4 VECTORS. 4.0 Introduction. Objectives. Activity 1


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1 4 VECTRS Chpter 4 Vectors jectives fter studying this chpter you should understnd the difference etween vectors nd sclrs; e le to find the mgnitude nd direction of vector; e le to dd vectors, nd multiply y sclr; pprecite the geometricl representtion of vectors; e le to evlute nd interpret the sclr nd vector products. 4.0 Introduction Some physicl quntities, such s temperture or time, re completely specified y numer given in pproprite units e.g. 20 C, hours. Quntities of this sort, which hve only mgnitude, re referred to s sclrs; wheres quntities for which it is lso necessry to give direction s well s mgnitude re clled vectors. Exmples include wind velocity force displcement. In this text we will print vectors in old type; for exmple,, etc. Unfortuntely, when writing vectors you cnnot distinguish them from sclrs in this wy. The stndrd wy of writing vectors is to underline them; for exmple ~, ~, etc. It is very importnt for you to conform to this nottion. lwys rememer to underline your vectors, otherwise gret confusion will rise! ctivity 1 Mke list of sclr nd vector quntities, distinguishing etween them. 89
2 Chpter 4 Vectors 4.1 Vector representtion vector cn e represented y section of stright line, whose length is equl to the mgnitude of the vector, nd whose direction represents the direction of the vector. ll the lines shown opposite hve the sme mgnitude nd direction, nd so ll represent the sme vector. Sometimes the nottion is used to represent the vector. Modulus The modulus of vector is its mgnitude. It is written s, nd is equl to the length of the line representing the vector. Equl vectors The vectors nd re equl if, nd only if, = nd nd re in the sme direction. How re the vectors shown opposite, = nd = PQ, relted? Q P Since =, nd they re in opposite directions, we sy tht =. Multipliction of vector y sclr If λ is positive rel numer, then λ is vector in the sme direction s nd is of mgnitude λ. If λ = 0, then λ is 2 the zero vector 0, nd if λ < 0 then λ is the vector in the opposite direction to nd of mgnitude λ. Does the definition mke sense when λ = 1? 90
3 Chpter 4 Vectors ctivity 2 Drw ny vector on sheet of pper, nd then lso drw () () (c) (d) 2, 3, , 1 2 c ctivity 3 h d The digrm opposite shows numer of vectors. Express vectors c, d, e, f, g nd h in terms of or. 4.2 Vector ddition f e g If the sides nd C of tringle C represent the vectors p nd q, then the third side, C, is defined s the vector sum of p nd q; tht is = p, C = q C = p+ q p q p + q C This definition of vector ddition is referred to s the tringle lw of ddition. You cn then sutrct vectors, for simply mens + ( ). For exmple = C C. s n exmple, consider the displcement vectors. If you wlk in the direction North for 5 miles, then Est for 10 miles, you cn represent these two displcements s vectors, nd, s shown opposite. The vector ddition of this indeed is the finl displcement from. nd is, nd 5 10 Exmple In the tringle C, represents, C represents. If D is the midpoint of, express C, C nd DC in terms of nd. 91
4 Chpter 4 Vectors C = + C = + C C = C = DC = D + C = D + C ) = 1 2 = ( +) (since D is the midpoint of D C Exmple C CD is prllelogrm. represents nd C represents. If M is the midpoint of C, nd N is the midpoint of D, find M nd N in terms of nd, nd hence show tht M nd N re coincident. D C = + C = + so M = 1 ( 2 + ). Similrly, D = + D = + C = +, so N = 1 ( 2 + ) nd N = + N = + 1 ( 2 + ) = = 1 2 ( + ) = M. Since N = M, N nd M must e coincident. 92
