THREE DIMENSIONAL GEOMETRY 463
The moving power of mathematical invention is not
reasoning but imagination. – A.DEMORGAN
11.1 Introduction
In Class XI, while studying Analytical Geometry in two
dimensions, and the introduction to three dimensional
geometry, we confined to the Cartesian methods only. In
the previous chapter of this book, we have studied some
basic concepts of vectors. We will now use vector algebra
to three dimensional geometry. The purpose of this
approach to 3-dimensional geometry is that it makes the
study simple and elegant*.
In this chapter, we shall study the direction cosines
and direction ratios of a line joining two points and also
discuss about the equations of lines and planes in space
under different conditions, angle between two lines, two
planes, a line and a plane, shortest distance between two
skew lines and distance of a point from a plane. Most of
the above results are obtained in vector form. Nevertheless, we shall also translate
these results in the Cartesian form which, at times, presents a more clear geometric
and analytic picture of the situation.
11.2 Direction Cosines and Direction Ratios of a Line
From Chapter 10, recall that if a directed line L passing through the origin makes
angles α, β and γ with x, y and z-axes, respectively, called direction angles, then cosine
of these angles, namely, cos α, cos β and cos γ are called direction cosines of the
directed line L.
If we reverse the direction of L, then the direction angles are replaced by their supplements,
i.e.,
πα
−
,
πβ
−
and
πγ
−
. Thus, the signs of the direction cosines are reversed.
Chapter
11
THREE DIMENSIONAL GEOMETRY
* For various activities in three dimensional geometry, one may refer to the Book
“A Hand Book for designing Mathematics Laboratory in Schools”, NCERT, 2005
Leonhard Euler
(1707-1783)
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MATHEMATICS
464
Note that a given line in space can be extended in two opposite directions and so it
has two sets of direction cosines. In order to have a unique set of direction cosines for
a given line in space, we must take the given line as a directed line. These unique
direction cosines are denoted by l, m and n.
Remark If the given line in space does not pass through the origin, then, in order to find
its direction cosines, we draw a line through the origin and parallel to the given line.
Now take one of the directed lines from the origin and find its direction cosines as two
parallel line have same set of direction cosines.
Any three numbers which are proportional to the direction cosines of a line are
called the direction ratios of the line. If l, m, n are direction cosines and a, b, c are
direction ratios of a line, then a = λl, b=λm and c = λn, for any nonzero λ ∈ R.
Note Some authors also call direction ratios as direction numbers.
Let a, b, c be direction ratios of a line and let l, m and n be the direction cosines
(d.c’s) of the line. Then
l
a
=
m
b
=
n
k
c
=
(say), k being a constant.
Therefore l = ak, m = bk, n = ck ... (1)
But l
2
+ m
2
+ n
2
= 1
Therefore k
2
(a
2
+ b
2
+ c
2
) = 1
or k =
222
1
abc
±
++
Fig 11.1
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THREE DIMENSIONAL GEOMETRY 465
Hence, from (1), the d.c.’s of the line are
222 222 222
, ,
a b c
l m n
abc abc abc
=± =± =±
++ ++ ++
where, depending on the desired sign of k, either a positive or a negative sign is to be
taken for l, m and n.
For any line, if a, b, c are direction ratios of a line, then ka, kb, kc; k ≠ 0 is also a
set of direction ratios. So, any two sets of direction ratios of a line are also proportional.
Also, for any line there are infinitely many sets of direction ratios.
11.2.1 Relation between the direction cosines of a line
Consider a line RS with direction cosines l, m, n. Through
the origin draw a line parallel to the given line and take a
point P(x, y, z) on this line. From P draw a perpendicular
PA on the x-axis (Fig. 11.2).
Let OP = r. Then
OA
cos
OP
α=
x
r
=
. This gives x = lr.
Similarly, y = mr and z = nr
Thus x
2
+ y
2
+ z
2
= r
2
(l
2
+ m
2
+ n
2
)
But x
2
+ y
2
+ z
2
= r
2
Hence l
2
+ m
2
+ n
2
= 1
11.2.2 Direction cosines of a line passing through two points
Since one and only one line passes through two given points, we can determine the
direction cosines of a line passing through the given points P(x
1
, y
1
, z
1
) and Q(x
2
, y
2
, z
2
)
as follows (Fig 11.3 (a)).
Fig 11.3
r
Z
X
Y
R
S
P
(
,
,
)
x
y
z
A
O
A
O
P
x
Fig 11.2
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466
Let l, m, n be the direction cosines of the line PQ and let it makes angles α, β and γ
with the x, y and z-axis, respectively.
Draw perpendiculars from P and Q to XY-plane to meet at R and S. Draw a
perpendicular from P to QS to meet at N. Now, in right angle triangle PNQ, ∠PQN=
γ (Fig 11.3 (b).
Therefore, cos γ =
21
NQ
PQ PQ
zz
−
=
Similarly cos α =
21 21
and cos
PQ
PQ
xx yy
− −
β=
Hence, the direction cosines of the line segment joining the points P(x
1
, y
1
, z
1
) and
Q(x
2
, y
2
, z
2
) are
21
PQ
xx
−
,
21
PQ
yy
−
,
21
PQ
zz
−
where PQ =
()
2
2 2
21 2 1 21
( )( )
x x yy zz
− +− +−
Note The direction ratios of the line segment joining P(x
1
, y
1
, z
1
) and Q(x
2
, y
2
, z
2
)
may be taken as
x
2
– x
1
, y
2
– y
1
, z
2
– z
1
or x
1
– x
2
, y
1
– y
2
, z
1
– z
2
Example 1 If a line makes angle 90°, 60° and 30° with the positive direction of x, y and
z-axis respectively, find its direction cosines.
Solution Let the d . c .'s of the lines be l , m, n. Then l = cos 90
0
= 0, m = cos 60
0
=
1
2
,
n = cos 30
0
=
2
3
.
Example 2 If a line has direction ratios 2, – 1, – 2, determine its direction cosines.
Solution Direction cosines are
222
) 2 ( ) 1 (2
2
−+−+
,
222
) 2() 1(2
1
−+−+
−
,
()
2
2
2
) 2 ( 12
2
− +− +
−
or
212
,
,
33 3
−−
Example 3 Find the direction cosines of the line passing through the two points
(– 2, 4, – 5) and (1, 2, 3).
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THREE DIMENSIONAL GEOMETRY 467
Solution We know the direction cosines of the line passing through two points
P(x
1
, y
1
, z
1
) and Q(x
2
, y
2
, z
2
) are given by
21 2 121
,
,
PQ PQ PQ
x xy yz z
−−−
where PQ =
()
2
12
2
12
2
12
) ( )( z z y yx x − + − +−
Here P is (– 2, 4, – 5) and Q is (1, 2, 3).
So PQ =
2 2 2
(1 ( 2)) (2 4) (3 ( 5))
−− + − + −−
=
77
Thus, the direction cosines of the line joining two points is
3 28
,,
77 77 77
−
Example 4 Find the direction cosines of x, y and z-axis.
Solution The x-axis makes angles 0°, 90° and 90° respectively with x, y and z-axis.
Therefore, the direction cosines of x-axis are cos 0°, cos 90°, cos 90° i.e., 1,0,0.
