ) i ( 04 ) j+ ( 00 ) k =0 i (4 ) j+0 k w u =4 j ( w u ) v =( 4 j ) ( 2 j )=( 4 ) ( 2 ) ( w u ) v =8 Then: 2. ( v w ) u =8= ( w u ) v u=i j+ k v =2 i + j2 k w =i +2 j k ( v w ) u ( v w ) u or | || i j k 1 2 2 2 2 1 v w= 2 1 2 = i j+ k 2 1 1 1 1 2 1 2 1 | | | | | v w=[ ( 1 ) (1 ) (2 )( 2 ) ] i [ ( 2 )(1 )(2 )(1 ) ] j+ [ ( 2 )( 2 ) ( 1 )(1 ) ] k v w=(1+4 ) i (22 ) j+ ( 4 +1 ) k =3 i (4 ) j+5 k v w =3 i + 4 j +5 k ( v w ) u =( 3 i + 4 j +5 k ) ( i j+ k )=( 3 ) ( 1 )+ ( 4 )(1 ) + ( 5 ) (1 ) =34+5 ( v w ) u =4 ( w u ) v | || j k i w u = 1 2 1 = 2 1 i 1 1 j + 1 2 k 1 1 1 1 1 1 1 1 1 || || | w u =[ ( 2 ) ( 1 )(1 )(1 ) ] i [ (1 ) (1 ) (1 ) ( 1 ) ] j+ [ (1 ) (1 ) ( 2 )( 1 ) ] k w u =( 21 ) i (1+1 ) j+ ( 12 ) k =i 0 j+ (1 ) k w u =i k ( w u ) v =( i k ) (2 i + j2 k )=( 1 ) ( 2 )+ ( 0 )( 1 ) + (1 ) (2 )=2+ 0+2 ( w u ) v =4 Then: 3. ( v w ) u =4=( w u ) v u= 2,1,0 v = 2,1,1 w = 1,2,0 ( v w ) u | || i j k 1 1 2 1 2 1 v w= 2 1 1 = i j+ k 2 0 1 0 1 2 1 2 0 | | | | | v w=[ (1 ) ( 0 )( 1 ) ( 2 ) ] i [ ( 2 ) ( 0 ) (1 )( 1 ) ] j+ [ ( 2 )( 2 ) (1 ) ( 1 ) ] k v w=( 02 ) i ( 01 ) j+ ( 4+1 ) k =2 i (1 ) j+5 k =2 i + j+5 k v w= 2,1,5 ( v w ) u =2,1,5 2,1,0 =(2 ) ( 2 )+ (1 )( 1 ) + ( 5 ) ( 0 )=4 +1+0 ( v w ) u =3 ( w u ) v | || i j k w u = 1 2 0 = 2 0 i 1 0 j + 1 2 k 1 0 2 0 2 1 2 1 0 || || | w u =[ ( 2 ) ( 0 )( 0 )( 1 ) ] i [ ( 1 )( 0 )( 0 ) ( 2 ) ] j+ [ ( 1 )( 1 ) (2 )( 2 ) ] k w u =( 00 ) i ( 00 ) j+ ( 14 ) k =0 i 0 j + (3 ) k =3 k w u = 0,0,3 ( w u ) v = 0,0,3 2,1,1 =( 0 )( 2 ) + ( 0 )(1 ) + (3 )( 1 )=0+03 ( w u ) v =3 Then: ( v w ) u =3=( w u ) v 2 Conceptual Problems Problem 8 (Functions). Are f ( x , y , z )=x 2+ y 2 and f ( x , y ) =x2 + y 2 the same function? Explain your reasoning. No, because the function f ( x , y , z )=x 2+ y 2 is a function of three variables that in this case depends on only two variables, and it maps in a fourth dimension, while the function f ( x , y ) =x2 + y 2 is a function of two variables and it maps in a third dimension. Problem 9 (Cauchy-Schwarz Inequality). We already saw the triangle inequality in class. Here you will prove another inequality involving dot products of vectors, known as the Cauchy-Schwarz Inequality. Recall the definition of the dot product from lecture: v w =|v|| w|cos where for any pair of vectors v v is the angle between and w and w . Show that |v w||v|| w| (in the plane or in 