5 Chpter 4 Vectors Exercise 4 1. Four points,,, nd C re such tht = 10, = 5, C = Find nd C in terms of nd nd hence show tht, nd C re colliner. 4. C is tetrhedron with =, =, C = c. C 2. CD is qudrilterl. Find single vector which is equivlent to () + C Hence deduce tht () C + CD + C + CD + D = If, nd c re represented y in the cue shown, find terms of, nd c, the vectors represented y other edges. (c) + C + CD, D G F nd F in H C E Find C, nd C hence show tht + C + C = 0. in terms of, nd c nd 5. In regulr hexgon CDEF, = nd C =. Find expressions for DE D in terms of nd., DC, D nd 6. Given tht + C = + D, show tht the qudrilterl CD is prllelogrm. D 4.3 Position vectors In generl, vector hs no specific loction in spce. However, if =, where is fixed origin, then is referred to s the position vector of, reltive to. Is the position vector unique? The use of position vectors for solving geometry prolems hs lredy een illustrted in the previous section. Here, you will see mny more exmples of this type, ut first n importnt result. Theorem If nd re points with position vectors nd nd point C divides nd in the rtio λ:µ, then C hs position vector µ + λ λ + µ ( ) 93
6 Chpter 4 Vectors Proof If C divides in the rtio λ :µ, then C = λ ( λ + µ )? C (nd C = µ ( λ + µ ) ) Hence since C C = nd λ ( λ + µ ) re in the sme direction. Thus C λ = λ + µ ( ) = λ ( λ + µ ) ( ) nd C = + C = + λ ( λ + µ ) ( ) = ( λ + µ λ )+λ λ +µ ( ) = µ+ λ, s required. ( λ + µ ) Note tht, in the specil cse λ = µ, when C is the midpoint of, then C = 1 ( 2 + ) C nd this result hs lredy een developed in the lst section. Exmple In the tringle, = nd =. If C divides the line in the rtio 1:2 nd D divides the line in the rtio 1:2, find DC nd hence show tht DC is prllel to. C D 94
7 Chpter 4 Vectors From the result in the theorem ove C = 2+ 3 = Clerly D = 1 3, so DC = D + C = = 2 3 Since DC is multiple of, it is in the sme direction s ; tht is, DC is prllel to. Exmple In the figure opposite, X nd Y re the midpoints of nd respectively. If = nd = find the position vector of the point Z, the intersection of X nd Y. X Z X = 1 nd Y = Thus X = X + Y = = 1 2 Since Z lies on X, for some sclr numer, sy t, XZ = t X = t 1 2 nd Z = X + XZ = t 1 2 = 1 2 ( 1 t)+t. 95
8 Chpter 4 Vectors Similrly, for some sclr numer s, using the fct tht Z lies on Y gives Z = s+ 1 2 ( 1 s). ut these vectors re equl, so equting coefficients of nd, 1 ( 2 1 t )= s 2s+t =1 t = 1 ( 2 1 s ) s+2t =1 s =t = 1 3. Hence Z = 1 3 ( + ) Exmple 4 1. In the digrm T = 2T = = C = 2 () Find in terms of nd (i) (ii) T (iii) C (iv) T (v) TC. () From (iv) nd (v), wht cn you deduce out, T nd C? 2. From n origin the points, nd C hve position vectors, nd 2 respectively. The points, nd re not colliner. The midpoint of is M, nd the point of trisection of C nerer to is T. Drw digrm to show,,, C, M, T. Find, in terms of nd, the position vectors of M nd T. Use your results to prove tht, M nd T re colliner, nd find the rtio in which M divides T. 3. The vertices, nd C of tringle hve position vectors,, c respectively reltive to n origin. The point P is on C such tht P:PC = 3:1; the point Q is on C such tht CQ: Q = 2:3; the point R is on produced such tht R: R = 2:1. T C 2 The position vectors of P, Q nd R re p, q nd r respectively. Show tht q cn e expressed in terms of p nd r nd hence, or otherwise, show tht P, Q nd R re colliner. Stte the rtio of the lengths of the line segments PQ nd QR. 4. The points, nd C hve position vectors,, c respectively referred to n origin. () Given tht the point X lies on produced so tht : X = 2:1, find x, the position vector of X, in terms of nd. () If Y lies on C, etween nd C so tht Y: YC = 1:3, find y, the position vector of Y, in terms of nd c. (c) Given tht Z is the midpoint of C, show tht X, Y nd Z re colliner. (d) Clculte XY: YZ. 5. The position vectors of three points, nd C reltive to n origin re p, 3q p, nd 9q 5p respectively. Show tht the points, nd C lie on the sme stright line, nd stte the rtio : C. Given tht CD is prllelogrm nd tht E is the point such tht D = 1 3 DE, find the position vectors of D nd E reltive to. 96
