40.66
40.67
40.68
40.69
40.70
40.71
40.72
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CHAPTER 40
by Problem 40.62. By virtue of the fact that V, is perpendicular to VjXV!, we have (V 3 x V,) -V, =0,
and, therefore,
K 2 x K 3 = (l/c
2 )[(V 3 x V,)-V 2 ]V,.
By Problem 40.64,
(V, x V,)-V 2 = V 2 -(V 3 x V,) =
V(V 2 xV 3 ) = c. Hence, K 2 xK 3 = (l/c)V,. Therefore,
Show that K,. 1 V;. for i^j. (For the notation, see Problem 40.65.)
since V 2 is perpendicular to V 2 x V 3 .
perpendicular to V 2 xV 3 . The other cases are handled similarly.
since V 3 is
Show that K,-V, = 1 for i = 1,2,3. (For the notation, see Problem 40.65.)
The other cases are similar.
If A = (2,-3,1) is normal to one plane S
>
1 and B = (-l,4, -2) is normal to another plane & 2 , show that 0>,
and £?, intersect and find a vector parallel to the line If 0>, and 2P 2 were parallel, s£ and 38 would have to be parallel. But A and B are not parallel (since A is not a
scalar multiple of B). Therefore, 3f l and SP 2 intersect. Since A is perpendicular to plane 0>,, A is perpendicular
to the line ,2" in 0*,. Likewise, B is perpendicular to 2£. Hence, .SCis parallel to A x B.
Find the vector representation, the parametric equations, and the rectangular equations for the line through the
points P(l, -2, 5) and 0(3, 4,6).
If R is any point on the line, then for some scalar / (see Fig. 40-6). Hence, OR= OP+ PR=
OP+tPQ.Thus, (1, 2, —5) + f(2, 6,1) is a vector function that generates the line. A parametric form
is x = 1 + 2t, y = 2 + 6r, z = — 5 + /. If we eliminate t from these equations, we obtain the rectangular
equations
Fig. 40-6
Find the points at which the line of Problem 40.69 cuts the coordinate planes.
By Problem 40.69, the parametric equations are x = 1 + 2t, y = 2 + 6t, z = — 5 + t. To find the intersection with the jty-plane, set z = 0. Then / = 5, *=11, y = 32. So, the intersection point is (11, 32, 0).
To find the intersection with the jez-plane, set y = 0. Then
So, the intersection
To find the intersection with the yz-plane, set x = 0. Then
So, the intersection is
Find the vector representation, parametric equations, and rectangular equations for the line through the points
P(3,2,l)andG(-l,2,4).
So, the line is generated by the vector function (3,2,1) +/(-4,0, 3). The parametric
The rectangular equations are
equations are x = 3 — 4t, y = 2, z = 1 + 3f.
Write equations for the line through the point (1,2, -6) and parallel to the vector (4,1, 3).
The line is generated by the vector function (1, 2, -6) + t(4,1, 3). Parametric equations are x = 1 + 4t,
y = 2 + t, z = — 6 + 3t. A set of rectangular equations is
PC = (-4,0,3).
y=2.
is