Note that the area of the triangle with sides A
and B = -||AXB|. Fig. 5-20
i j k 5.15. If A = Aii + A
2j + Ask and B = B¿ + B
2j + Bak, prove that A x B = Ai A
2A
3.
BI BZ By
A X B = (Ati + A2j + A3k) X (J5,i + B£ + B3k)
= Aji X (B,i + B2j + B3k) + AHJ X (Bji + Eg + B3k) + A3k X (Bji + B2j + B3k) - AjBji X i + A^i X j + A^gi x k + A2B¡j X i + A2B^ x j + A2B3j x k
+ A3B,k X i + A3£2k X j + A3B3k X k
i j k
= (A2B3-A3£2)i + (A3B1-A1B8)j + (AjB2-A2B,)k = A^ A2 A3 Bi Bn B»
5.18. Find the area of the triangle with vertices at P(2,3,5), Q(4,2,-l), ¿2(3,6,4).
PQ = (4-2)i + (2-3)j + (-l-5)k = 2i - j - 6k PB = (3-2)i + (6-3)j + (4-6)k = i + 3j - k Area of triangle = £ | P Q X P R | = £ | (2i-j-6k) X (i +3j-k) |
TRIPLE PRODUCTS
5.19. Show that A • (B x C) is in absolute value equal to the volume of a paral- lelepiped with sides A, B and C. See Fig. 5-21.
Let n be a unit normal to parallelogram I, having the direction of B X C, and let h be the height of the terminal point of A above
the parallelogram I. Fig. 5-21 Volume of parallelepiped = (height A)(area of parallelogram 7)
= (A-n)(|BXC|)
= A - { | B X C | n } = A - ( B X C )
If A, B and C do not form a right-handed system, A - n < 0 and the volume = |A • (B X C)|.
5.20. If A = Aii + A2J + A3k, B = B¿ + B
2j + Bsk, C = Cii + C
2j + Cak show that
AI AZ A3
A - ( B x C ) = Bi B
2B
3 Cl C/2 C/3i J k A - ( B X C ) = A- B! BZ B3
Ci GZ C3
= (A ti + A-j + A3k) • [(B2C3 - B3C2)i + (B3C, - B^s)] + (BjC2 - B^k]
AI AZ A3
= Aí(BíCs-BaCz)+Az(B3Cl-BlCí) + AA(BlCí-B.iCl) = Bl B2 B3 Ci C% Cg
5.21. Find the volume of a parallelepiped with sides A = 3i - j, B = j + 2k, C = i + 5j + 4k.
3-1 0 By Problems 5.19 and 5.20, volume of parallelepiped = |A • (B X C)| = | 0 1 2 |
1 5 4
= |-20| = 20
5.22. Prove that A- (B x C) = (A x B) • C, i.e. the dot and cross can be interchanged.
¿1 ¿2 ¿3 Cl C2 C3
By Problem 5.20: A - ( B X C ) = B^ B2 B3 , (A X B) • C = C ' ( A X B ) = Ax Az A3
C*i C2 C3 BI B2 B3
Since the two determinants are equal, the required result follows.
5.23. Let ri = Xii + yij + Zik, r
2= xà + j/
2j + z
2k and r
s= xd + 2/sj + Zsk be the position vectors of points PI(XI, y\, Zi), P
2(£C
2, y
2, z
2) and Ps(xs,ys,zs). Find an equation for the plane passing through Pi,Pz and Pa.
See Fig. 5-22.
We assume that P1( P2 and P3 do not lie in the same straight line; hence they determine a plane.
Let r = xi + yj + zk denote the position vector of any point P(x, y, z) in the plane. Con- sider vectors PiP2 — r2 ~ ri> PI PS = r3 ~~ ri and PjP = r — rx which all lie in the plane, Thel1
or (r - rx) • (ra - rt) X (i3 - rx) = 0
In terms of rectangular coordinates this becomes Fig. 5-22
[(* - *i)i + (V- Vl)j + (Z- 2j)k] • [(*2 - Xjl + (j/2 - 2/j)j + (Z2 - 2j)k]
X [(xa - xji + (2/3 - y j j + (z3 - «!)k] = 0 or, using Problem 5.20,
X — %!
* 2 - * l
«3 — Xl
y - 3/1
Vz-Vi
2/3-2/1
Z - Z !
* 2 - « l Z3- Z ,
= 0
5.24. Find an equation for the plane passing through the points Pi (3,1, — 2), P
2(-l,2,4), P,(2,-l,l).
The position vectors of P1( P2, P3 and any point P(x, y, z) on the plane are respectively rj = 3i + j - 2k, r2 = -i + 2j + 4k, r3 = 2i - j + k, r = xi + y} + zk Then P P j ^ r - r j , P2P1 = r2 —r,, P3P, =r3 — rlt all lie in the required plane and so the required equation is (r — rj) • (r2 —1¡) X (rs — rt) = 0, i.e.,
{(*-3)i+ (2/-l)j + (z + 2)k}-{-4i + j + 6k> X {-i-2j + 3k> = 0 {(* - 3)i + (y- l)j + (z + 2)k> • {15i + 6j + 9k} = 0
1B(* - 3) + 6(j/ - 1) + 9(z + 2) = 0 or 5x + 2y + 3z = 11 Another method. By Problem 5.23, the required equation is
x-B y-1 2+2
-1-3 2-1 4 + 2 =0 or 2 - 3 - 1 - 1 1 + 2
5x + 2y + 3z = 11
5.25. If A = i + j, B = 2i-3j + k, C = 4j-3k, find (a) (AxB)xC, (6) Ax(BxC).
(a) A X B =
(6) B X C =
i J k 1 1 0 2 - 3 1 i J k 2 - 3 1
= i - j - 5k. Then (A X B) X C = i J k 1 -1 -5 0 4 - 3
= 23i + 3j + 4k.