9 Chpter 4 Vectors 4.4 Components of vector In this section, the ide of unit vector is first introduced. This is vector which hs unit mgnitude. So if is ny vector, its mgnitude is, nd (provided tht 0) the vector â = is unit vector, prllel to. Introducing crtesin xes xyz in the usul wy, unit vectors in the direction x, y nd z re represented y z i, j nd k. Suppose P is ny point with coordintes (x, y, z) reltive to s illustrted. i k j y Wht is the mgnitude nd direction of the vector x = xi? Clerly = xi ( = x) x z D nd + = xi+ C = xi+ yj nd + + P = xi+ yj+ D = xi+ yj+ zk Wht is the vector + + P? x P (x, y, z) r k j i C y Thus r= P = xi+ yi+ zk This vector is often written s 3 1 column mtrix x r= y z nd the nottion r= ( x, y, z) is lso sometimes used. 97
10 Chpter 4 Vectors ctivity 4 y For vectors with two dimensions, using unit vector i nd j s shown opposite, express,, c, d nd e in terms of i nd j. j c From the ctivity ove, you should hve shown tht, for exmple, c= i+ jnd e= 2i+ j. Then the vector c+ e is simply 3i+ 2j: dding vectors in form is just mtter of dding components. For vectors in 2 dimensions, in generl r= P = xi+ yj f i d e x nd its mgnitude is given y the length P, where P = x 2 + y 2 r = x 2 + y 2 y y P (x, y) Exmple If = i+ 3j, = 2i 5j, c= 2i+4j, find j x () the component form of the vectors (i) + (ii) + c (iii) (iv) + c (v) 3+ 2; i x () the mgnitude of the vectors in (); (c) unit vectors in the direction of + nd + c. () (i) + = ( i+ 3j)+ ( 2i 5j)=3i 2j (ii) + c= ( 2i 5j)+ ( 2i+4j)=0i j= j (iii) = ( i+ 3j) ( 2i 5j)= i+8j (iv) + c= ( + ) c=3i 2j ( 2i+4j)=5i 6j (v) 3+ 2 = 3( i+ 3j)+22i 5j ( )=3i+9j+4i 10 j= 7i j () (i) + = ( 2) 2 = 9+4 = 13 (ii) + c = 1 (iii) = ( 1) = 65 (iv) + + c = 25 + ( 6) 2 = 61 98
11 Chpter 4 Vectors (v) 3+ 2 = ( 4) 2 = 65 (c) If n = +, ˆn = 1 ( 3i 2j 13 )= 3 13 i 2 13 j If n = + c, ˆn = 1 ( 1 j)= j For vectors in 3 dimensions, the position vector of the point P with coordintes (x, y, z) is given y r= xi+ yj+ zk nd its mgnitude is given y r = x 2 + y 2 + z 2 Why is this result true? You cn perform lgeric opertions in the usul wy; for exmple, if thus = 3i+ 2j+ k = i 2j+ k + = ( 3i+ 2 j+ k)+ ( i 2j+k) = 4i+ 2k = ( 3i+ 2 j+ k) ( i 2j+k) = 2i+ 4j = = 14 = ( 2 2 )+1 2 = 6 Note tht two vectors, nd re prllel if = λ for some nonzero λ ; furthermore, if λ > 0 they re ctully in the sme direction. lso note tht two vectors = 1 i+ 2 j+ 3 k nd = 1 i+ 2 j+ 3 k re equl if, nd only if, their components re equl, i.e. 1 = 1, 2 = 2, 3 = 3. 99
12 Chpter 4 Vectors ctivity 5 If = 2i+ j k, = i 2j+ 3k, which of the vectors elow re prllel to or? (i) (iii) (v) 2i j+ k (ii) 5i 10 j+15k 4i 2j 2k (iv) 6i+ 3j 3k 2i+ 4j 6k (vi) 4i 2j+ 2k Exercise 4C 1. Write in the form xi+ yj+ zk, the vectors represented y P if P is the point () (1, 1, 1) () (2, 1, 1) (c) (1, 1, 0) 2. P represents the vector r. Write down the coordintes of P if () r= 3i 4 j+ k () r= i+ 2 j k (c) r= 4k 3. Find the mgnitude of the vectors () = 6i+ 2 j+ 3k () = 2i j 2k 4. If = 2i+ 5j k, = i+ j+ 2k, c= 2i+3j k find () + (d) 5 +11c () ( + ) + c (c) + ( + c) 5. Show tht the points (4, 2, 