Similarly, direction cosines of y-axis and z-axis are 0, 1, 0 and 0, 0, 1 respectively.
Example 5 Show that the points A (2, 3, – 4), B (1, – 2, 3) and C (3, 8, – 11) are
collinear.
Solution Direction ratios of line joining A and B are
1 – 2, – 2 – 3, 3 + 4 i.e., – 1, – 5, 7.
The direction ratios of line joining B and C are
3 –1, 8 + 2, – 11 – 3, i.e., 2, 10, – 14.
It is clear that direction ratios of AB and BC are proportional, hence, AB is parallel
to BC. But point B is common to both AB and BC. Therefore, A, B, C are
collinear points.
EXERCISE 11.1
1. If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its
direction cosines.
2. Find the direction cosines of a line which makes equal angles with the coordinate
axes.
3. If a line has the direction ratios –18, 12, – 4, then what are its direction cosines ?
4. Show that the points (2, 3, 4), (– 1, – 2, 1), (5, 8, 7) are collinear.
5. Find the direction cosines of the sides of the triangle whose vertices are
(3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2).
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468
11.3 Equation of a Line in Space
We have studied equation of lines in two dimensions in Class XI, we shall now study
the vector and cartesian equations of a line in space.
A line is uniquely determined if
(i) it passes through a given point and has given direction, or
(ii) it passes through two given points.
11.3.1 Equation of a line through a given point and parallel to a given vector
b
Let
a
be the position vector of the given point
A with respect to the origin O of the
rectangular coordinate system. Let l be the
line which passes through the point A and is
parallel to a given vector
b
. Let
r
be the
position vector of an arbitrary point P on the
line (Fig 11.4).
Then
AP
is parallel to the vector
b
, i.e.,
AP
= λ
b
, where λ is some real number.
But
AP
=
OP – OA
i.e. λ
b
=
ra
−
Conversely, for each value of the parameter λ, this equation gives the position
vector of a point P on the line. Hence, the vector equation of the line is given by
r
=
a+ b
... (1)
Remark If
ˆ
ˆˆ
b ai bj ck
= ++
, then a, b, c are direction ratios of the line and conversely,
if a, b, c are direction ratios of a line, then
ˆ
ˆˆ
= ++
b ai bj ck
will be the parallel to
the line. Here, b should not be confused with |
b
|
.
Derivation of cartesian form from vector form
Let the coordinates of the given point A be (x
1
, y
1
, z
1
) and the direction ratios of
the line be a, b, c. Consider the coordinates of any point P be (x, y, z). Then
kzjyixr
ˆ
ˆˆ
+ +=
;
k z j yi xa
ˆ
ˆˆ
111
++ =
and
ˆ
ˆˆ
b ai b j ck
=++
Substituting these values in (1) and equating the coefficients of
ˆˆ
,
ij
and
k
ˆ
, we get
x = x
1
+ λ a; y = y
1
+ λ b; z = z
1
+ λ c ... (2)
Fig 11.4
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THREE DIMENSIONAL GEOMETRY 469
These are parametric equations of the line. Eliminating the parameter λ from (2),
we get
1
x–x
a
=
11
y– y z–z
=
bc
... (3)
This is the Cartesian equation of the line.
Note If l, m, n are the direction cosines of the line, the equation of the line is
1
x–x
l
=
11
y– y z–z
=
mn
Example 6 Find the vector and the Cartesian equations of the line through the point
(5, 2, – 4) and which is parallel to the vector
ˆ
ˆˆ
3 28
i jk
+−
.
Solution We have
a
=
ˆ ˆ
ˆˆ ˆˆ
5 2 4 and 3 2 8
ijk bijk
+− =+−
Therefore, the vector equation of the line is
r
=
ˆ ˆ
ˆˆ ˆˆ
5 2 4 (3 2 8 )
ijk ijk
+−+λ +−
Now,
r
is the position vector of any point P(x, y, z) on the line.
Therefore,
ˆ
ˆˆ
xi y j z k
++
=
ˆ ˆ
ˆˆ ˆˆ
5 2 4 (3 2 8 )
+−+λ + −
i jk i j k
=
ˆ
(5 3) (2 2) ( 4 8)
+λ + + λ +− −λ
ij k
Eliminating λ , we get
5
3
x
−
=
24
28
yz
−+
=
−
which is the equation of the line in Cartesian form.
11.3.2 Equation of a line passing through two given points
Let
a
and
b
be the position vectors of two
points A (x
1
, y
1
, z
1
) and B (x
2
, y
2
, z
2
),
respectively that are lying on a line (Fig 11.5).
Let
r
be the position vector of an
arbitrary point P(x, y, z), then P is a point on
the line if and only if
AP
ra
=−
and
AB
ba
=−
are collinear vectors. Therefore,
P is on the line if and only if
()
ra ba
− =λ −
Fig 11.5
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MATHEMATICS
470
or
()
r a ba
= +λ −
= +λ −= +λ −
= +λ −
,
λ ∈λ ∈
λ ∈λ ∈
λ ∈ R. ... (1)
This is the vector equation of the line.
Derivation of cartesian form from vector form
We have
111
ˆ ˆ
ˆˆ ˆ ˆ
,
rxi yjzkaxi yjzk
=+ + = + +
and
22 2
ˆ
ˆˆ
,
b xi y j z k
=+ +
Substituting these values in (1), we get
11 1 21 2 1 21
[( ) ( ) ( ) ]
xiyjzkxiyjzk x xi y yj z zk
+ + = + + +λ − + − + −
Equating the like coefficients of
kji
ˆ
,
ˆ
,
ˆ
, we get
x = x
1
+ λ (x
2
– x
1
); y = y
1
+ λ (y
2
– y
1
); z = z
1
+ λ (z
2
– z
1
)
On eliminating λ , we obtain
1 11
21 2121
xx yy zz
xx yyzz
− −−
− −−− −−
− −−
==
====
==
− −−
− −−− −−
− −−
which is the equation of the line in Cartesian form.
Example 7 Find the vector equation for the line passing through the points (–1, 0, 2)
and (3, 4, 6).
Solution Let
a
and
b
be the position vectors of the point A(– 1, 0, 2) and B (3, 4, 6).
Then
ˆ
ˆ
2
ai k
=− +
and
ˆ
ˆˆ
346
bi jk
=+ +
Therefore
ˆ
ˆˆ
444
ba i j k
−= + +
Let
r
be the position vector of any point on the line. Then the vector equation of
the line is
ˆ ˆ
ˆ ˆˆ
2 (4 4 4 )
r ik i jk
=− + +λ + +
Example 8 The Cartesian equation of a line is
356
242
xyz
+−+
==
Find the vector equation for the line.
Solution Comparing the given equation with the standard form
1 11
xx yy zz
a bc
− −−
==
We observe that x
1
= – 3, y
1
= 5, z
1
= – 6; a = 2, b = 4, c = 2.