= 5i + 6j + 8k. Then A X (B X C) = 0 4 - 3
It follows that, in general, (A X B) X C ¥- A X (B X C).
i J k 1 1 0
5 6 8 = 8i-8j + k.
0
5.29. If A = a;2 sin y i + z2 cosy j - xy2k, find dA.
Method 1.
If x = 1, y - -2, z = -1, this becomes -12i - 12j + 2k.
5.28. If $(x, y, z) - x
2yz and A = 3x
2yi + yz*j - xzk, find at the point (1, —2,1).
0A = (x*yz)(3x*yi + yz2j - xzk) - 3x*y*zi + x2y2zaj - x^yz^i Method 2. Let A = Ati + A2j + A3k, B = Bti + B2j + B3k. Then
Method 1.
5.27. Prove that
tions of u. where A and B are differentiable func-
If t represents time, these represent respectively the velocity, magnitude of the velocity, acceleration and magnitude of the acceleration at í = 0 of a particle moving along the space curve x = t3 + Zt, y = —3e~2t, z = 2 sin 5i.
(d) From (6) From
DERIVATIVES
5.26. If r = (í
3+ 2í)i-3e-
2tj + 2sin5ík, find t = 0 and give a possible physical significance.
(a)
(c)
GRADIENT, DIVERGENCE AND CURL
5.31. If A = x
2yz
3and A = xzi - y*j + 2x
2yk, find (d) div
The components d-v/dt and v2/p in the direction of T and N are called the tangential and normal components of the acceleration, the latter being1 sometimes called the centripetal acceleration. The quantities p and K are respectively the radius of curvature and curvature of the space curve.
Defining p = 1/rc, (2) becomes dT/ds = N/p. Thus from (Í) we have, as required,
where « is the magnitude of dT/ds. Now since X = dr/ds [see equation (9), page 126], we have dT/ds = eZV<fe2. Hence
from which it follows that dT/ds is perpendicular to T. Denoting by N the unit vector in the direc- tion of dT/ds, and called the principal normal to the space curve, we have
Since T has unit magnitude, we have T • T = 1. Then differentiating with respect to s, The velocity of the particle is given by v = vT. Then the acceleration is given by
where T and N are unit tangent and normal vectors to the space curve and
5.30. A particle moves along a space curve r = r(f), where t is the time measured from some initial time. If v = \dr/dt\ = ds/dt is the magnitude of the velocity of the particle (s is the arc length along the space curve measured from the initial position), prove that the acceleration a of the particle is given by
Method 2.
dA
Laplacian of <f>
5.35.
find and
Another unit normal to the surface at P is Then a unit normal to the surface at P is
5.34. Find a unit normal to the surface 2x2 + 4yz - 5z* = -10 at the point P(B, —1,2).
By Problem 5.33, a vector normal to the surface is
is perpendicular to ar and therefore to the surface.
Let r = xi + yj + zk be the position vector to any point P(x, y, z) on the surface.
Then dr = dx i + dy j + dz k lies in the plane tangent to the surface at P. But
5.33. Prove that is a vector perpendicular to the surface ¿>(x, y, z) = c, where c is a constant.
5.32. Prove (e) curl (0A) (d) div(0A)
5.38. Find equations for (a) the tangent line, (6) the normal plane to a space curve
* = f(u), y = g(u), z = h(u) at the point where u = u
0. See Fig. 5-24.
Pig. 5-23 Fig. 5-24
which in rectangular form is
(6) If r is the vector drawn from O in Pig. 5-23 to any point (x,y,e) on the normal line, then r — r0 is collinear with N0 and so
since r — r0 is perpendicular to N0. In rectangular form this is
5.37. Find equations for (a) the tangent plane and (6) the normal line to the surface F(x,y,z) = 0 at the point P(xo,y
0,Zo). See Fig. 5-23.
(a) A vector normal to the surface at P is N0 = VF\P. Then if r0 and r are the vectors drawn respectively from O to P(x0, y0, z0) and Q(x,y,z) on the plane, the equation of the plane is assuming that A has continuous second partial derivatives so that the order of differentiation is immaterial.
div curl A
5.36. Prove div curl A = 0.
Another method.
(a) If R = f(u)i + g(u)j + h(u)k, then a vector tangent to the curve C at point P is given by T0 = -T- . Then if r0 and r are the vectors drawn respectively from O to P and Q on the tangent line, we have ,
(r-r0)XT0 = (r-r0)xg p = 0 since r — r0 is collinear with T0. In rectangular form this becomes
(6) If r is the vector from O to any point (x, y, z) on the normal plane, it follows that r — r0 is perpendicular to T0. Then the required equation is
(r-r0)'T0 = ( r - r 0 ) ' = 0 or in rectangular form
f'(uo)(x - x0) + ff'(u0)(y - Vo) + h'(u0)(z - z0) = 0