3-space). Under what circumstances is the inequality actually an equality? We know that: 1 cos 1 . If we take absolute value of the cosine function, we have: 0 |cos | 1 , then |cos | 1 |v|| w| : If we multiply both sides of this inequality by w|)(|cos | 1 ) |v|| w||cos ||v|| w| (|v|| and, if a 0b 0, thenab|c|=|abc|. In this case: a=|v|, b=| w|c=cos , then |v|| w||cos |=||v|| w|cos | . If we replace this in the equation above: |v|| w||cos ||v|| w|||v|| w| cos ||v|| w| w|cos =v w , then: and |v|| ||v|| w|cos ||v|| w||v w||v|| w| Under what circumstances is the inequality actually an equality? The inequality is an equality when: 1 cos =1 =cos 1 =0 ; that means the angle between v and w () is equal to 0. Problem 10 (Cancellation for dot products). When we take products of real numbers and w , if we know that conclude that w ? i.e. if u 0 and uv=uw then we can always cancel the v =w . Is the same true when we take dot products of vectors u v =u w , can we \"cancel\" u to conclude that Counter-example: u to and v = w ? If so, prove it. If not, give a counter-example. No, it is not the same true when we take dot products of vectors u , v u,v u , v and w . u= 1 ,1 , v = 3 , 2 ; w= 4 ,3 u v = 1,1 3, 2 =( 1 )( 3 ) + (1 ) (2 )=32 u v =1 u w= 1,1 4 , 3 =( 1 ) ( 4 )+ (1 ) ( 3 )=43 u w =1 Then: u v =1=u w , but , v = 3, 2 4,3 = w Homework Assignment 1 Multivariable functions and vector geometry 1 Computational Problems Problem 1 (Functions). f 1. Suppose f is a function defined by the formula f ( x , y , z , w )=x + yzw . Evaluate ( x , y , z , w )=( 4, 2 , 8 , 3 ) . whenever x=4, y= 2 , z= 8 , w=3 f ( 4, 2 , 8 ,3 ) =4 + 2 83=4+ 2 83 f ( 4, 2 , 8 , 3 ) =4 + 163=4 +43 f ( 4, 2 , 8 , 3 ) =5 g 2. Suppose is a function that returns a value 1 when all its parameters are integers and a value of 0 otherwise. What is about g ( g ( tan ( 30 ) ln ( 8 ) + ln (2 ) , 3 tan ( 30 ) , ln ( e ) + ln ( e/4 ) 3 tan ( 60 ) ln ( 8 )+ ln ( 2 ) , 3 tan ( 60 ) , ln ( e )+ ln ( 4 /e ) 3 ) ? Note that there could not possibly a formula for this function ( ( )) tan ( 30 ) ln ( 8 ) + ln ( 2 ) g , 3 tan ( 30 ) , =g 3 e ln ( e ) + ln 4 ( g ( 2 ) ( g ! ) 3 3 3 3 ln ( 2 )+ ln ( 2 ) , 3 , = 3 1+ln ( e )ln ( 4 ) 3 3 1 , ( 3 ) , 3 ln ( 2 ) + ln ( 2 ) =g 1 , 3 , 4 ln (2 ) =g 1 ,1, 4 ln ( 2 ) = 3 3 3 3 3 22 ln ( 2 ) 3 2 [ 1ln ( 2 ) ] 1+1ln ( 22 ) ) ( ) ( 4 ( 