VECTORS IN SPACE. LINES AND PLANES
353
40.57
40.58
40.59
40.60
40.61
40.62
40.63
40.64
40.65
Show that (A + B) • ((B + C) x (C + A)) = 2A • (B x C).
(A + B)-((B + C)x(C + A)) = (A + B)-((BXC) + (BXA) + (CXC) + (CXA)) = (A+ B)-((B x C) +
(BxA) + (CxA)) = A-(BxC) + A-(BxA) + A-(Cx A)+ B-(B x C) + B-(B x A) + B-(CxA)
=
A-(BxC) + B-(CxA), since, for all D and E, D-(DxE) = 0 and D-(ExD) = 0. Hence, we obtain
2A-(BxC), since B-(C x A) = (BxC)-A by Problem 40.55.
Prove (A x B) • (C x D) = (A • C)(B • D) - (A • D)(B • C).
A x B = (a2b, - a3b2, a3bl - atb,, atb2 - a2bt) and C x D = (c2d3 - c3d2, c3dl - ctd3, ctd2 - c2dt).
So, (A x B) • (C x D) = (a2b3 - a3b2)(c2d3 - c3d2) + (a3b, - a,fc3)(c3d1 - <:, O 2 b 3 c 2 d 3 -a 2 b 3 c 3 d 2 ~a 3 b 2 c 2 d 3 + a 3 b 2 c 3 d 2 + a 3 & 1 c 3 d, - fljVA - a t b 3 c 3 d, + a t b 3 c l d 3 + a l b 2 c l d 2 -
a l b 2 c 1 ,d l -a 2 b l c l d 2 + a 2 6,c 2 d,. On the other hand, (A-C)(B-D) - (A-D)(B-C) = (a,c, + a 2 c 2 + a 3 c 3 ) x
(b l d l + b 2 d 2 + b 3 d 3 ) - (a l d l + a 2 d 2 + a 3 d 3 )(b l c 1 + b 2 c 2 + b 3 c 3 )
=
a^.c.d, + a,b 2 c,d 2 + a l b 3 c,d 3 +
a 2 b 2 c 2 d 2 + a 2 b 3 c 2 d 3 + a 3 b l c 3 d l + a 3 b 2 c 3 d 2 + a 3 b 3 c 3 d 3 - a l b l c l d l - a l b 2 c 2 d l - a l b 3 c } d l - a 2 b 1 c l d 2 -
a 2b2c2d2 - a2b3c3d2 - a3btcld3 - a3b2c2d3 - a3b3c3d3. Hence, the two sides are equal.
Prove (A - B) x (A + B) = 2(A x B).
(A-B)x(A + B) = (Ax A) + (AxB)-(BxA)-(BxB). Since AxA = BxB = 0 and BxA =
-(A x B), the result is 2(A x B).
Show that C x (A x B) is a linear combination of A and B; namely, C x (A x B) = (C • B)A - (C • A)B.
A x B = (a2b3 - a3b2, a3br - a,b3, a,b2 - a2ft,). So, C x (A x B) = (c2(a,b2 - a^,) - c3(a3bl - a,63),
C3(a2b3 - a3b2) - c,(a,b2 - a2bt), c^b, - atb3) - c2(a2b3 - a3b2)) = (qfc, + c2b2 + c3b3) (a,, a2, a3) -
(« 1 c 1 + fl 2 c 2 + fl J c J )(ft 1 ,/> 2 ,fr,) = (C-B)A-(C-A)B.
Show that Ax(AxB) = (A-B)A-(A-A)B.
In Problem 40.60, let C = A.
Show that (A x B) x (C x D) = ((A x B) • D)C - ((A x B) • C)D.
In Problem 40.60, substitute C for A, D for B, and A x B for C.
Show that the points (0,0, 0), (a,, a 2 , a 3 ), (b^, b 2 , b 3 ), and (c t , c 2 , c 3 ) are coplanar if and only if
Let A = (a,,a 2 ,a 3 ), B = (6,, b 2 , fc 3 ), C = (c,, c 2 , c 3 ). By Problems 40.45 and 40.49, the determinant
above is equal to A • (B x C), which is equal in magnitude to the nonzero volume of the parallelepiped formed by
A, B, C, when the given points are not coplanar. If those points are coplanar, either B x C = 0, and,
therefore, A-(BxC) = 0; or BxC^O and A is perpendicular to B x C (because A is in the plane of B and
C), so that again A • (B X C) = 0.
Show that A-(BxC) = B-(CxA).
This follows at once from Problem 40.55 (exchange dot and cross on the right side).
(Reciprocal crystal lattices). Assume V,, V 2 , V 3 are noncoplanar vectors. Let
Show that K, • (K2 x K3) = 1 /[V, • (V2 x V3)].
