Thomas Calculus 13th [Solutions]

922 Chapter 12 Vectors and the Geometry of Space
(c) d cos wcos cos ( )
| T2 | d w
; d | T
d
d sin ( ) 2
2 | 0 wcos cos ( ) 0
sin ( ) d
cos( ) 0 ;
2 2
d
d
2
wcos cos ( ) wcos cos ( ) 1
| T2
| d
;
d sin ( ) sin ( )
2 2 3
2
d
d
2
2
| T 2 | w cos 0 local minimum when
2
2
7. (a) If P( x, y, z ) is a point in the plane determined by the three points P1 ( x1 , y1 , z1),
P2 ( x2, y2, z 2 ) and
P3 ( x3 , y3, z 3),
then the vectors PP1 , PP 2 and 3 PP1 PP2 PP3 0
x1 x y1 y z1
z
x2 x y2 y z2
z 0 by the determinant formula for the triple scalar product in Section 12.4.
x3 x y3 y z3
z
(b) Subtract row 1 from rows 2, 3, and 4 and evaluate the resulting determinant (which has the same value as
the given determinant) by cofactor expansion about column 4. This expansion is exactly the determinant
in part (a) so we have all points P( x, y, z ) in the plane determined by P1 ( x1 , y1 , z1 ), P2 ( x2, y2, z 2),
and
P3 ( x3 , y3, z3).
8. Let L1 : x a1s b1 , y a2s b2 , z a3s b 3 and L2: x c1t d1, y c2t d2, z c3t d 3.
If L1 || L 2,
then for
a1 c1 b1 d1 kc1 c1 b1 d1
some k, ai
kci
, i 1, 2, 3 and the determinant a2 c2 b2 d2 kc2 c2 b2 d2
0, since the first
a3 c3 b3 d3 kc3 c3 b3 d3
column is a multiple of the second column. The lines L 1 and L 2 intersect if and only if the system
a1s c1t b1 d1
0
a2s c2t b2 d2
0 has a nontrivial solution the determinant of the coefficients is zero.
a3s c3t b3 d3
0
9. (a) Place the tetrahedron so that A is at (0, 0, 0), the point P is on the y -axis, and ABC lies in the
xy -plane. Since ABC is an equilateral triangle, all the angles in the triangle are 60 and since AP
bisects BC ABP is a 30 60 90 triangle. Thus the coordinates of P are 0, 3, 0 , the
coordinates of B are 1, 3, 0 , and the coordinates of C are 1, 3, 0 . Let the coordinates of D be
given by ( a, b, c ). Since all of the faces are equilateral triangles all the angles in each of the triangles
3
are 60 cos( ) cos(60 ) a b
DAB AD AB
1 a b
| || | (2)(2) 2
3 2 and cos( DAC) cos(60 )
AD AB
AD AC a b 3 1 a b 3 2. Add the two equations to obtain: 2b
3 4 b 2 .
| AD|| AC|
(2)(2) 2
3
Substituting this value for b in the first equation gives us: a 2 3 2 a 0. Since
3
2
2 2 2 2 2 2 2 2
2 2
| AD | a b c 2 0 c 4 c . Thus the coordinates of D are 0, 2 , .
3 3
3 3
cos cos( 2 1
DAP ) AD AP
cos 1 57.74
| AD|| AP| 2 3 3
(b) Since ABC lies in the xy -plane the normal to the face given by ABC is n1 k . The face given by
BCD is an adjacent face. The vectors
Copyright
1 2 2
2 2
DB i j k and DC i 1 j k both lie in the
3 3
3 3
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