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THREE DIMENSIONAL GEOMETRY 471
Thus, the required line passes through the point (– 3, 5, – 6) and is parallel to the
vector
ˆ
ˆˆ
24 2
i jk
++
. Let
r
be the position vector of any point on the line, then the
vector equation of the line is given by
ˆ
ˆˆ
(356)
r i jk
=−+−
+ λ
ˆ
ˆˆ
(2 4 2 )
i jk
++
11.4 Angle between Two Lines
Let L
1
and L
2
be two lines passing through the origin
and with direction ratios a
1
, b
1
, c
1
and a
2
, b
2
, c
2
,
respectively. Let P be a point on L
1
and Q be a point
on L
2
. Consider the directed lines OP and OQ as
given in Fig 11.6. Let θ be the acute angle between
OP and OQ. Now recall that the directed line
segments OP and OQ are vectors with components
a
1
, b
1
, c
1
and a
2
, b
2
, c
2
, respectively. Therefore, the
angle θ between them is given by
cos θ =
12 12 12
222 222
111 222
aa bb cc
abcabc
++
++++
++
++ ++
++ ++++ ++
++ ++
... (1)
The angle between the lines in terms of sin θ is given by
sin θ =
2
1 cos
−θ
=
()()
2
1 2 12 12
222 222
111 222
()
1
aa bb cc
abcabc
++
−
++ ++
=
(
)
(
)
( )
()
()
2
222 222
1 1 1 2 2 2 1 2 12 12
222 222
111 222
abc abc aabbcc
abc abc
++ ++ − + +
++ ++
=
2 2 2
12 21 12 21 1 2 2 1
2 2 2 2 22
111 222
( )( )( )
− +− +−
++ ++
ab a b bc b c ca c a
abc abc
... (2)
Note In case the lines L
1
and L
2
do not pass through the origin, we may take
lines
12
L and L
′′
which are parallel to L
1
and L
2
respectively and pass through
the origin.
Fig 11.6
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MATHEMATICS
472
If instead of direction ratios for the lines L
1
and L
2
, direction cosines, namely,
l
1
, m
1
, n
1
for L
1
and l
2
, m
2
, n
2
for L
2
are given, then (1) and (2) takes the following form:
cos θ = | l
1
l
2
+ m
1
m
2
+ n
1
n
2
| (as
2 22
1 11
1
lmn
+ +=
2 22
2 22
lmn
=+ +
) ... (3)
and sin θ =
()
2
2
2
1 2 2 1 1 2 2 1 12 21
( )( )
lm l m mn m n nl n l
− − − +−
... (4)
Two lines with direction ratios a
1
, b
1
, c
1
and a
2
, b
2
, c
2
are
(i) perpendicular i.e. if θ = 90° by (1)
a
1
a
2
+ b
1
b
2
+ c
1
c
2
= 0
(ii) parallel i.e. if θ = 0 by (2)
1
2
a
a
=
11
22
bc
bc
=
==
=
Now, we find the angle between two lines when their equations are given. If θ is
acute the angle between the lines
r
=
11
ab
+λ
and
r
=
22
ab
+µ
then cosθ =
12
12
bb
bb
⋅
In Cartesian form, if θ is the angle between the lines
1
1
xx
a
−
=
11
11
yy zz
bc
−−
=
... (1)
and
2
2
xx
a
−
=
22
22
yy zz
bc
−−
=
... (2)
where, a
1
, b
1,
c
1
and a
2,
b
2
, c
2
are the direction ratios of the lines (1) and (2), respectively,
then
cos θ =
1 2 12 12
222 222
111 222
aa bb cc
abcabc
++
++ ++
Example 9 Find the angle between the pair of lines given by
r
=
ˆ ˆ
ˆˆ ˆˆ
3 2 4 ( 2 2)
ijkijk
+−+λ++
and
r
=
ˆ
ˆˆ ˆˆ
52 (32 6)
i j i jk
− +µ + +
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THREE DIMENSIONAL GEOMETRY 473
Solution Here
1
b
=
ˆ
ˆˆ
22
i jk
++
and
2
b
=
ˆ
ˆˆ
32 6
i jk
++
The angle θ between the two lines is given by
cos θ =
12
12
ˆ
ˆ
ˆˆ ˆˆ
( 2 2 ) (3 2 6 )
1449436
bb
i jk i jk
bb
⋅
++ ⋅ ++
=
++ ++
=
3 4 12 19
3 7 21
++
=
×
Hence θ = cos
–1
19
21
Example 10 Find the angle between the pair of lines
3
3
x
+
=
13
54
yz
−+
=
and
1
1
x
+
=
45
12
yz
−−
=
Solution The direction ratios of the first line are 3, 5, 4 and the direction ratios of the
second line are 1, 1, 2. If θ is the angle between them, then
cos θ =
222222
3.1 5 .1 4. 2 16 16 8 3
15
506 526
354112
++
===
+ + ++
Hence, the required angle is cos
–1
83
15
.
11.5 Shortest Distance between Two Lines
If two lines in space intersect at a point, then the shortest distance between them is
zero. Also, if two lines in space are parallel,
then the shortest distance between them
will be the perpendicular distance, i.e. the
length of the perpendicular drawn from a
point on one line onto the other line.
Further, in a space, there are lines which
are neither intersecting nor parallel. In fact,
such pair of lines are non coplanar and
are called skew lines. For example, let us
consider a room of size 1, 3, 2 units along
x, y and z-axes respectively Fig 11.7.
Fig 11.7
2019-20
MATHEMATICS
474
l
2
S
T
Q
P
l
1
The line GE that goes diagonally across the ceiling and the line DB passes through
one corner of the ceiling directly above A and goes diagonally down the wall. These
lines are skew because they are not parallel and also never meet.
By the shortest distance between two lines we mean the join of a point in one line
with one point on the other line so that the length of the segment so obtained is the
smallest.
For skew lines, the line of the shortest distance will be perpendicular to both
the lines.
11.5.1 Distance between two skew lines
We now determine the shortest distance between two skew lines in the following way:
Let l
1
and l
2
be two skew lines with equations (Fig. 11.8)
r
=
11
ab
+λ
... (1)
and
r
=
22
ab
+µ
... (2)
Take any point S on l
1
with position vector
1
a
and T on l
2
, with position vector
2
a
.
Then the magnitude of the shortest distance vector
will be equal to that of the projection of ST along the
direction of the line of shortest distance (See 10.6.2).
If
PQ
is the shortest distance vector between
l
1
and l
2
, then it being perpendicular to both
1
b
and
2
b
, the unit vector
n
ˆ
along
PQ
would therefore be
ˆ
n
=
12
12
||
bb
bb
×
×
... (3)
Then
PQ
= d
n
ˆ
where, d is the magnitude of the shortest distance vector. Let θ be the angle between
ST
and
PQ
. Then
PQ = ST | cos θ |
But cos θ =
PQ ST
| PQ | | ST |
⋅
=
21
ˆ
()
ST
dn a a
d
⋅−
(since
21
ST )
aa
=−
=
1 2 21
12
( )( )
ST
bbaa
bb
×⋅−
×
[From (3)]
Fig 11.8
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THREE DIMENSIONAL GEOMETRY 475
Hence, the required shortest distance is
d = PQ = ST |cos θ|
or d =
Cartesian form
The shortest distance between the lines
l
1
:
xx
a
−
1
1
=
yy
b
zz
c
−
=
−
1
1
1
1
and l
2
:
xx
a
−
2
2
=
yy
b
zz
c
−
=
−
2
2
2
2
is
x xy yz z
a bc
a bc
bc bc ca ca ab
2 12 12 1
1 1 1
2 2 2
12 21
2
1 2 21
2
12
−− −− −−
−− ++ −− ++
( )( )(
−−−−
ab
21
2
)
11.5.2 Distance between parallel lines
If two lines l
1
and l
2
are parallel, then they are coplanar. Let the lines be given by
... (1)
and
… (2)
where,
1
a
is the position vector of a point S on l
1
and
2
a
is the position vector of a point T on l
2
Fig 11.9.