Let c=V,-(V, XV,). Then
353
40.57
40.58
40.59
40.60
40.61
40.62
40.63
40.64
40.65
Show that (A + B) • ((B + C) x (C + A)) = 2A • (B x C).
(A + B)-((B + C)x(C + A)) = (A + B)-((BXC) + (BXA) + (CXC) + (CXA)) = (A+ B)-((B x C) +
(BxA) + (CxA)) = A-(BxC) + A-(BxA) + A-(Cx A)+ B-(B x C) + B-(B x A) + B-(CxA)
=
A-(BxC) + B-(CxA), since, for all D and E, D-(DxE) = 0 and D-(ExD) = 0. Hence, we obtain
2A-(BxC), since B-(C x A) = (BxC)-A by Problem 40.55.
Prove (A x B) • (C x D) = (A • C)(B • D) - (A • D)(B • C).
A x B = (a2b, - a3b2, a3bl - atb,, atb2 - a2bt) and C x D = (c2d3 - c3d2, c3dl - ctd3, ctd2 - c2dt).
So, (A x B) • (C x D) = (a2b3 - a3b2)(c2d3 - c3d2) + (a3b, - a,fc3)(c3d1 - <:, O 2 b 3 c 2 d 3 -a 2 b 3 c 3 d 2 ~a 3 b 2 c 2 d 3 + a 3 b 2 c 3 d 2 + a 3 & 1 c 3 d, - fljVA - a t b 3 c 3 d, + a t b 3 c l d 3 + a l b 2 c l d 2 -
a l b 2 c 1 ,d l -a 2 b l c l d 2 + a 2 6,c 2 d,. On the other hand, (A-C)(B-D) - (A-D)(B-C) = (a,c, + a 2 c 2 + a 3 c 3 ) x
(b l d l + b 2 d 2 + b 3 d 3 ) - (a l d l + a 2 d 2 + a 3 d 3 )(b l c 1 + b 2 c 2 + b 3 c 3 )
=
a^.c.d, + a,b 2 c,d 2 + a l b 3 c,d 3 +
a 2 b 2 c 2 d 2 + a 2 b 3 c 2 d 3 + a 3 b l c 3 d l + a 3 b 2 c 3 d 2 + a 3 b 3 c 3 d 3 - a l b l c l d l - a l b 2 c 2 d l - a l b 3 c } d l - a 2 b 1 c l d 2 -
a 2b2c2d2 - a2b3c3d2 - a3btcld3 - a3b2c2d3 - a3b3c3d3. Hence, the two sides are equal.
Prove (A - B) x (A + B) = 2(A x B).
(A-B)x(A + B) = (Ax A) + (AxB)-(BxA)-(BxB). Since AxA = BxB = 0 and BxA =
-(A x B), the result is 2(A x B).
Show that C x (A x B) is a linear combination of A and B; namely, C x (A x B) = (C • B)A - (C • A)B.
A x B = (a2b3 - a3b2, a3br - a,b3, a,b2 - a2ft,). So, C x (A x B) = (c2(a,b2 - a^,) - c3(a3bl - a,63),
C3(a2b3 - a3b2) - c,(a,b2 - a2bt), c^b, - atb3) - c2(a2b3 - a3b2)) = (qfc, + c2b2 + c3b3) (a,, a2, a3) -
(« 1 c 1 + fl 2 c 2 + fl J c J )(ft 1 ,/> 2 ,fr,) = (C-B)A-(C-A)B.
Show that Ax(AxB) = (A-B)A-(A-A)B.
In Problem 40.60, let C = A.
Show that (A x B) x (C x D) = ((A x B) • D)C - ((A x B) • C)D.
In Problem 40.60, substitute C for A, D for B, and A x B for C.
Show that the points (0,0, 0), (a,, a 2 , a 3 ), (b^, b 2 , b 3 ), and (c t , c 2 , c 3 ) are coplanar if and only if
Let A = (a,,a 2 ,a 3 ), B = (6,, b 2 , fc 3 ), C = (c,, c 2 , c 3 ). By Problems 40.45 and 40.49, the determinant
above is equal to A • (B x C), which is equal in magnitude to the nonzero volume of the parallelepiped formed by
A, B, C, when the given points are not coplanar. If those points are coplanar, either B x C = 0, and,
therefore, A-(BxC) = 0; or BxC^O and A is perpendicular to B x C (because A is in the plane of B and
C), so that again A • (B X C) = 0.
Show that A-(BxC) = B-(CxA).
This follows at once from Problem 40.55 (exchange dot and cross on the right side).
(Reciprocal crystal lattices). Assume V,, V 2 , V 3 are noncoplanar vectors. Let
Show that K, • (K2 x K3) = 1 /[V, • (V2 x V3)].
Let c=V,-(V, XV,). Then