0.693147181 ) 2.772588722 =g 0.333333333,1, = 2 ( 10.693147181 ) 2 ( 0.306852819 ) ( 2.772588722 =g ( 0.333333333,1, 4.517782707 ) 0.613705639 g 0.333333333,1, g 0.333333333,1, ) ( ) ) ) ? What g g ( tan ( 30 ) ln ( 8 ) + ln ( 2 ) , 3 tan ( 30 ) , =0 e 3 ln ( e ) + ln 4 ( tan ( 60 ) ln ( 8 )+ ln (2 ) , 3 tan ( 60 ) , =g 3 4 ln ( e )+ ln e () ( )) ( 3 ln ( 2 )+ ln ( 2 ) 2 g 1, ( 3 ) , g ( ) 2 1+ln ( 2 )1 ) ( =g 1,3, ( 3 3 , 3 3 , ln ( 2 ) +ln ( 2 ) = 1+ln ( 4 )ln ( e ) 3 ) 4 ln ( 2 ) =g ( 1, 3, 2 ) 2 ln ( 2 ) ) tan ( 60 ) ln ( 8 )+ ln (2 ) , 3 tan ( 60 ) , =g ( 1, 3,2 ) =1 4 3 ln ( e )+ ln e 3. Suppose () P ) is a function that only takes integers as parameters and returns a value of 1 if all parameters are even, -1 if all parameters are odd and 0 otherwise. What is the value of P ( P ( 1,0,2 ) +1,2 P ( 1,1,1 )P ( 2,2,0 ) , P ( 0,2,0 )1 ) ? 0 is an even number. P (1,0,2 ) =0 ; P ( 1,1,1 )=1; P ( 2,2,0 )=1 ; P ( 0,2,0 )=1 P ( P ( 1,0,2 ) +1,2 P ( 1,1,1 )P ( 2,2,0 ) , P ( 0,2,0 )1 )=P ( 0+1, 2 (1 )1, 21 ) = P (1,21, 1 )=P ( 1,3, 1 )=1 P ( P ( 1,0,2 ) +1,2 P ( 1,1,1 )P ( 2,2,0 ) , P ( 0,2,0 )1 )=1 Problem 2 (Contours). Draw three or four contours in the plane for each of the following functions. Don't forget to label each contour with its corresponding function value. 1. 2 f ( x , y ) =x +2 y 2 2. f ( x , y ) =sin ( x+ y ) 3. f ( x , y ) =e xy 4. f ( x , y ) =x e y y e x 5. f ( x , y ) =x2 + y 3 6. f ( x , y ) =2 x y 7. f ( x , y )= y3 8. f ( x , y ) =cos ( x )+ cos ( y ) Problem 3 (Level Surfaces) Match the following equations with the plotted surfaces (if this takes you a lot of time, you may wish to get some ideas from http://wolframalpha.com/ but just make sure you cite it in your bibliography if you use it - see \"Homework Guidelines\"): 1. 2. 3 x2 +2 y 2+ z2 =5 2 2 2 2 2 2 3x 2 y z 5 x y z + + = + + =1 5 5 5 5 5 5 5 3 2 y 2=z3 x 2 y 2+ z=3 x 2 z=3 x 2 y 2 z = 2 x 2 y 2 x2 y2 x2 y2 (1 ) z= z= + + =z ; z 0 1 1 1 1 1 1 3 3 3 ( 3. 4. ) x 2+ y 2 +3 z 2=5 2 x=z y 2 2 2 2 2 2 2 2 x y 1 1 3 x y 3z 5 x y z + + = + + =1 5 5 5 5 5 5 5 3 2 2 2 2 5. z z z =2 ( 1+ x 2 y 2) =1+ x 2 y 2 + y 2x 2=1 6. x+ y 2+ z 2=0 x= y 2z 2 (1 ) ( x= y 2z 2 ) x= y 2 + z 2 2 2 2 y + z =x , x 0 7. 2 x 22 y 2 +2 z2=1 2 x 2 +2 z 22 y 2=1 x2 z2 y2 + =1 1 1 1 2 2 2 8. x 2 3 z 2 10 x 2 z2 x =103 z x + 3 z =10 + = + =1 10 10 10 10 10 3 9. x= y 2z 2 2 2 2 2 z2 3 y 2 8 z 2 y 2 z +3 y =8 + = + =1 8 8 8 8 8 10. 