As l
1
, l
2
are coplanar, if the foot of the perpendicular
from T on the line l
1
is P, then the distance between the
lines l
1
and l
2
= | TP |.
Let θ be the angle between the vectors
ST
and
b
.
Then
ST
×=
b
... (3)
where
n
ˆ
is the unit vector perpendicular to the plane of the lines l
1
and l
2.
But
ST
=
21
aa
−
Fig 11.9
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MATHEMATICS
476
Therefore, from (3), we get
21
()
baa
×−
=
ˆ
| | PT
bn
(since PT = ST sin θ)
i.e.,
21
| ( )|
baa
×−
=
| | PT 1
b
⋅
(as |
|
ˆ
n
= 1)
Hence, the distance between the given parallel lines is
d =
Example 11 Find the shortest distance between the lines l
1
and l
2
whose vector
equations are
r
=
ˆ
ˆ ˆ ˆˆ
(2 )
i j i jk
+ +λ − +
... (1)
and
r
=
ˆ ˆ
ˆˆ ˆ ˆ
2 (3 5 2 )
i jk i j k
+ − +µ − +
... (2)
Solution Comparing (1) and (2) with
r
=
11
ab
+λ
and
22
ba r
µ
+=
respectively,
we get
1
a
=
1
ˆ
ˆˆ ˆˆ
,2
i jb i jk
+ = −+
2
a
= 2
ˆ
i
+
ˆ
j
–
ˆ
k
and
2
b
= 3
ˆ
i
– 5
ˆ
j
+ 2
ˆ
k
Therefore
21
aa
−
=
ˆ
ˆ
ik
−
and
12
bb
×
=
ˆ ˆ
ˆˆ ˆ ˆ
(2 ) (3 5 2 )
i jk i j k
−+ × − +
=
ˆ
ˆˆ
ˆ
ˆˆ
2 11 3 7
3 52
i jk
ij k
− = −−
−
So
12
||
bb
×
=
9 1 49 59
++ =
Hence, the shortest distance between the given lines is given by
d =
||
)( . ) (
21
12
21
bb
a ab b
×
−×
59
10
59
| 70 3 |
=
+−
=
Example 12 Find the distance between the lines l
1
and l
2
given by
r
=
ˆ ˆ
ˆˆ ˆˆ
2 4 (2 3 6 )
i jk ijk
+−+λ ++
and
r
=
ˆ ˆ
ˆˆ ˆˆ
3 3 5 (2 3 6 )
i jk ijk
+−+µ ++
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THREE DIMENSIONAL GEOMETRY 477
Solution The two lines are parallel (Why? ) We have
1
a
=
ˆ
ˆˆ
24
i jk
+−
,
2
a
=
ˆ
ˆˆ
335
i jk
+−
and
b
=
ˆ
ˆˆ
236
i jk
++
Therefore, the distance between the lines is given by
d =
21
()
||
baa
b
×−
=
ˆ
ˆˆ
23 6
21 1
4936
ijk
−
++
or =
ˆ
ˆˆ
| 9 14 4 | 293 293
7
49 49
i jk−+ −
==
EXERCISE 11.2
1. Show that the three lines with direction cosines
12 3 4 4 12 3 3 4 12
, , ; , ,; , ,
13 13 13 13 13 13 13 13 13
−− −
are mutually perpendicular.
2. Show that the line through the points (1, – 1, 2), (3, 4, – 2) is perpendicular to the
line through the points (0, 3, 2) and (3, 5, 6).
3. Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line
through the points (– 1, – 2, 1), (1, 2, 5).
4. Find the equation of the line which passes through the point (1, 2, 3) and is
parallel to the vector
ˆ
ˆˆ
32 2
i jk
+−
.
5. Find the equation of the line in vector and in cartesian form that passes through
the point with position vector
ˆ
ˆ
24
ij k
−+
and is in the direction
ˆ
ˆˆ
2
i jk
+−
.
6. Find the cartesian equation of the line which passes through the point (– 2, 4, – 5)
and parallel to the line given by
348
356
xyz
+−+
==
.
7. The cartesian equation of a line is
546
372
xyz
−+−
==
. Write its vector form.
8. Find the vector and the cartesian equations of the lines that passes through the
origin and (5, – 2, 3).
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9. Find the vector and the cartesian equations of the line that passes through the
points (3, – 2, – 5), (3, – 2, 6).
10. Find the angle between the following pairs of lines:
(i)
ˆ ˆ
ˆˆ ˆˆ
25 (32 6)
r i jk i j k
= − + +λ + +
and
ˆ ˆ
ˆ ˆˆ
7 6 ( 2 2)
rik i jk
= − +µ + +
(ii)
ˆ ˆ
ˆˆ ˆˆ
3 2 ( 2)
rijk ijk
= + − +λ − −
and
ˆ ˆ
ˆˆ ˆˆ
2 56 (3 5 4 )
r ij k i j k
= − − +µ − −
11. Find the angle between the following pair of lines:
(i)
213 2 45
and
2 5 3 18 4
xyz xyz
− −+ + − −
==
==
−−
(ii)
523
and
221 4 1 8
xyzxyz
−−−
== = =
12. Find the values of p so that the lines
1 7 14 3
32 2
xy z
p
− −−
==
and
77 56
3 15
xy z
p
− −−
==
are at right angles.
13. Show that the lines
52
7 51
xyz
−+
==
−
and
123
xyz
==
are perpendicular to
each other.
14. Find the shortest distance between the lines
ˆ
ˆˆ
(2 )
r i jk
=+ +
+
ˆ
ˆˆ
()
i jk
λ −+
and
ˆ ˆ
ˆˆ ˆˆ
2 (2 2 )
r i jk i j k
= −−+µ ++
15. Find the shortest distance between the lines
111
7 61
xyz
+++
==
−
and
357
1 21
xyz
−−−
==
−
16. Find the shortest distance between the lines whose vector equations are
ˆ
ˆˆ
( 2 3)
ri j k
=+ +
+
ˆ
ˆˆ
( 3 2)
i jk
λ− +
and
ˆ ˆ
ˆˆ ˆˆ
4 5 6 (2 3 )
r i j k i jk
= + + +µ + +
17. Find the shortest distance between the lines whose vector equations are
ˆ
ˆˆ
(1 ) ( 2 ) (3 2 )
r ti t j tk
=− +− +−
and
ˆ
ˆˆ
( 1) (2 1) (2 1)
rs i s j s k
=++−−+
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THREE DIMENSIONAL GEOMETRY 479
11.6 Plane
A plane is determined uniquely if any one of the following is known:
(i) the normal to the plane and its distance from the origin is given, i.e., equation of
a plane in normal form.
(ii) it passes through a point and is perpendicular to a given direction.
(iii) it passes through three given non collinear points.
Now we shall find vector and Cartesian equations of the planes.
