12.4
1. a=<2,3,0> b=<1,0,5> axb = (3(5)-(0)(0))-(2(5)-(0)(1))+(2(0)-(3)(1)) = <15,-10,-3>
(axb)·a=30-30+0= 0 so it is orthogonal to a
(axb)·b=15+0-15= 0 so it is orthogonal to b
2. a=<4,3,-2> b=<2,-1,1> axb = (3(1)-(-2)(-1))-(4(1)-(-2)(2))+(4(-1)-(3)(2)) = <1,-8,-10>
(axb)·a=4-24+20= 0 so it is orthogonal to a
(axb)·b=2+8--10= 0 so it is orthogonal to b
3. a=2j-4k b=-i+3j+k axb = (2(1)-(-4)(3))-(0(1)-(-4)(1))+(0(3)-(2)(-1)) = <14,4,2>
(axb)·a=0+8-8= 0 so it is orthogonal to a
(axb)·b=-14+12+2= 0 so it is orthogonal to b
4. a=3i+3j-3k b=3i-3j+3k axb = (3(3)-(-3)(-3))-(3(3)-(-3)(3)+(3(-3)-(3)(3)) = <0,-18,-18>
(axb)·a=0-54+54= 0 so it is orthogonal to a
(axb)·b=-0+54-54= 0 so it is orthogonal to b
5. a=(1/2)i+(1/3)j+(1/4)k b=i+2j-3k axb = ((1/3)(-3)-(1/4)(2))-((1/2)(-3)-(1/4)(1)+((1/2)(2)-(1/3)(1)) =
<-(3/2),(7/4),(2/3)>
(axb)·a=(-3/4)+(7/12)+(2/12)= 0 so it is orthogonal to a
(axb)·b=(-3/2)+(14/4)-2= 0 so it is orthogonal to b
6. a=ti+costj+sintk b=i-sintj+costk axb = (cost(cost)-(sint)(-sint))-(t(cost)-(sint)(1)+(t(-sint)-(cost)(1))
= <1,sint-tcost,-tsint-cost>
(axb)·a=t+sintcost-tcostcost-sintsint-costsint= 0 so it is orthogonal to a
(axb)·b=1-sintaint+tcostsint-tsintcost-costcost= 0 so it is orthogonal to b
7. a=<t,1,1/t> b=<t^2,t^2,1> axb = (1(1)-(1/t)(t^2))-(t(1)-(1/t)(t^2))+(t(t^2)-(1)(t^2)) =
<1-t,0,t^3-t^2>
(axb)·a=t-t^2+t^2-t= 0 so it is orthogonal to a
(axb)·b=t^2-t^3+t^3-t^2= 0 so it is orthogonal to b
8. a=i-2k b=j+k axb= (0(1)-(-2(1))-(1(1)-(-2)(0))+(1(1)-(0)(0)) =
<2,-1,1>
9. (ixj)xk ixj=k jxk=i kxi=j
10. Kx(i-2j) kx(i+(-2)j
(k)xk=0
(kxi)+(-2)(kxj)
j+(-2)(-i) = j+2i=2i+j
11. (j-k)x(k-i) = jx(k-i)-kx(k-i) = jxk-jxi-kxk+kxi = i-(-k)-0+j = i+j+k
12. (i+j)x(i-j) = ix(i-j) + jx(i-j) = ixi-ixj+jxi-jxj = 0-k+(-k)+0 = -2k
13. Bxc Is a vector, so a·(bxc) is a scalar and meaningful
b. b·c is a scalar, so , since the cross product is defined for only two vectors, ax(b·c) is meaningless
c. bxc is a vector, and ax(bxc) results in a vector, so it has meaning
d. b·c is a scalar, so , since the dot product is only defined for two vectors, a·(b·c) is meainingless.
e. Since (a·b) and (c·d) are both scalars, (a·b)x(c·d) is meaningless
f. axb and cxd are both vectors, so (axb) ·(cxd) is a scalar and meaningful
14. /uxv/ = /u//v/sinθ = 4(5)sin45 = 20((2^(1/2))/2) = 10(2^(1/2)) using the right hand rule uxv goes out
of the page
15. . /uxv/ = /u//v/sinθ = 12(16)sin120 = 12(16)((3^(1/2))/2) = 96(2^(1/2)) using the right hand rule and
using the acute angle uxv goes into the page
16. a. /axb/ = /a//b/sinθ = 3(2)sin90 = 6
b. because axb is always orthogonal to b which lies on the z-axis, vector axb must be in the xy plane. So,
the z-component must be 0. From the right hand rule, we know that x-component is positive and ycomponent is negative.