16), (0, 10, 4 ) nd C( 6, 22, 14) re colliner. (c) c= + (d) d = lso find unit vectors in the direction of nd. 4.5 Products of vectors The 'lger' of vectors hs een developed in previous sections. You cn dd nd sutrct vectors, multiply vector y sclr ( λ), ut s yet not 'multiply' vectors. There re, in fct, two wys of multiplying vectors: one, the sclr product leding to sclr quntity; the other, the vector product, eing vector. Sclr product For ny two vectors nd, the sclr product, denoted y. is defined y. = cosθ Here is the modulus of, is the modulus of, nd θ is the ngle etween the direction of the two vectors. (Some texts refer to the sclr product s the 'dot' product, nd you sy ' dot ' for.) 100
13 Chpter 4 Vectors Exmple If = 2i, = 5j nd c= i+ j, find (). ().c (c).c (). = cos90 = = 0 (). c= c cos45 = ( since c = 2 ) 2 θ = 2 (c). c= c cos45 y = = 5 From the definition of the sclr product: (i) If nd re perpendiculr (s in () ove), then θ = 90 nd cosθ = 0, which gives. = 0. j c (ii) If, for nonzero vectors nd,. = 0, then cosθ = 0 cosθ = 0, since 0, 0 ; then θ = 90 nd nd re perpendiculr. i x To summrise, for nonzero vectors nd. = 0, perpendiculr lso it is cler tht. = 2 ctivity 6 Evlute the sclr products () () (c) (d) i.i, i. j, i.k j.i, j. j, j.k k.i, k. j, k.k ( i+ j).j, ( 2i+k).k 101
14 Chpter 4 Vectors Check, in (d), tht for exmple, ( 2i+ k).k=2i.k+k.k ssuming tht the sclr product lwys ehves in this nturl wy, deduce formul for. when nd re expressed in component form = 1 i+ 2 j+ 3 k, = 1 i+ 2 j+ 3 k You should hve found in ctivity 6 tht i.i= j. j= k.k = 1, (i, j, k re unit vectors) i. j= j.k = k.i= 0 (i, j, k re mutully perpendiculr) So if nd re expressed in component form So. = ( 1 i+ 2 j+ 3 k). ( 1 i+ 2 j+ 3 k) = 1 ( 1 i.i+ 2 i. j+ 3 i.k ) + 2 ( 1 j.i+ 2 j. j+ 3 j.k ) + 3 ( 1 k.i+ 2 k. j+ 3 k.k) = (using the results ove). = Exmple If = 2i+ j+ 3k, = 3i+j 2k, find. nd the cosine of the ngle etween nd.. = ( 2i+ j+ 3k). ( 3i+ j 2k) = 2 ( 3) ( 2)= 6+1 6= 11 So. = cosθ = cosθ = 11 cosθ =
15 Chpter 4 Vectors Exmple Show tht the vectors, = i+ 2 j k nd = 2i 2 j 2k, re perpendiculr.. = ( i+ 2j k).2i 2j 2k ( ) = ( 2)+ ( 1) ( 2) = = 0 Hence vectors nd re perpendiculr. ctivity 7 For the vector x = 3i+ 2j, y = i+ mj, determine the vlues of m for which () x is perpendiculr to y () x is prllel to y (c) the ngle etween x nd y is 30. Exmple If = 3i j+ 2k nd = mi 2j 3k, find the vlue of m for which nd re perpendiculr.. = ( 3i j+ 2k). ( mi 2j 3k) = 3m + ( 1) ( 2)+2( 3) = 3m = 3m 4 = 0 m = 4 3 So nd re perpendiculr when m =
16 Chpter 4 Vectors Vector product For ny two vectors, nd, the vector product, denoted y (or ) is defined y n turn from to = sinθ ˆn Here is the mgnitude of, is the mgnitude of, nd ˆn is unit vector, perpendiculr to oth nd nd in the sense of direction of liner motion when screw turns from to s illustrted. In the figure if nd re in horizontl plne, then n is verticl. direction of screw motion This implies tht, nd ˆn form righthnded system similr to the i, j, k system. n Wht is the mgnitude of the vectors nd? Wht is the direction of the vector? To follow the direction of screw's motion turning from to gives the direction ˆn, tht is n = =. sin θ ( ˆn ) n turn from to So + = 0 nd the vector product is not, in generl, commuttive ( ). y Exmple If = 2i, = 5i nd c= i+ j, find () () (c) c (d) () = 2 5 sin90 k = 10k (k is perpendiculr to nd ) () c= 5 2 sin 45 ( k ) = 5k j c (c) c= 2 2 sin 45 ( k )=2k i x (d) = 2 2 sin 0 =0 =0 In similr wy, you cn see tht 104 i j= i j sin 90 k = k (since k is perpendiculr to i nd j, nd i, j, k form righthnded system).