11.6.1 Equation of a plane in normal form
Consider a plane whose perpendicular distance from the origin is d (d ≠ 0). Fig 11.10.
If
ON
is the normal from the origin to the plane, and
n
ˆ
is the unit normal vector
along
ON
. Then
ON
= d
n
ˆ
. Let P be any
point on the plane. Therefore,
NP
is
perpendicular to
ON
.
Therefore,
NP ON
⋅
= 0 ... (1)
Let
r
be the position vector of the point P,
then
NP
=
n dr
ˆ
−
(as
ON NP OP
+=
)
Therefore, (1) becomes
()
r dn dn
∧∧
−⋅
= 0
or
()
r dn n
∧∧
−⋅
= 0 (d ≠ 0)
or
r n dn n
∧ ∧∧
⋅− ⋅
= 0
i.e.,
rn
∧
∧∧
∧
⋅
⋅⋅
⋅
= d
(as 1)
nn
∧∧
⋅=
… (2)
This is the vector form of the equation of the plane.
Cartesian form
Equation (2) gives the vector equation of a plane, where
n
ˆ
is the unit vector normal to
the plane. Let P(x, y, z) be any point on the plane. Then
OP
=
ˆ
ˆˆ
r xi y j zk
=+ +
Let l, m, n be the direction cosines of
n
ˆ
. Then
ˆ
n
=
ˆ
ˆˆ
li m j nk
++
X
Y
Z
N
P( )x,y,z
r
d
O
Fig 11.10
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480
Therefore, (2) gives
ˆ ˆ
ˆˆ ˆ ˆ
( )( )
xi y j zk li m j nk d
++ ⋅+ + =
i.e., lx + my + nz = d ... (3)
This is the cartesian equation of the plane in the normal form.
Note Equation (3) shows that if
ˆ
ˆˆ
( )
r ai b j ck
⋅++
= d is the vector equation
of a plane, then ax + by + cz = d is the Cartesian equation of the plane, where a, b
and c are the direction ratios of the normal to the plane.
Example 13 Find the vector equation of the plane which is at a distance of
29
6
from the origin and its normal vector from the origin is
ˆ
ˆˆ
23 4
i jk
−+
. Also find its
cartesian form.
Solution Let
k j in
ˆ
4
ˆ
3
ˆ
2
+−=
. Then
||
ˆ
n
n
n
=
=
ˆ ˆ
ˆˆ ˆˆ
23 4 234
4 9 16 29
ijk ijk
− + −+
=
++
Hence, the required equation of the plane is
2 34 6
ˆ
ˆˆ
29 29 29 29
r i jk
−
⋅ ++ =
Example 14 Find the direction cosines of the unit vector perpendicular to the plane
ˆ
ˆˆ
(6 3 2 ) 1
ri j k
⋅ −− +
= 0 passing through the origin.
Solution The given equation can be written as
ˆ
ˆˆ
(6 3 2
⋅− + +
r i jk
) = 1 ... (1)
Now
ˆ
ˆˆ
|6 3 2|
i jk
−++
=
36 9 4 7
++=
Therefore, dividing both sides of (1) by 7, we get
632
ˆ
ˆˆ
777
r i jk
⋅− + +
=
1
7
which is the equation of the plane in the form
ˆ
rn d
⋅=
.
This shows that
k j in
ˆ
7
2
ˆ
7
3
ˆ
7
6
ˆ
++− =
is a unit vector perpendicular to the
plane through the origin. Hence, the direction cosines of
n
ˆ
are
7
2
,
7
3
,
7
6
−
.
2019-20
THREE DIMENSIONAL GEOMETRY 481
Z
Y
X
O
P
(
x
1
,
y
1
,
z
1
)
Example 15 Find the distance of the plane 2x – 3y + 4z – 6 = 0 from the origin.
Solution Since the direction ratios of the normal to the plane are 2, –3, 4; the direction
cosines of it are
2 22 2 22 2 22
2 3 4
, ,
2 (3) 4 2 (3) 4 2 (3) 4
−
+− + +− + +− +
, i.e.,
2 34
,,
29 29 29
−
Hence, dividing the equation 2x – 3y + 4z – 6 = 0 i.e., 2x – 3y + 4z = 6 throughout by
29
, we get
2 346
29 29 29 29
x yz
−
+ +=
This is of the form lx + my + nz = d, where d is the distance of the plane from the
origin. So, the distance of the plane from the origin is
29
6
.
Example 16 Find the coordinates of the foot of the perpendicular drawn from the
origin to the plane 2x – 3y + 4z – 6 = 0.
Solution Let the coordinates of the foot of the perpendicular P from the origin to the
plane is (x
1
, y
1
, z
1
) (Fig 11.11).
Then, the direction ratios of the line OP are
x
1
, y
1
, z
1
.
Writing the equation of the plane in the normal
form, we have
23 4 6
29 29 29 29
xy z
−+=
where,
2 34
,
,
29 29 29
−
are the direction
cosines of the OP.
Since d.c.’s and direction ratios of a line are proportional, we have
1
2
29
x
=
11
34
29 29
yz
=
−
= k
i.e., x
1
=
29
2k
, y
1
=
1
34
,
29 29
kk
z
−
=
Fig 11.11
2019-20
MATHEMATICS
482
Substituting these in the equation of the plane, we get k =
29
6
.
Hence, the foot of the perpendicular is
12 18 24
,,
29 29 29
−
.
Note If d is the distance from the origin and l, m, n are the direction cosines of
the normal to the plane through the origin, then the foot of the perpendicular is
(ld, md, nd).
11.6.2 Equation of a plane perpendicular to a
given vector and passing through a given point
In the space, there can be many planes that are
perpendicular to the given vector, but through a given
point P(x
1
, y
1
, z
1
), only one such plane exists (see
Fig 11.12).
Let a plane pass through a point A with position
vector
a
and perpendicular to the vector
N
.
Let
r
be the position vector of any point P(x, y, z) in the plane. (Fig 11.13).
Then the point P lies in the plane if and only if
AP
is perpendicular to
N
. i.e.,
AP
.
N
= 0. But
AP
ra
=−
. Therefore,
( )N 0
ra
− ⋅=
− ⋅=− ⋅=
− ⋅=
… (1)
This is the vector equation of the plane.
Cartesian form
Let the given point A be (x
1
, y
1
, z
1
), P be (x, y, z)
and direction ratios of
N
are A, B and C. Then,
111
ˆ ˆ
ˆ ˆ ˆˆ
,
axi yjzkrxi yjzk
= + + =+ +
and
ˆ
ˆˆ
NA B C
i jk
=++
Now
( – ) N=0
ra⋅
So
( )( )( )
1 1 1
ˆ ˆ
ˆˆ ˆˆ
(A B C ) 0
xxi y y j z zk i j k
− + − +− ⋅ + + =
i.e. A (x – x
1
) + B (y – y
1
) + C (z – z
1
) = 0
Example 17 Find the vector and cartesian equations of the plane which passes through
the point (5, 2, – 4) and perpendicular to the line with direction ratios 2, 3, – 1.
