Vectors C For each problem, calculate the vector product in the bracet first and then perform the scalar product on the answer. a b c= 3 0 4 =4i j 3 a.(b c=(5i+j.(4i j 3 =0 +3= b c a = 3 0 4 5 = 8i+3j+6 b.(c a=(i+j+.( 8i+3j+6 = 8+3+6= c a b= 5 =3i 6j+3 c.(a b=(3i+4.(3i 6j+3 =9+= b c= 3 5 = 8i+8j 8 a.(b c=(i j.( 8i+8j 8 = 8 8+6=0 If a.(b c=0 then a is perpendicular to b c. This means that a is parallel to the plane containing b and c (in fact 3 a= b+ c. 8 8 3 AB AD= 3 0 = i+ j+ 6 0 Volume of the parallelepiped AE ( AB AD =. = ( i+ j+ 3.( i+ j+ 6 = ++8 =7 Alternatively, volume = e e e 3 b = d d d 3 3 3 0 0 =(0 (0 +3(6 0=7 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.
4 Let the vertices A, B, D and E have position vectors a, b, d and e respectively. AB= b a= 4i AD= d a= 3i+ j AE = e a= 3i+ j+ Using the scalar triple product 3 AE. AB AD = 4 0 = 3(0+ 4 ( 4+ 6 + (8 0 = + 8= 8 3 So volume of parallelpiped = AE. AB AD = 8 5 Let the vertices A, B, C and D have position vectors a, b, c and d respectively. AB= b a= 3i+ j+ AC = c a= j AD= d a= i j+ 3 AB. AC AD = 0 = 3( (0+ 4 + (0 = 9 9 3 Volume of tetrahedron = AB. ( AC AD = = 6 6 6 a Let the vertices A, B, C and D have position vectors a, b, c and d respectively. BC = c b= i+ j+ BD= d b= j+ BC BD= = i+ 4j 4 0 36 Area of face BCD= BC BD= + 4 + ( 4 = = 3 b The normal to the face BCD is in the direction of BC BD, i.e. in the direction i+ 4j 4 As i+4j 4 = +4 +( 4 =6 The unit vector normal to the face is 6 (i+4j 4= 3 (i+j Multiplying by also gives a vector normal to the face BCD, so (i+j is a solution. 3 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free.
6 c BA= a b= i+ 3j BA. BC BD = i+ j. i+ j = + + = Volume of tetrahedron = AB. ( AC AD = 4 = 7 ( 3 ( 4 4 4 4 7 a Let the vertices A, B, C and D have position vectors a, b, c and d respectively. (Note a = 0. Now find the length of all edges, as a tetrahedron is regular if all of its edges are the same length. AB= b a= b, so AB = AC = c a= c, so AC = + ( 3 = 3 6 3 4 AD= d a= d = + + = + + = 3 3 9 9 BC = c b= i+ 3 j, so BC = ( + ( 3 = BD, soad 3 6 3 6 3 4 d b 3 3 + = + + = 3 3 9 9 = = + +, so BD = ( + 3 6 3 6 4 8 CD= d c= j+, so CD = + = + = 3 3 3 3 3 3 All 6 edges have the same length and so the tetrahedron is regular. 0 0 6 3 4 8 = 3 0 = = = 4 3 3 3 6 3 3 Volume of tetrahedron = AB. ( AC AD = 4 = b AB. ( AC AD 8 a AB= ( i+ j+ ( i+ j = i j+ 3 AC= ( i j+ ( i+ j = i 3j+ AB AC = 3 = 7i+ 7j+ 7 3 b Area of triangle ABC 47 7 3 = AB AC = 7 + 7 + 7 = = c AO. ( AB AC = ( i j+.(7i+ 7j+ 7 = 7 4+ 7= 4 Volume of tetrahedron = AO. ( AB AC = 4 = 7 Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 3