17 Chpter 4 Vectors ctivity 8 Determine ll the vector products () i i, i j, i k () j i, j j, j k (c) k i, k j, k k You should hve found tht wheres nd i j= k, j k = i, k i= j j i= k, k j= i, i k = j i i= j j= k k = 0 gin ssuming tht ddition nd sutrction ehve in nturl wy, you cn use these results to find formul for in terms of their components. If then it cn e shown tht ctivity 9 = 1 i+ 2 j+ 3 k, = 1 i+ 2 j+ 3 k = ( )i+( )j+( )k Prove the formul ove for. Writing out n rry i j k work out n esy wy of rememering the formul for. Exmple If = i+ j+ k, = 2i+ 3j k, find. 105
18 Chpter 4 Vectors = ( i+ j+ k) ( 2i+3j k) = i ( 2i+ 3j+ k) j2i+3j k ( )+k ( 2i+3j k) = 20+ 3k ( j)+ ( 2k)+30 i+2j 3i 0 = 4i+3j+k lterntively you cn quickly evlute the vector product using the formul from ctivity 9; this gives = ( 1 3)i+ ( 2 ( 1) )j+( 3 2)k = 4i+3j+k Note tht if two vectors nd re prllel (or ntiprllel) then θ = 0 or π, nd = 0 ˆn = 0 Conversly, for nonzero vectors nd, Hence = 0 sin θ = 0 θ = 0, π = 0 = 0 or = 0 or, prllel Exmple If = i 3j+ 2k nd = 2i+6j 4k, find. Wht cn you sy out nd? = ( i 3j+ 2k) ( 2i+6j 4k) = i ( 2i+6j 4k) 3j ( 2i+6j 4k)+2k ( 2i+6j 4k) = 20+6k+4j 6k i 4j 12i 80 = 0 Hence nd re prllel. In fct you cn redily see tht =
19 Chpter 4 Vectors Exercise 4D 1. If = 2i+ j 2k nd = 3i+4k, find (). () the cute ngle etween these vectors (to the nerest degree) (c) unit vector which is perpendiculr to oth nd. 2. For nd in Question 1, find. Use this to find the ngle etween these vectors (to the nerest degree). 5. Given the vectors u = 3i+ 2 j nd v = 2i+ λj, determine the vlue of λ so tht () u nd v re t right ngles () u nd v re prllel (c) the cute ngle etween u nd v is The ngle etween the vectors i+ j nd i+ j+ λk is 45. Find the possile vlues of λ. 7. Given tht = 2i+ k, = i 2 j+ 3k clculte 3. Let = i 2 j+ k, = 2i+ j k. Given tht c= λ+ µ nd tht c is perpendiculr to, find the rtio of λ () the sclr product. () the vector product to µ. 8. The vectors u nd v re given 4. Find the vlue of λ for which the vectors 2i 3j+ k nd 3i+ 6 j+ λk re perpendiculr. y u = 2i j+ 2k, v = pi+ qk. Given tht u v = i+ sk, find p, q nd s. Find lso the cosine of the ngle etween u nd v. 4.6 pplictions In Chpter 5 you will see how vectors cn e used to solve prolems in 3dimensionl spce concerned with lines nd plnes. Using vectors for these prolems is very convenient ut it is not the principl ppliction of vectors, which is for solving prolems in mechnics. These pplictions, nd tht of vector clculus to prolems in fluid mechnics, re eyond the scope of this text, ut if you pursue mthemtics in Higher Eduction you will pprecite their importnce. Here we look t some simpler pplictions. Projection of vector Let P e the foot of the perpendiculr from to the line. is clled the projection of onto the line. P Note tht P = cosθ, where = nd θ is ngle P P If i is unit vector in the direction, then.i= i cosθ = cosθ = P So Projection of onto the line =.i 107
20 Chpter 4 Vectors Exmple The force F = 10ĉ, where ĉ is unit vector in direction mking n ngle of 45 with ech of the positive x nd y xes. Find the projection of F on the xxis. y F ĉ = 1 2 i+ j ( ) j 45 o x So F = 10 ( 2 i+ j) i nd F.i= 10 ( 2 i+ j).i = = 10 2 re of tringle For the tringle, let =, = So = sin θ θ P = ( sinθ) = P ( = se height ) So re of = 1 2 Exmple If is the point (5, 0), is the point (3, 0), find the re of the tringle. = 5i, = 3i+ 6j re of tringle = 1 2 = 1 2 5i ( 3i+ 6j) = k = k = =15 108
