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Select Answers to Section 1.5 Exercises
of Worldwide Multivariable Calculus
i j k
1. v×w= 0 1 0 =i.
0 0 1
i j k
3. v×w= 1 0 1 =2i−3j−2k.
0 −2 3
5. v×w=0since v=−2w, i.e. they are parallel vectors.
i j k
7. v×w= 3 −2 5 =8i−3j−6k.
0 −2 1
9. The area spanned by j and k is |i| = 1, so a unit vector orthogonal to j and k is i.
11. The area spanned by v = (1,0,1) and w = (0,−2,3) is |v×w| = |(2,−3,−2)| =
√17, so a unit vector orthogonal to both v and w is
v×w 2 3 2
|v ×w| = √ ,−√ ,−√ .
17 17 17
13. The area spanned by v = (−4,2,6) and w = (2,−1,−3) is 0, because they are
parallel.
15. The area spanned by v = 3i−2j+5k and w = −2j+k is |v×w| = |(8,−3,−6)| =
√109. Thus a unit vector orthogonal to both v and w is
v×w =√8 ,−√3 ,−√6 .
|v ×w| 109 109 109
17. a×a=0.
19. a×(3b−2c)=3(a×b)−2(a×c)=(−8,−38,28).
21. (c×a)·(1,1,1) = (−7,−13,11)·(1,1,1) = −9.
23. (b×a)·(a×b)=(−2,4,−2)·(2,−4,2)=−24.
1
25. (b×a)×(b×c)=(−2,4,−2)×(1,7,1)=(18,0,−18).
27. If a = (0,0,0), b = (1,2,3), c = (1,0,0), then the area of the triangle with these
points as vertices is
√
i j k
1 1 1 13
|(b −a)×(c−a)|= 1 2 3 = |(0,3,−2)|= .
2 2 2 2
1 0 0
Astandard equation for the plane containing the triangle is then 3y −2z = 0.
29. If a = (1,0,−1),b = (2,5,3),c = (3,2,1), then the area of the triangle with these
points as vertices is
i j k √ √
1 1 1
|(b −a)×(c−a)|= 1 5 4 = |(2,6,−8)| = ( 104)/2 = 26.
2 2 2
2 2 2
Astandard equation for the plane containing the triangle is then 2(x−1)+6y −
8(z +1) = 0 or 2x+6y−8z−10=0; any non-zero scalar multiple of this is also
correct, e.g., x + 3y − 4z − 5 = 0.
31. If (x,y,z) = (1,0,1)+t(1,1,1) and p = (3,2,1), then the vectors v = (1,1,1) and
w=(3,2,1)−(1,0,1) = (2,2,0) are not parallel, and a normal vector to both v
and w is
i j k
v×w= 1 1 1 =(−2,2,0),
2 2 0
so a standard equation for the plane containing the line and the point is −2(x −
1)+2y =0, which you could simplify to −x+y +1 = 0.
33. If (x,y,z) = (4,−1,5) + t(2,0,7) and p = (1,1,1), then the vectors v = (2,0,7)
and w = (1,1,1)−(4,−1,5) = (−3,2,−4) are not parallel, and a normal vector
to both is
i j k
v×w= 2 0 7 = (−14,−13,4),
−3 2 −4
so a standard equation for the plane containing the line and the point is −14(x−
1) −13(y −1)+4(z−1)=0, i.e., −14x−13y+4z+23=0.
35. Since the rod is about the origin, d = (0,1) and the force F = (3,4) acts on the
rod at (0,1), the torque produced is
i j k
d×F= 0 1 0 = −3k Newton-meters.
3 4 0
Since the sign is negative, the torque vector points into the page.
2
37. Since the rod is about the origin, d = (1,2) and the force F = (3,4) acts on the
rod at (1,2), the torque produced is
i j k
d×F= 1 2 0 = −2k Newton-meters.
3 4 0
Since the sign is negative, the torque vector points into the page.
39. The volume of the parallepiped spanned by the vectors (1,0,1),(2,−1,2), and
(5,3,0) is
1 0 1
2 −1 2 = 5.
5 3 0
41. Since we are given four points, a = (1,0,1),a = (7,1,4),a = (1,1,1), and
1 2 3
a =(3,2,1) that are vertices of the parallelpiped (with the points a all adjacent
4 i
to a ) we are then looking for the volume of the parallelpiped spanned by the
1
vectors a −a (for i = 2,3,4). This volume is given by
i 1
6 1 3
0 1 0 =6.
2 2 0
43. In an analogous way to calculate the area of a triangle spanned by two vectors,
the volume of the tetrahedron spanned by three vectors i,j and k will be
1 0 0
1 1
0 1 0 = .
4 4
0 0 1
45. The 4-dimensional volume spanned by these vectors will be
1 0 1 0
2 −1 2 0 2 2
0 2 −1 2
= 5 3 0 + 0 5 0 =62−10=52.
0 5 3 0
2 3 4 1 2 4
1 2 3 4
3
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