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Vectors in Space. Lines and Planes solved questions
Vectors in Space. Lines and Planes solved questions
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CHAPTER 40Vectors in Space. Lines and Planes
40.1 Find the distance between the points (2,4,7) and (1,5,10).
The distance between any two points (xlt y., z.) and (x2, y^, z,) is
Hence, the distance between (2,4,7) and (1,5,10) is
40.2 Find the distance between (-1, 2, 0) and (4, 3, -5).
The distance is
40.3 Find the distance between a point (x, y, z) and the origin (0, 0, 0).
The distance is
40.4 Find the distance between (2, —1,4) and the origin.
By Problem 40.3, the distance is
40.5 Find the midpoint of the line segment connecting the points (4, 3,1) and (—2, 5, 7).
The coordinates of the midpoint are the averages of the coordinates of the endpoints. In this case, the
midpoint is
40.6 Find the equation of a sphere y oi radius r and center (a,b,c).
A point (x, y, z) is on y if and only if its distance from (a,b,c) is r, that is, if and only if
or, equivalently, (x - a)2 + (y - b)2 + (z — c)2 = r2.
40.7 Find the equation of the sphere with radius 5 and center (2, —1, 3).
By Problem 40.6, the equation is (x - 2)2 + (y + I)2 + (z - 3)2 = 25.
40.8 Describe the surface with the equation x2 + y2 + z2 = 49,
By Problem 40.6, this is the sphere with center (0,0, 0) and radius 7.
40.9 Describe the graph of the equation x2 + 4x + y2 + z2 — Sz = 5
Complete the square in x and in z: (x + 2)2 + y2 + (z - 4)2 = 5 + 4 + 16 = 25. This is the equation of the
sphere with center (—2,0, 4) and radius 5.
40.10 Show that every sphere has an equation of the form x2 + y2 + z2 + Ax + By + Cz + D = 0.
The sphere with center (a, b, c) and radius r has the equation (x — a)2 + (y — b)2 + (z — c)2 = r2. Expand-
ing, we obtain x2 — 2ax + a2 + y2 — 2by + b2 + z2 — 2cz + c2 = r2, which is equivalent to x2 + y2 + z2 —
2ax - 2by - 2cz + (a2 + b2 + c2 - r 2 ) = 0.
40.11 When does an equation x2 + y2 + z 2 + Ax + By + Cz + D = 0 represent a sphere?
Complete the squares: This is a sphere if and
only if the right side is positive; that is, if and only if A2 + B2 + C2 - 4D > 0. In that case, the sphere has
radius and center When A2 + B2+C2-4D=0, the graph
is a single point When A2 + B2 + C2-4D<0, there are no points on the graph at all.
347
, 348 CHAPTER 40
40.12 Find an equation of the sphere with the points P(7, -1, -2) and G(3,4,6) as the ends of a diameter.
The center is the midpoint of the segment PQ, namely, (5, |,4). The length of the diameter is
So, the radius is jV"57. Thus, an equation of the
sphere is (x - 5)2 + (y - §)2 + (z - 4)2 = f.
40.13 Show that the three points P(l, 2, 3), Q(4, -5,2), and R(0,0,0) are the vertices of a right triangle.
Thus, Hence, by the converse of the Pythagorean theorem, &.PQR is a right triangle with
ight angle at R.
40.14 If line L passes through point (1, 2,3) and is perpendicular to the xy-plane, what are the coordinates of the points
on the line that are at a distance 7 from the point P(3, -1,5)?
The points on L have coordinates (1,2, z). Their distance from P is
Set this equal to 7. = 7, 13 + (5-z) 2 = 49, (5-z) 2 = 36, 5 - r = ±6,
z = -l orz = ll. So, the required points are (1,2, —1) and (1,2,11).
40.15 Show that the points P(2, -1,5), 2(6,0,6), and R(14,2, 8) are collinear.
So, Hence, the points are collinear. (If three points are not collinear, they form a
triangle; then the sum of two sides must be greater than the third side.) Another method
40.16 Find an equation of the sphere with center in the *z-plane and passing through the points P(0,8,0), Q(4, 6,2),
and R(0,12,4).
Let C(a, 0, c) be the center. Then So,
Hence, Equate the first
and third terms: 64 + c2 = 144 + (c - 4)2, c2 = 80 + c2 - 8c + 16, 8c = 96, c = 12. Substitute 12 for c in
the first equation: a2 + 64 + 144 = (a - 4)2 + 36 + 100, a2 + 72= a2 - 8a + 16, 8a = -56, a = -7. So, the
center is (-7,0,12). The radius
40.17 [f the midpoints of the sides of PQR are (5, -1,3), (4,2,1), and (2,1,0), find the vertices.
Let P = (*,, _y,, zj, Q = (x2, y2, z2), R = (x3, y3, z3). Assume (5,-1,3) is the midpoint of PQ,
(4,2,1) is the midpoint of QR, and (2,1,0) is the midpoint of PR. Then, (x1+x2)/2 = 5, (x2 + -v 3 )/2 = 4,
and (xl+x3)/2 = 2. So, xt+x2 = 10, x2 + x*, = 8, and .*! + *3 = 4. Subtract the second equation
from the first: x t — x3 = 2, and add this to the third equation: 2xt = 6, xl = 3. Hence, x2 = 7. *3 = 1.
Similarly, y,+y2 = -2, y2 + y, = 4, yl+y, = 2. Then, y,-y3 = -6, 2y, =-4, yl = -2. So,
>"2 = 0' y 3 = 4 - Finally, z, + z 2 = 6, z 2 + z3 = 2, z, + z, = 0. Therefore, z, - z, = 4, 2 z , = 4 . z t = 2 ,
z2=4, z3 = -2. Hence, P = (3, -2,2), Q = (7,0,4), "and /? = (!,4,-2).
40.18 Describe the intersection of the graphs of x2 + y2 = 1 and z = 2.
As shown in Fig. 40-1, *2 + _y 2 = l is a cylinder of radius 1 with the z-axis as its axis of symmetry. z = 2is
a plane two units above and parallel to the xy-plane. Hence, the intersection is a circle of radius 1 with center at
(0,0, 2) in the plane z = 2.
Fig. 40-1
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