9 a AB= b a= i j+, BC= c b= 3i j AB BC= = 5i j 7 3 BD= d b= i+, DC = c d= i j 4 BD DC= 0 = i 8j+ 4 b i AB BC= BA BC AB BC = BA BC So area of triangle ABC = BA BC = AB BC = 5i+ j+ 7 = 5+ + 49 5 3 = 75= b ii AB BC= BA BC BA BC= AB BC BD. ( BA = ( i+.( 5i+ j+ 7 = 5+ 4= 9 9 So volume of tetrahedron ABCD= BD.( BA = 6 6 0 a b c== 3 =i+j 4 5 As a= ( b c,op is perpendicular to OQ and to OR, i.e. OP is perpendicular to the plane OQR. b OP = a = + 4 = 0= 5 Area of OQR = b c = + = 5 Volume of tetrahedron = 3 base height= 3 5 5= 5 3 c Using the scalar triple product: 4 0 a.b ( c = 3 = ( 5+ 6 4(0 = + 8= 0 4 5 a.(b c is 6 volume of tetrahedron (from part b, so result verified. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 4
a OB OC= (i j 4 ( i+ 4j = 4 = 8i+ j+ 6 4 b OA. ( BA = (3i+ j+.(8i + j+ 6 = 54+ 4+ 6= 84 84 So volume in design units = OA.( OB OC = = 4 6 6 Prototype volume = volume in design units = 4 8= cm 3 3 Mass of a prototype model=.3= 7g (to the nearest gram By drawing a rough setch, the three non-parallel edges of the parallelepiped that join at the origin are given by the vectors 0.6i + 0.6j, 0.9i 0.9j, and 0.4i 0.4j.3 0.6 0.6 0 The volume in the scientist s scale = 0.9 0.9 0 = 0.6( 0.9.3 0.6(0.9.3 =.404 0.4 0.4.3 angstrom (Å = 0. nanometres, so nanometre = 0 Å, and cubic nanometre = 0 3 Å 3 So volume unit cell of crystal in cubic angstroms =.404 0 = 400Å ( s.f. 3 a The volume of the parallelepiped = AB. ( BD The volume of the tetrahedron = EC. ( EM NC Now from the diagram: EC = AB EM = BD NC= NB+ BC 6 Using these results and the fact that both the vector product and the scalar product are distributive over vector addition, this gives: Volume of the tetrahedron = AB. ( BD ( NB+ = AB. (( BD NB + ( BD as vector product distributive = AB. ( BD NB + AB. ( BD as scalar product distributive But BD NB is perpendicular to AB so AB. ( BD NB = 0 So the expression for the volume of the tetrahedron simplifies to AB ( BD. Hence the ratio of the two volumes is : b The ratio remains unchanged since the argument in part a does not use any data about N other than it lies on the line AB. Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 5
4 Split the pyramid into two tetrahedrons AEDB and CEDB so the volume of the pyramid is just the combined volume of the two tetrahedrons. AD= i+, AB= i+ j+, AE= 4i+ 0 AD. AB AE = = 6= 4 4 0 4 7 Volume of AEDB = = 4= = CB= i, CD= i j, CE= i j 0 CB. CD CE = = = 4 Volume of CEDB = 4 7 = 4 = = Hence the combined volume = 7 3 +7 3 =4 3 Challenge a Let a=a i+a j+a 3 b=b i+ j+ c=c i+c j+c 3 a a a 3 a.b ( c = b b b = a ( bc bc a ( bc bc + a ( bc bc 3 3 3 3 3 3 c c c 3 = a ( bc bc + a ( bc bc + a ( bc bc 3 3 3 3 3 a b=(a i+a j+a 3 (b i+ j+ =a a j a b +a i+a 3 b j a 3 i=(a a 3 i+(a 3 b a j+(a a b (a b.c=((a a 3 i+(a 3 b a j+(a a b.(c i+c j+c 3 =(a a 3 c +(a 3 b a c +(a a b c 3 =a ( c 3 c +a ( c b c 3 +a 3 (b c c So a.(b c=(a b.c b d.(a (b+c=(d a.(b+c =(d a.b+(d a.c =d.(a b+d.(a c =d.(a b+a c applying part a scalar product is distributive over vector addition applying part a scalar product is distributive over vector addition c Since the equality holds for any choice of vector d, it follows that a ( b+c= a b+a c Pearson Education Ltd 08. Copying permitted for purchasing institution only. This material is not copyright free. 6