Fig 11.12
Fig 11.13
2019-20
THREE DIMENSIONAL GEOMETRY 483
Y
Z
O
a
r
R
P
S
b
c
(RS RT)
X
X
T
Solution We have the position vector of point (5, 2, – 4) as
ˆ
ˆˆ
524
ai jk
=+ −
and the
normal vector
N
perpendicular to the plane as
ˆ
ˆˆ
N=2 +3
i jk
−
Therefore, the vector equation of the plane is given by
( ).N 0
ra
−=
or
ˆ ˆ
ˆˆ ˆˆ
[ (5 2 4 )] (2 3 ) 0
r i j k i jk
− +− ⋅ +−=
... (1)
Transforming (1) into Cartesian form, we have
ˆ ˆ
ˆˆ ˆˆ
[( – 5) ( 2) ( 4) ] (2 3 ) 0
x i y j z k i jk
+− ++ ⋅ + −=
or
2( 5) 3( 2) 1( 4) 0
xyz
−+ −− +=
i.e. 2x + 3y – z = 20
which is the cartesian equation of the plane.
11.6.3 Equation of a plane passing through three non collinear points
Let R, S and T be three non collinear points on the plane with position vectors
a
,
b
and
c
respectively (Fig 11.14).
Fig 11.14
The vectors
RS
and
RT
are in the given plane. Therefore, the vector
RS RT
×
is perpendicular to the plane containing points R, S and T. Let
r
be the position vector
of any point P in the plane. Therefore, the equation of the plane passing through R and
perpendicular to the vector
RS RT
×
is
( ) (RS RT)
ra−⋅ ×
= 0
or
( ).[( )×( )]
r–a b–a c–a
=0 … (1)
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MATHEMATICS
484
Fig 11.15
This is the equation of the plane in vector form passing through three noncollinear
points.
Note Why was it necessary to say that the three points
had to be non collinear? If the three points were on the same
line, then there will be many planes that will contain them
(Fig 11.15).
These planes will resemble the pages of a book where the
line containing the points R, S and T are members in the binding
of the book.
Cartesian form
Let (x
1
, y
1
, z
1
), (x
2
, y
2
, z
2
) and (x
3
,
y
3
, z
3
) be the coordinates of the points R, S and T
respectively. Let (x, y, z) be the coordinates of any point P on the plane with position
vector
r
. Then
RP
= (x – x
1
)
ˆ
i
+ (y – y
1
)
ˆ
j
+ (z – z
1
)
ˆ
k
RS
= (x
2
– x
1
)
ˆ
i
+ (y
2
– y
1
)
ˆ
j
+ (z
2
– z
1
)
ˆ
k
RT
= (x
3
– x
1
)
ˆ
i
+ (y
3
– y
1
)
ˆ
j
+ (z
3
– z
1
)
ˆ
k
Substituting these values in equation (1) of the vector form and expressing it in the
form of a determinant, we have
1
11
21 2121
31 3131
0
xx yy zz
xx yyzz
xx yyzz
− −−
− −−− −−
− −−
− − −=
− − −=− − −=
− − −=
− −−
− −−− −−
− −−
which is the equation of the plane in Cartesian form passing through three non collinear
points (x
1
, y
1
, z
1
), (x
2
, y
2
, z
2
) and (x
3
, y
3
, z
3
).
Example 18 Find the vector equations of the plane passing through the points
R (2, 5, – 3), S(– 2, – 3, 5) and T(5, 3,– 3).
Solution Let
ˆ
ˆˆ
25 3
ai jk
=+−
,
ˆ
ˆˆ
23 5
b i jk
=−−+
,
ˆ
ˆˆ
53 3
ci jk
=+ −
Then the vector equation of the plane passing through
a
,
b
and
c
and is
given by
( ) (RS RT)
ra−⋅ ×
= 0 (Why?)
or
()[()()]
ra ba ca
−⋅ −×−
= 0
i.e.
ˆ ˆ
ˆˆ ˆˆ ˆˆ
[ (2 5 3 )] [( 4 8 8 ) (3 2 )] 0
r i jk i jk i j− + − ⋅− − + × − =
R
S
T
2019-20
THREE DIMENSIONAL GEOMETRY 485
11.6.4 Intercept form of the equation of a plane
In this section, we shall deduce the equation of a plane in terms of the intercepts made
by the plane on the coordinate axes. Let the equation of the plane be
Ax + By + Cz + D = 0 (D ≠ 0) ... (1)
Let the plane make intercepts a, b, c on x, y and z axes, respectively (Fig 11.16).
Hence, the plane meets x, y and z-axes at (a, 0, 0),
(0, b, 0), (0, 0, c), respectively.
Therefore Aa + D = 0 or A =
D
a
−
Bb + D = 0 or B =
D
b
−
Cc + D = 0 or C =
D
c
−
Substituting these values in the equation (1) of the
plane and simplifying, we get
xyz
abc
++
= 1 ... (1)
which is the required equation of the plane in the intercept form.
Example 19 Find the equation of the plane with intercepts 2, 3 and 4 on the x, y and
z-axis respectively.
Solution Let the equation of the plane be
xyz
abc
++
= 1 ... (1)
Here a = 2, b = 3, c = 4.
Substituting the values of a, b and c in (1), we get the required equation of the
plane as
1
234
xyz
++=
or 6x + 4y + 3z = 12.
11.6.5 Plane passing through the intersection
of two given planes
Let π
1
and π
2
be two planes with equations
1
ˆ
rn
⋅
= d
1
and
2
ˆ
rn
⋅
= d
2
respectively. The position
vector of any point on the line of intersection must
satisfy both the equations (Fig 11.17).
Fig 11.16
Fig 11.17
2019-20
MATHEMATICS
486
If
t
is the position vector of a point on the line, then
1
ˆ
tn
⋅
= d
1
and
2
ˆ
tn
⋅
= d
2
Therefore, for all real values of λ, we have
12
ˆˆ
()
tn n
⋅ +λ
=
12
dd
+λ
Since
t
is arbitrary, it satisfies for any point on the line.
Hence, the equation
1 21 2
()rn n d d
⋅ +λ = +λ
represents a plane π
3
which is such
that if any vector
r
satisfies both the equations π
1
and π
2
, it also satisfies the equation
π
3
i.e., any plane passing through the intersection of the planes
1
rn
⋅
=
1 22
and
d rn d
⋅=
has the equation
12
()
rn n
⋅ +λ
⋅ +λ⋅ +λ
⋅ +λ
= d
1
+
λλ
λλ
λd
2
... (1)
Cartesian form
In Cartesian system, let
1
n
=
12 1
ˆ
ˆˆ
AB C
i jk
++
2
n
=
22 2
ˆ
ˆˆ
ABC
i jk
++
and
r
=
ˆ
ˆˆ
xi y j zk
++
Then (1) becomes
x (A
1
+ λA
2
) + y (B
1
+ λB
2
) + z (C
1
+ λC
2
) = d
1
+ λd
2
or (A
1
x + B
1
y + C
1
z – d
1
) +
λλ
λλ
λ(A
2
x + B
2
y + C
2
z – d
2
) = 0 ... (2)
which is the required Cartesian form of the equation of the plane passing through the
intersection of the given planes for each value of λ.
Example 20 Find the vector equation of the plane passing through the intersection of
the planes
ˆ
ˆ
ˆˆ ˆˆ
( ) 6 and (2 3 4 ) 5,
r i jk r i j k
⋅ + + = ⋅ + + =−
and the point (1, 1, 1).
