CHAPTER 40
Vectors in Space. Lines and Planes
40.1
40.2
40.3
40.4
40.5
40.6
40.7
40.8
40.9
40.10
40.11
Find the distance between the points (2,4,7) and (1,5,10).
The distance between any two points (x lt y., z.) and (x 2 , y^, z,) is
Hence, the distance between (2,4,7) and (1,5,10) is
Find the distance between (-1, 2, 0) and (4, 3, -5).
The distance is
Find the distance between a point (x, y, z) and the origin (0, 0, 0).
The distance is
Find the distance between (2, —1,4) and the origin.
By Problem 40.3, the distance is
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
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 = r
2 .
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.
Describe the surface with the equation x
2 + y
2 + z
2 = 49,
By Problem 40.6, this is the sphere with center (0,0, 0) and radius 7.
Describe the graph of the equation x
2 + 4x + y
2 + z
2 — Sz = 5
Complete the square in x and in z: (x + 2)
2 + y
2 + (z - 4)
2 = 5 + 4 + 16 = 25. This is the equation of the
sphere with center (—2,0, 4) and radius 5.
Show that every sphere has an equation of the form x
2 + y
2 + z
2 + 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 = r
2 . Expanding, we obtain x
2 — 2ax + a
2 + y
2 — 2by + b
2 + z
2 — 2cz + c
2 = r
2 , which is equivalent to x
2 + y
2 + z
2 —
2ax - 2by - 2cz + (a
2 + b
2 + c
2 - r
2 ) = 0.
When does an equation x
2 + y
2 + z
2 + Ax + By + Cz + D = 0 represent a sphere?
Complete the squares:
only if the right side is positive; that is, if and only if A
2 + B
2 + C
2 - 4D > 0.
This is a sphere if and
In that case, the sphere has
radius
and center
When A
2 + B
2 +C
2 -4D=0,
there are no points on the graph at all.
the graph
is a single point
When A
2 + B
2 + C
2 -4D<0,
347
Vectors in Space. Lines and Planes
40.1
40.2
40.3
40.4
40.5
40.6
40.7
40.8
40.9
40.10
40.11
Find the distance between the points (2,4,7) and (1,5,10).
The distance between any two points (x lt y., z.) and (x 2 , y^, z,) is
Hence, the distance between (2,4,7) and (1,5,10) is
Find the distance between (-1, 2, 0) and (4, 3, -5).
The distance is
Find the distance between a point (x, y, z) and the origin (0, 0, 0).
The distance is
Find the distance between (2, —1,4) and the origin.
By Problem 40.3, the distance is
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
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 = r
2 .
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.
Describe the surface with the equation x
2 + y
2 + z
2 = 49,
By Problem 40.6, this is the sphere with center (0,0, 0) and radius 7.
Describe the graph of the equation x
2 + 4x + y
2 + z
2 — Sz = 5
Complete the square in x and in z: (x + 2)
2 + y
2 + (z - 4)
2 = 5 + 4 + 16 = 25. This is the equation of the
sphere with center (—2,0, 4) and radius 5.
Show that every sphere has an equation of the form x
2 + y
2 + z
2 + 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 = r
2 . Expanding, we obtain x
2 — 2ax + a
2 + y
2 — 2by + b
2 + z
2 — 2cz + c
2 = r
2 , which is equivalent to x
2 + y
2 + z
2 —
2ax - 2by - 2cz + (a
2 + b
2 + c
2 - r
2 ) = 0.
When does an equation x
2 + y
2 + z
2 + Ax + By + Cz + D = 0 represent a sphere?
Complete the squares:
only if the right side is positive; that is, if and only if A
2 + B
2 + C
2 - 4D > 0.
This is a sphere if and
In that case, the sphere has
radius
and center
When A
2 + B
2 +C
2 -4D=0,
there are no points on the graph at all.
the graph
is a single point
When A
2 + B
2 + C
2 -4D<0,
347