3 2 2 2 2 2 11. x = y + z 2 2 2 2 2 2 2 2 2 2 2 2 12. x y =5 z x =5 z + y x +5 z = y y =x + 5 z Problem 4 (Dot product and projection). Find for the following pair of vectors: 1. v =2 i +4 j 6 k u =3 i + 4 j+ 6 k 2 v u =( 2 ) (3 )+ ( 4 )( 4 ) + ( 6 ) ( 6 )=6+16( 6 ) =6 +166 v u=4 (a) |v|= ( 2 )2+ ( 4 )2+ ( 6 ) = 4+16 +6 |v|= 26 2 |u|= (3 )2 + ( 4 )2 + ( 6 ) = 9+16+6 |u|= 31 2 (b) the cosine of the angle between v and u v u 4 4 4 = = cos = |v||u| ( 26 ) ( 31 ) 26 31 806 cos = v (c) the scalar component of v u u = 2 u 4 ( | | ) ( ( ) ) ( 31 2 ( 23 ) j ( 12 ) k u=i +6 j v = v u 4 = |u| 31 4 (3 i +4 j+ 6 k ) 3 i +4 j+ 6 k )= 31 ( 1231 ) i+( 1631 ) j +( 431 6 )k proju ( v )= 2. u= proju ( v ) . (d) the vector proju ( v )= in the direction of ( ) (a) ( 23 ) ( 6 )+( 12 ) ( 0) =0+ 123 0 v u=4 v u =( 0 )( 1 ) + |v|= 4 ( 4 )+ 9 (1 ) 2 2 1 2 4 1 16+9 25 25 + = 0+ + = = = = 3 2 9 4 36 36 36 36 ()( ) |v|= ( 0 )2 + 5 6 |u|= ( 1 ) + ( 6 ) + ( 0 ) = 1+ 36+0 |u|= 37 2 2 2 (b) the cosine of the angle between v and u v u 4 4 6 24 = = =4 cos = |v||u| 5 5 37 5 37 ( 37 ) 5 37 6 6 cos = () (c) the scalar component of in the direction of u = v u 4 = |u| 37 proju ( v ) . (d) the vector 3. v 4 4 v u ( i + 6 j )= ( i +6 j ) u = 2 2 37 |u| ( 37 ) proju ( v )= ( ) ( proju ( v )= ( 374 ) i+( 2437 ) j ) ( ) v = 2, (a) 1 u= 1,2 2 v u =( 2 ) ( 1 )+ |v|= ( 2 )2+ ( 12 ) ( 2)=2+ 22=2+ 1 v u =3 4 ( 4 )+ 1 1 2 1 16+1 17 17 17 = 4+ = = = = |v|= 2 4 4 4 4 2 4 () |u|= ( 1 ) + ( 2 ) = 1+ 4 |u|= 5 2 2 (b) the cosine of the angle between v u = |v||u| cos = and u 3 3 3 2 6 = = =3 cos = 85 85 17 ( 5 ) 17 5 85 2 2 2 ( ) (c) the scalar component of (d) the vector v v in the direction of u= v u 3 = |u| 5 proju ( v ) . ( ) ( ) v u 3 u = 1,2 = 3 1,2 = 3 ( 1 ) , 3 ( 2 ) 2 2 5 5 5 |u| ( 5) proju ( v )= ( ) ( ) proju ( v )= () 3 6 , 5 5 Problem 5 (Angle between vectors). 1. Find the angles between the following pairs of vectors: (a) u=i +2 j k v =2 i + j u v =( 1 ) ( 2 )+ ( 2 )( 1 ) + (1 )( 0 )=2+20 u v =4 |u|= ( 1 ) + ( 2 ) + (1 ) = 1+ 4+ 1|u|= 6 2 2 2 |v|= ( 2 )2+ ( 1 )2+ ( 0 )2= 4+ 1+ 0 |v|= 5 u v 4 4 4 4 = = = = cos =0.730296743 5.477225575 |u||v| ( 6 ) ( 5 ) 6 5 30 cos = =cos1 ( 0.730296743 ) =43.08872313 43.09 (b) u=3 i +4 j v =2 i 2 j + k u v =( 3 ) ( 2 ) + ( 4 )(2 ) + ( 0 ) ( 1 )=68+ 0 u v =2 |u|= ( 