Solution Here,
1
ˆ
ˆˆ
n i jk
=++
and
2
n
=
ˆ
ˆˆ
2 3 4;
i jk
++
and d
1
= 6 and d
2
= –5
Hence, using the relation
1 21 2
()
rn n d d
⋅ +λ = +λ
, we get
ˆ ˆ
ˆˆ ˆ ˆ
[ (2 3 4 )]
ri jk i j k
⋅ + + +λ + +
=
65
−λ
or
ˆ
ˆˆ
[(1 2 ) (1 3 ) (1 4 ) ]
r i jk
⋅ +λ + +λ + +λ
=
65
−λ
… (1)
where, λ is some real number.
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THREE DIMENSIONAL GEOMETRY 487
Taking
ˆ
ˆˆ
r xi y j zk
=+ +
, we get
ˆ ˆ
ˆˆ ˆˆ
( ) [(1 2 ) (1 3 ) (1 4 ) ] 6 5
xi y j zk i j k
+ + ⋅ +λ + +λ + +λ =−λ
or (1 + 2λ ) x + (1 + 3λ) y + (1 + 4λ) z = 6 – 5λ
or (x + y + z – 6 ) + λ (2x + 3y + 4 z + 5) = 0 ... (2)
Given that the plane passes through the point (1,1,1), it must satisfy (2), i.e.
(1 + 1 + 1 – 6) + λ (2 + 3 + 4 + 5) = 0
or λ =
3
14
Putting the values of λ in (1), we get
39 6
ˆ
ˆˆ
11 1
7 14 7
ri j k
+ ++ ++
=
15
6
14
−
or
10 23 13
ˆ
ˆˆ
7 14 7
ri jk
++
=
69
14
or
ˆ
ˆˆ
(20 23 26 )
ri jk
⋅ ++
= 69
which is the required vector equation of the plane.
11.7 Coplanarity of Two Lines
Let the given lines be
r
=
11
ab
+λ
... (1)
and
r
=
22
ab
+µ
... (2)
The line (1) passes through the point, say A, with position vector
1
a
and is parallel
to
1
b
. The line (2) passes through the point, say B with position vector
2
a
and is parallel
to
2
b
.
Thus,
AB
=
21
aa
−
The given lines are coplanar if and only if
AB
is perpendicular to
12
bb
×
.
i.e.
12
AB.( )
bb
×
= 0 or
2 1 12
( )( )
a a bb
− ⋅×
= 0
Cartesian form
Let (x
1
, y
1
, z
1
) and (x
2
, y
2
, z
2
) be the coordinates of the points A and B respectively.
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Let a
1
, b
1
, c
1
and a
2
, b
2
, c
2
be the direction ratios of
1
b
and
2
b
, respectively. Then
21 2 1 21
ˆ
ˆˆ
AB ( ) ( ) ( )
x xi y y j z zk
=− +− +−
11 1 1 22 2 2
ˆ ˆ
ˆˆ ˆˆ
and
b ai b j ck b ai b j c k
=++ = + +
The given lines are coplanar if and only if
(
)
12
AB 0
bb
⋅× =
. In the cartesian form,
it can be expressed as
21 2 121
111
222
0
xxy yz z
abc
abc
−−−
=
... (4)
Example 21 Show that the lines
+3 1 5
–3 1 5
x yz
−−
==
and
+1 2 5
–1 2 5
xy z
−−
==
are coplanar.
Solution Here, x
1
= – 3, y
1
= 1, z
1
= 5, a
1
= – 3, b
1
= 1, c
1
= 5
x
2
= – 1, y
2
= 2, z
2
= 5, a
2
= –1, b
2
= 2, c
2
= 5
Now, consider the determinant
21 2 121
111
222
210
315 0
125
x xy yz z
abc
abc
−−−
=− =
−
Therefore, lines are coplanar.
11.8 Angle between Two Planes
Definition 2 The angle between two planes is defined as the angle between their
normals (Fig 11.18 (a)). Observe that if θ is an angle between the two planes, then so
is 180 – θ (Fig 11.18 (b)). We shall take the acute angle as the angles between
two planes.
Fig 11.18
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THREE DIMENSIONAL GEOMETRY 489
If
1
n
and
2
n
are normals to the planes and θ be the angle between the planes
1
rn
⋅
= d
1
and
22
.
dn r
=
.
Then θ is the angle between the normals to the planes drawn from some common
point.
We have, cos
θθ
θθ
θ =
Note The planes are perpendicular to each other if
1
n
.
2
n
= 0 and parallel if
1
n
is parallel to
2
n
.
Cartesian form Let θ be the angle between the planes,
A
1
x + B
1
y + C
1
z + D
1
= 0 and A
2
x + B
2
y + C
2
z + D
2
= 0
The direction ratios of the normal to the planes are A
1
, B
1
, C
1
and A
2
, B
2
, C
2
respectively.
Therefore, cos
θθ
θθ
θ =
AA BB CC
A B C ABC
121212
1
2
1
2
1
2
2
2
2
2
2
2
++ ++
++ ++ ++ ++
Note
1. If the planes are at right angles, then θ = 90
o
and so cos θ = 0.
Hence, cos θ = A
1
A
2
+ B
1
B
2
+ C
1
C
2
= 0.
2. If the planes are parallel, then
1 11
2 22
A BC
A BC
==
.
Example 22 Find the angle between the two planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7
using vector method.
Solution The angle between two planes is the angle between their normals. From the
equation of the planes, the normal vectors are
1
N
=
ˆ
ˆˆ
22
ij k
+−
and
2
ˆ
ˆˆ
N3 6 2
i jk
=− −
Therefore cos θ = =
4
21
Hence θ = cos
– 1
21
4
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Example 23 Find the angle between the two planes 3x – 6y + 2z = 7 and 2x + 2y – 2z =5.
Solution Comparing the given equations of the planes with the equations
A
1
x + B
1
y + C
1
z + D
1
= 0 and A
2
x + B
2
y + C
2
z + D
2
= 0
We get A
1
= 3, B
1
= – 6, C
1
= 2
A
2
= 2, B
2
= 2, C
2
= – 2
cos θ =
( )( )
2 2 2 22 2
3 2 ( 6) (2) (2) ( 2)
3 ( 6) ( 2) 2 2 ( 2)
× +− + −
+− +− + +−
=
10 5 5 3
21
7 23 73
−
==
×
Therefore, θ = cos
-1
53
21
11.9 Distance of a Point from a Plane
Vector form
Consider a point P with position vector
a
and a plane π
1
whose equation is
ˆ
rn
⋅
= d (Fig 11.19).
Consider a plane π
2
through P parallel to the plane π
1
. The unit vector normal to
π
2
is
n
ˆ
. Hence, its equation is
ˆ
( )0
r an
− ⋅=
i.e.,
ˆ
rn
⋅
=
ˆ
an
⋅
Thus, the distance ON′ of this plane from the origin is
ˆ
||
an
⋅
. Therefore, the distance
PQ from the plane π
1
is (Fig. 11.21 (a))
i.e., ON – ON′ = | d –
ˆ
|
an
⋅
Fig 11.19
(a)
a
Z
X
Y
d
N’
P
O
1
2
N
Q
1
(b)
P
X
Z
Y
O
d
N’
N
a
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THREE DIMENSIONAL GEOMETRY 491
which is the length of the perpendicular from a point to the given plane.
