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94 CHAPTER 6. VECTOR PRODUCTS

Therefore, in terms of the given vectors ui, the angular velocity vector is 120πu3. Thevelocity of the given point is∣∣∣∣∣∣∣

u1 u2 u3

0 0 120π

2 3 −1

∣∣∣∣∣∣∣=−360πu1 +240πu2

in meters per minute. Note how this gives the answer in terms of these vectors which arefixed in the body, not in space. Since ui depends on t, this shows the answer in this casedoes also. Of course this is right. Just think of what is going on with the wheel rotating.Those vectors which are fixed in the wheel are moving in space. The velocity of a point inthe wheel should be constantly changing. However, its speed will not change. The speedwill be the magnitude of the velocity and this is√

(−360πu1 +240πu2) · (−360πu1 +240πu2)

which from the properties of the dot product equals√(−360π)2 +(240π)2 = 120

√13π

because the ui are given to be orthogonal.

6.6 Vector Identities And NotationTo begin with consider u× (v×w) and it is desired to simplify this quantity. It turnsout this is an important quantity which comes up in many different contexts. Let u =(u1,u2,u3) and let v and w be defined similarly.

v×w =

∣∣∣∣∣∣∣i j k

v1 v2 v3

w1 w2 w3

∣∣∣∣∣∣∣= (v2w3− v3w2) i+ (w1v3− v1w3)j+ (v1w2− v2w1)k

Next consider u×(v×w) which is given by

u×(v×w) =

∣∣∣∣∣∣∣i j k

u1 u2 u3

(v2w3− v3w2) (w1v3− v1w3) (v1w2− v2w1)

∣∣∣∣∣∣∣ .When you multiply this out, you get

i(v1u2w2 +u3v1w3−w1u2v2−u3w1v3)+j (v2u1w1 + v2w3u3−w2u1v1−u3w2v3)

+k (u1w1v3 + v3w2u2−u1v1w3− v2w3u2)

and if you are clever, you see right away that

(iv1 +jv2 +kv3)(u1w1 +u2w2 +u3w3)− (iw1 +jw2 +kw3)(u1v1 +u2v2 +u3v3) .

Thusu×(v×w) = v (u ·w)−w (u ·v) . (6.32)

94 CHAPTER 6. VECTOR PRODUCTSTherefore, in terms of the given vectors u;, the angular velocity vector is 12073. Thevelocity of the given point isUU; U2 U30 O 120m | = —3607u + 2407U22 3 —1in meters per minute. Note how this gives the answer in terms of these vectors which arefixed in the body, not in space. Since wu; depends on f, this shows the answer in this casedoes also. Of course this is right. Just think of what is going on with the wheel rotating.Those vectors which are fixed in the wheel are moving in space. The velocity of a point inthe wheel should be constantly changing. However, its speed will not change. The speedwill be the magnitude of the velocity and this isJ (—3607u; + 2402u2) - (—3607u; + 2402u2)which from the properties of the dot product equalsvi (—3602)? + (2402)? = 120/132because the wu; are given to be orthogonal.6.6 Vector Identities And NotationTo begin with consider u x (v x w) and it is desired to simplify this quantity. It turnsout this is an important quantity which comes up in many different contexts. Let wu =(uj, ,U2,u3) and let v and w be defined similarly.i jkvUxw=l] vy vo v3 | = (v2w3—Vv3w2) t+ (wiv3 —v1W3) J+ (Viw2 — v2W1) kW, W2 W3Next consider ux (v x w) which is given bya j kux (vx w) = uy uz u3(vow3 —V¥3W2) (wiv3—V1W3) (Viw2 — v2W1)When you multiply this out, you get1 (viu2w2 + U3V1W3 — W{U2V2 — u3W1V3) +] (vourw1 + V2W3U3 — W2uy, V1 — U3W2V3)+k (ujw1Vv3 + V3W2U2 — UV, W3 — V2W3U2 )and if you are clever, you see right away that(tv, + jvo + kv3) (uywy + ugw2 + u3w3) — (tw) +Jw2 + kw3) (uv) + u2v2 + u3V3).Thusux (vx w)=v(u-w)—w(Uu-v). (6.32)