4.6. LINE AND PLANE IN SPACE 129
R
R
R
R
Mxyz(,,)
M
1
1
1
2
2
Figure 4.40. Plane passing through a point and parallel to two straight lines.
4.6.1-9. Equation of plane passing through point and parallel to two straight lines.
The equation of the plane passing through a point M
1
(x
1
, y
1
, z
1
) and parallel to two straight
lines with direction vectors R
1
=(l
1
, m
1
, n
1

n
2





= 0,or

(r – r
1
)R
1
R
2

= 0,(4.6.1.13)
where r and r
1
are the position vectors of the points M(x, y, z)andM
1
(x
1
, y
1
, z
1
), respec-
tively.
Example 7. Let us find the equation of the plane passing through the point M

1
, z
1
)andM
2
(x
2
, y
2
, z
2
)and
perpendicular to the plane given by the equation Ax + By + Cz + D = 0 is determined by
the equation





x – x
1
y – y
1
z – z
1
x
2
– x
1
y

(x
1
, y
1
, z
1
), and
M
2
(x
2
, y
2
, z
2
), respectively.
Remark. If the straight line passing through points M
1
(x
1
, y
1
, z
1
)andM
2
(x
2
, y
2




= 0,
whence
2x – y – 3z + 7 = 0.
4.6.1-11. Plane passing through point and perpendicular to two planes.
The plane (see Fig. 4.42) passing through a point M
1
(x
1
, y
1
, z
1
) and perpendicular to two
(nonparallel) planes A
1
x + B
1
y + C
1
z + D
1
= 0 and A
2
x + B
2
y + C
2



= 0,or

(r – r
1
)N
1
N
2

= 0,(4.6.1.15)
where N
1
=(A
1
, B
1
, C
1
)andN
2
=(A
2
, B
2
, C
2
) are the normals to the given planes and r
and r




x – 0 y – 1 z – 2
1 –11
–11 1





= 0,
whence
x + y – 1 = 0.
4.6.1-12. Equation of plane passing through line of intersection of planes.
The planes passing through the line of intersection of the planes A
1
x +B
1
y + C
1
z + D
1
= 0
and A
2
x + B
2
y + C
2

+ λ(A
2
x + B
2
y + C
2
z + D
2
)=0.(4.6.1.17)
By varying the parameter λ from –∞ to +∞, we obtain all the planes in the pencil. For
λ =
1, we obtain equations of the planes that bisect the angles between the given planes
provided that the equations of the latter are given in normalized form.
Remark. The passage from equation (4.6.1.16) to equation (4.6.1.17) excludes the case α = 0. Equa-
tion (4.6.1.17) does not define the plane A
2
x + B
2
y + C
2
z + D
2
= 0; i.e., equation (4.6.1.17) for various λ
defines all the planes in the pencil but one (the second of the two given planes).
4.6.2. Line in Space
4.6.2-1. Parametric equation of straight line.
The parametric equation of the line that passes through a point M
1
(x
1

, the coefficients
l, m, n are the cosines of the angles α, β,andγ formed by this straight line (the direction of
the vector R
0
) with the coordinate axes OX, OY ,andOZ. These cosines can be expressed
via the coordinates of the direction vector R as
cos α =
l
√
l
2
+ m
2
+ n
2
,cosβ =
m
√
l
2
+ m
2
+ n
2
,cosγ =
n
√
l
2
+ m

y – y
1
m
=
z – z
1
n
,or(r – r
1
) × R = 0,(4.6.2.3)
is called the canonical equation of the straight line passing through the point M
1
(x
1
, y
1
, z
1
)
with position vector r
1
=(x
1
, y
1
, z
1
) and parallel to the direction vector R =(l, m, n).
Remark 1. One can obtain the canonical equation (4.6.2.3) from the parametric equations (4.6.2.1) by
eliminating the parameter t.

1
z + D
1
= 0,
A
2
x + B
2
y + C
2
z + D
2
= 0,
or
r ⋅ N
1
+ D
1
= 0,
r ⋅ N
2
+ D
2
= 0,
(4.6.2.4)
where N
1
=(A
1
, B

B
1
C
1
B
2
C
2



, m =



C
1
A
1
C
2
A
2



, n =




= 0 (and only in this case), the line passes through the origin.
Example 4. Let us reduce the equation of the straight line
x + 2y – z + 1 = 0, x – y + z + 3 = 0
to canonical form.
We choose one of the coordinates arbitrarily; say, x = 0.Then
2y – z + 1 = 0,–y + z + 3 = 0,
and hence y =–4, z =–7. Thus the desired line contains the point M(0,–4,–7). We find the cross product of the
vectors N
1
=(1, 2,–1)andN
2
=(1,–1, 1) and, according to (4.6.2.5), obtain the direction vector R =(1,–2,–3)
of the desired line. Therefore, with (4.6.2.3) taken into account, the equation of the line becomes
x
1
=
y + 4
–2
=
z + 7
–3
.
4.6.2-4. Equation of line in projections.
The equation of a line in projections can be obtained by eliminating first z and then y from
the general equations (4.6.2.4):
y = kx + a, z = hx + b.(4.6.2.7)
Each of two equations (4.6.2.7) defines a plane projecting the straight line onto the planes
OXY and OXZ (see Fig. 4.45).
Remark 1. For straight lines parallel to the plane OYZ, this form of the equations cannot be used; one
should take the projections onto some other pair of coordinate planes.

, y
2
, z
2
)is
x – x
1
x
2
– x
1
=
y – y
1
y
2
– y
1
=
z – z
1
z
2
– z
1
,or(r – r
1
) × (r
2
– r

, z
1
)
and M
2
(x
2
, y
2
, z
2
) in the rectangular Cartesian coordinate system OXY Z can be written as
x = x
1
(1 – t)+x
2
t,
y = y
1
(1 – t)+y
2
t,
z = z
1
(1 – t)+z
2
t,
or r =(1 – t)r
1
+ tr

0
, z
0
) and perpendicular
to the plane given by the equation Ax + By+ Cz+ D = 0,orr ⋅ N + D = 0 (see Fig. 4.47), is
x – x
0
A
=
y – y
0
B
=
z – z
0
C
,(4.6.2.11)
where N =(A, B, C) is the normal to the plane.
4.6. LINE AND PLANE IN SPACE 135
4.6.3. Mutual Arrangement of Points, Lines, and Planes
4.6.3-1. Angles between lines in space.
Consider two straight lines determined by vector parametric equations r = r
1
+ tR
1
and
r = r
2
+ tR
2

y – y
1
m
1
=
z – z
1
n
1
and
x – x
2
l
2
=
y – y
2
m
2
=
z – z
2
n
2
,
then the angle ϕ between the lines can be found from the formulas
cos ϕ =
l
1
l

sin ϕ =




m
1
n
1
m
2
n
2



2
+



n
1
l
1
n
2
l
2


l
2
2
+ m
2
2
+ n
2
2
,
(4.6.3.1)
which coincide with formulas (4.6.3.1) written in coordinate form.
R
R
2
1
φ
Figure 4.48. Angles between lines in space.
Example 1. Let us find the angle between the lines
x
1
=
y – 2
2
=
z + 1
2
and
x
0

Two straight lines given by vector parametric equations r = r
1
+ tR
1
and r = r
2
+ tR
2
are
parallel if
R
2
= λR
1
or R
2
× R
1
= 0,