21 Chpter 4 Vectors Exmple The tringle C is defined y the points (0, 1, 2), (1, 5, 5) nd C(2, 3, 1). Find the re of C. = ( 1, 4, 3) nd C = ( 2, 2, 1). So you cn think of clculte C s s i+ 4j+ 3k nd C s 2i+ 2j k nd ( 4. ( 1) 3.2)i+ ( ( 1) )j+( )k or 10i+ 7j 6k. Hence re of C = 1 2 ( 10, 7, 6) = = Work done y force Work is done when force moves prticle through distnce. If F is the constnt force eing pplied, nd the prticle is moved from to where = d, then work done = F.d Exmple R lock slides down n inclined plne from to. Ignoring friction, the forces cting on the lock re its weight, W nd norml rection R. Clculte the work done y the forces in terms of h. h W W = Wj, d = hj+i W.d = Wj. ( hj+i)=wh nd R.d = 0 since R nd d re perpendiculr. So the work done is simply Wh. 109
22 Chpter 4 Vectors 4.7 Miscellneous Exercises 1. Find the sum of the vectors 2i+ j k, i+ 3j+ k, 3i+ 2 j. 2. Find the mgnitude of the vector = 3i 2 j+ 6k. 3. If ( + 2)i+ ( 1)j nd ( 1)i j re equl vectors, find the vlues of nd. 4. If λi 4 j is prllel to 2i 6 j, find the vlue of λ. 5. Find the unit vector in the direction of 2i j+ 2k. 6. Find the vector with mgnitude three nd prllel to 6i 3j+ 2k. 7. If C = 4i+14 j 5k, = i+ 2 j+ 7k, nd = 2i+ 6 j+ 37k, show tht the vectors C, C re prllel. Hence deduce tht the points, nd C re colliner. 8. QP = p, R = 3p, Q = q. M is the midpoint of QR. () Express P nd RQ in terms of p nd q. () Express MQ S in terms of p nd q. (c) If S lies on QP produced so tht QS = k QP, express MS in terms of p, q nd k. (d) Find the vlue of k if MS is prllel to Q. 9. Show tht 3i+ 7j+ 2k is perpendiculr to 5i j 4k. 10. The points, nd C hve coordintes (2, 1, 1), (1, 7, 3) nd ( 2, 5, 1) respectively. Find the re of the tringle C. 11. If L, M, N nd P re the midpoints of D, D, C nd C respectively, show tht LM is prllel to NP. P R M Q 12. The position vectors of points P nd R re 2i 3j+ 7k nd 4i+ 5j+ 3k respectively. Given tht R divides PQ in the rtio 2:1, find the position vector of Q if () R divides PQ internlly () R divides PQ externlly. 13. Given tht = i+ j, = 5i+ 7j, find the position vectors of the other two vertices of the squre of which nd re one pir of opposite vertices. 14. Given tht p = t 2 i+ ( 2t +1)j+ k nd ( )k where t is sclr q = ( t 1)i+ 3tj t 2 + 3t vrile, determine () the vlues of t for which p nd q re perpendiculr. () the ngle etween the vectors p nd q when t = 1, giving your nswer to the nerest 0.1. (E) 15. The point P hs position vector ( 1+ µ )i+ ( 3 2µ )j+ ( 4 + 2µ )k where µ is vrile prmeter. The point Q hs position vector 4i+ 2 j+ 3k. () The points P 0 nd P 1 re the positions of P when µ = 0 nd µ = 1 respectively. Clculte the size of ngle P 0 QP 1, giving your nswer to the nerest degree. () Show tht PQ 2 = ( 3µ 1) nd hence, or otherwise, find the position vector of P when it is closest to Q. (E) 16. Referred to fixed origin, the points, nd C hve position vectors i 2j+ 2k, 3i k nd i+ j+ 4k respectively. Clculte the cosine of the ngle C. Hence, or otherwise, find the re of the tringle C, giving your nswer to three significnt figures. (E) 110
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