17. a=<2,-1,3> b=<4,2,1> axb = (-1(1)-(3)(2))-(2(1)-(3)(4))+(2(2)-(-1)(4)) = <-7,10,8>
bxa = <7,-10,-8>
18. a=<1,0,1> b=<2,1,-1> axb = (0(-1)-(1)(1))-(1(-1)-(1)(2))+(1(1)-(0)(2)) = <-1,3,1>
c= <0,1,3> (axb)xc = (3(3)-(1)(1))-(-1(3)-(1)(0))+(-1(1)-(3)(0)) = <8,3,-1>
bxc = (1(3)-(-1)(1))-(2(3)-(-1)(0))+(2(1)-(1)(0)) = <4,-6,2>
ax(bxc) = (0(2)-(1)(-6))-(1(2)-(1)(4))+(1(-6)-(0)(4)) = <6,2,-6>
(axb)xc does not equal ax(bxc)
19. a=<3,2,1> b=<-1,1,0> axb = (2(0)-(1(1))-(3(0)-(1)(-1))+(3(1)-(2)(-1)) = <-1,-1,5>
/axb/ = 27^(1/2) = 3*(3^(1/2)) ( axb)/(/axb/) = <-1/(27^(1/2)),-1/(27^(1/2)),5/(27^(1/2))>
Bxa = <1,1,-5> /bxa/ = 27^(1/2) (bxa)/(/bxa/) = <1/(27^(1/2)),1/(27^(1/2)),-5/(27^(1/2))>
20. a =j-k b =i+j axb = (1(0)-(-1(1))-(0(0)-(-1)(1))+(0(1)-(1)(1)) = <1,-1,-1>
/axb/ = 3^(1/2)
(axb)/ (/axb/) = <1/(3^(1/2)),-1/(3^(1/2)),-1/(3^(1/2))>
Bxa = <-1,1,1> /bxa/ =3^(1/2)) (bxa)/(/bxa/) = <-1/(3^(1/2)),1/(3^(1/2)),1/(3^(1/2))>
21. a= <x,y,z> b= <0,0,0>=0 ax0= (y(0)-(z(0))-(x(0)-(z)(0))+(x(0)-(y)(0)) = 0
0xa = (0(z)-(0(y))-(0(z)-(0)(x))+(0(y)-(0)(x)) = 0
Ax0=0=0xa
22. a=<a1,a2,a3> b= <b1,b2,b3> (axb)·b = <a2b3-a3b2,-a1b3+a3b1,a1b2-a2b1> · <b1,b2,b3> =
B1(a2b3-a3b2)+b2(-a1b3+a3b1)+b3(a1b2-a2b1) = 0
27. A(-3,0) B(-1,3) C(5,2) D(3,-1)
AB =<6,-1>
AD =<2,3>
ABxAD = 20
28. P(1,0,2) Q(3,3,3) R(7,5,8) S(5,2,7) PQ = <2,3,1> PS = <4,2,5>
PQxPS = <13,-6,-8> /PQxPS/ = (169+36+64)^(1/2) = 269^(1/2)
29. P(1,0,1) Q(-2,1,3) R(4,2,5)
PQ = <-3,1,2> PR = <3,2,4>
PQxPR = <0,18,-9> or PRxPQ = <0,-18,9>
b. /PQxPR/ = 9*(5^(1/2))
(/PQxPR/)/2 = area of triangle = (9*(5^(1/2)))/2
30. . P(0,0,-3) Q(4,2,0) R(3,3,1)
PQ = <4,2,3> PR = <3,3,4>
PQxPR = <-1,-7,6> or PRxPQ = <1,7,-6>
b. /PQxPR/ = 86^(1/2)
31.
(/PQxPR/)/2 = area of triangle = = (86^(1/2) )/2
P(0,-2,0) Q(4,1,-2) R(5,3,1)
PQ = <4,3,-2> PR = <5,5,1>
PQxPR = <13,-14,5> or PRxPQ = <-13,14,-5>
b. /PQxPR/ = 390^(1/2)
(/PQxPR/)/2 = area of triangle = = (390^(1/2) )/2
32. P(2,-3,4) Q(-1,-2,2) R(3,1,-3)
PQ = <3,1,-2> PR = <1,4,-7>
PQxPR = <1,-23,-13> or PRxPQ = <-1,23,13>
b. /PQxPR/ = 699^(1/2)
(/PQxPR/)/2 = area of triangle = = (699^(1/2) )/2