3 ) + ( 4 ) + ( 0 ) = 9+16+0= 25 |u|=5 2 2 2 |v|= ( 2 ) + (2 ) + ( 1 ) = 4 +4 +1= 9 |v|=3 2 2 2 u v 2 2 2 = = = cos =0.133333333 |u||v| ( 5 ) ( 3 ) 15 15 cos = 1 =cos (0.133333333 ) =97.66225566 97.66 (c) u= 7 i +2 j k v = 7 i j 2 u v =( 7 ) ( 7 ) + ( 2 ) (1 ) + (1 ) ( 0 )=( 7 ) 20=72 u v =5 |u|= ( 7 ) + ( 2 )2+ (1 )2= 7+ 4+1=12= 4 3= 4 3|u|=2 3 2 |v|= ( 7 ) + (1 )2 + ( 0 )2= 7+1+0= 8= 4 2= 4 2|v|=2 2 2 u v 5 5 5 5 5 = = = = = |u||v| ( 2 3 )( 2 2 ) 4 3 2 4 6 4 ( 2.449489743 ) 9.797958971 cos = 1 cos =0.510310363 =cos ( 0.510310363 ) =59.31549491 59.32 2. Using what you have learnt in class, compute the angles of the triangle in the plane whose vertices are P=( 2,1 ) , Q=( 1,2 ) and R=(1,0 ) . Graph: The angle in vertex R RP RQ cos = |RP||RQ| ( ) is the angle between the vectors RP and RQ , then: RP=PR=( 2,1 )(1,0 )= 2(1 ) , 10 = 2+1,1 RP= 3,1 | RP|= ( 3 ) + ( 1 ) = 9+1 | RP|= 10 2 2 RQ=QR=( 1,2 ) (1,0 ) = 1(1 ) ,20 = 1+1,2 RQ= 2,2 | RQ|= ( 2 ) + (2 ) = 4+ 4 | RQ|= 8 2 2 RP RQ= 3,1 2,2 =( 3 ) ( 2 )+ (1 )(2 )=62 RP RQ=4 cos = RP RQ 4 4 4 4 4 4 = = = = = = | RP|| RQ| 10 8 10 8 5 2 8 5 16 5 16 5 ( 4 ) cos = 1 1 = cos =0.447213596 =cos1 ( 0.447213596 ) 2.236067978 5 =63.43494882 63.435 The angle in vertex P ( ) is the angle between the vectors P R P R PQ cos = |P R||P Q| and PQ , then: PR= RP= 3,1 PR= 3,1 | PR|=| RP|| PR|= 10 PQ=QP=( 1,2 ) (2 , 1 )= 12 ,21 P Q= 1,3 | P Q|= (1 ) + (3 ) = 1+9 | P Q|= 10 2 2 P R PQ=3,1 1 ,3 =(3 )(1 )+ (1 ) (3 )=3+3 P R P Q=6 cos = P R PQ 6 6 6 3 = = = cos = =0.6 =cos1 ( 0. 6 ) 2 5 | P R|| P Q| 10 10 ( 10 ) 10 =53.13010235 5 3.130 The angle in vertex Q QR QP cos = | Q R|| Q P| QR ( ) is the angle between the vectors and Q R= RQ= 2,2 = 2,(2 ) Q R=2 , 2 | QR|=| RQ|| Q R|= 8 Q P= PQ=1 ,3 = (1 ) ,(3 ) Q P= 1 , 3 | Q P|=| P Q|| QP|= 10 Q R Q P=2 , 2 1 ,3 =(2 ) ( 1 )+ (2 )( 3 ) =2+6 Q R Q P=4 Q R QP 4 4 4 4 4 4 = = = = = = |Q R||Q P| 8 10 8 10 8 2 5 16 5 16 5 4 5 cos = Q P , then: cos = 1 1 = cos =0.447213596 =cos1 ( 0.447213596 ) 2.236067978 5 =63.43494882 63.435 To check, the sum of the internal angles of any triangle must be equal to 180: + +=63.435 + 53.130 +63.435 + + =180 Ok 3. Similarly, compute the angles between the diagonals of the rectangle in the plane whose vertices are A= (1,0 ) , B=( 0,3 ) , C=( 3,4 ) and D=( 4,1 ) . Graph: and . The angle is the angle between the vectors DB