We may establish the similar results for (Fig 11.19 (b)).
Note
1. If the equation of the plane π
2
is in the form
N
rd
⋅=
, where
N
is normal
to the plane, then the perpendicular distance is
|N |
|N|
ad
⋅−
.
2. The length of the perpendicular from origin O to the plane
N
⋅=
rd
is
||
|N|
d
(since
a
= 0).
Cartesian form
Let P(x
1
, y
1
, z
1
) be the given point with position vector
a
and
Ax + By + Cz = D
be the Cartesian equation of the given plane. Then
a
=
111
ˆ
ˆˆ
xi y j zk
++
N
=
ˆ
ˆˆ
ABC
i jk
++
Hence, from Note 1, the perpendicular from P to the plane is
111
222
ˆ ˆ
ˆˆ ˆˆ
( ) (A B C ) D
ABC
xi y j zk i j k+ + ⋅ ++ −
++
=
1 11
222
ABCD
ABC
xyz
++−
++
Example 24 Find the distance of a point (2, 5, – 3) from the plane
ˆ
ˆˆ
(632)
rijk
⋅ −+
= 4
Solution Here,
ˆ ˆ
ˆˆ ˆˆ
25 3,N63 2
=+ − =−+
ai jk i jk
and d = 4.
Therefore, the distance of the point (2, 5, – 3) from the given plane is
ˆ
ˆ
ˆˆ ˆ ˆ
| (2 5 3 ) (6 3 2 ) 4|
ˆ
ˆˆ
|6 3 2 |
i jk i jk
i jk
+− ⋅ − + −
−+
=
| 12 15 6 4 | 13
7
36 9 4
− −−
=
++
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11.10 Angle between a Line and a Plane
Definition 3 The angle between a line and a plane is
the complement of the angle between the line and
normal to the plane (Fig 11.20).
Vector form If the equation of the line is
ba r
λ
+=
and the equation of the plane is
rn d
⋅=
. Then the angle θ between the line and the
normal to the plane is
cos θ =
||||
bn
bn
⋅
⋅
and so the angle φ between the line and the plane is given by 90 – θ, i.e.,
sin (90 – θ) = cos θ
i.e. sin φ =
||||
bn
bn
⋅
or φ =
–1
sin
bn
bn
⋅
Example 25 Find the angle between the line
1
2
x
+
=
3
36
yz
−
=
and the plane 10 x + 2y – 11 z = 3.
Solution Let θ be the angle between the line and the normal to the plane. Converting the
given equations into vector form, we have
r
=
ˆ ˆ
ˆ ˆˆ
( – 3 ) (2 3 6 )
ik i jk
+ +λ + +
and
ˆ
ˆˆ
( 10 2 11 )
r ijk
⋅ +−
= 3
Here
b
=
ˆ
ˆˆ
236
i jk
++
and
kj in
ˆ
11
ˆ
2
ˆ
10
− +=
sin φ =
2 22 22 2
ˆ
ˆ
ˆˆ ˆ ˆ
(2 3 6 ) (10 2 11 )
2 3 6 10 2 11
ijk i j k
++ ⋅ + −
++ ++
=
40
7 15
−
×
=
8
21
−
=
8
21
or φ =
1
8
sin
21
−
Fig 11.20
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THREE DIMENSIONAL GEOMETRY 493
EXERCISE 11.3
1. In each of the following cases, determine the direction cosines of the normal to
the plane and the distance from the origin.
(a) z = 2 (b) x + y + z = 1
(c) 2x + 3y – z = 5 (d) 5y + 8 = 0
2. Find the vector equation of a plane which is at a distance of 7 units from the
origin and normal to the vector
k
ji
ˆ
6
ˆ
5
ˆ
3
−+
.
3. Find the Cartesian equation of the following planes:
(a)
ˆ
ˆˆ
( )2
ri j k
⋅+− =
(b)
ˆ
ˆˆ
(2 3 4 ) 1
ri j k
⋅ +− =
(c)
ˆ
ˆˆ
[( 2 ) (3 ) (2 ) ] 15
r s ti t j s t k
⋅ − +− + + =
4. In the following cases, find the coordinates of the foot of the perpendicular
drawn from the origin.
(a) 2x + 3y + 4z – 12 = 0 (b) 3y + 4z – 6 = 0
(c) x + y + z = 1 (d) 5y + 8 = 0
5. Find the vector and cartesian equations of the planes
(a) that passes through the point (1, 0, – 2) and the normal to the plane is
ˆ
ˆˆ
.
i jk
+−
(b) that passes through the point (1,4, 6) and the normal vector to the plane is
ˆ
ˆˆ
2.
i jk
−+
6. Find the equations of the planes that passes through three points.
(a) (1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3)
(b) (1, 1, 0), (1, 2, 1), (– 2, 2, – 1)
7. Find the intercepts cut off by the plane 2x + y – z = 5.
8. Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX
plane.
9. Find the equation of the plane through the intersection of the planes
3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).
10. Find the vector equation of the plane passing through the intersection of the
planes
ˆ
ˆˆ
.(2 2 3 ) 7
ri jk
+− =
,
ˆ
ˆˆ
.(2 5 3 ) 9
ri jk
++ =
and through the point
(2, 1, 3).
11. Find the equation of the plane through the line of intersection of the
planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane
x – y + z = 0.
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12. Find the angle between the planes whose vector equations are
ˆ
ˆˆ
(2 2 3 ) 5
ri j k
⋅+− =
and
ˆ
ˆˆ
(3 3 5 ) 3
ri j k
⋅−+ =
.
13. In the following cases, determine whether the given planes are parallel or
perpendicular, and in case they are neither, find the angles between them.
(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0
(b) 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0
(c) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0
(d) 2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0
(e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0
14. In the following cases, find the distance of each of the given points from the
corresponding given plane.
Point Plane
(a) (0, 0, 0) 3x – 4y + 12 z = 3
(b) (3, – 2, 1) 2x – y + 2z + 3 = 0
(c) (2, 3, – 5) x + 2y – 2z = 9
(d) (– 6, 0, 0) 2x – 3y + 6z – 2 = 0
Miscellaneous Examples
Example 26 A line makes angles α, β, γ and δ with the diagonals of a cube, prove that
cos
2
α + cos
2
β + cos
2
γ + cos
2
δ =
4
3
Solution A cube is a rectangular parallelopiped having equal length, breadth and height.
Let OADBFEGC be the cube with each side of length a units. (Fig 11.21)
The four diagonals are OE, AF, BG and CD.
The direction cosines of the diagonal OE which
is the line joining two points O and E are
222 222 222
0 0 0
, ,
a a a
aaa aaa aaa
−
− −
++ ++ ++
i.e.,
3
1
,
3
1
,
3
1
A( ,0,0)a
B(0, ,0)a
C(0,0, )a
G
( ,0, )aa
F(0, , )aa
E
(,,)aaa
X
D( , ,0)aa
Y
Z
O
Fig 11.21
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THREE DIMENSIONAL GEOMETRY 495
Similarly, the direction cosines of AF, BG and CD are
–1
3
,
3
1
,
3
1
;
3
1
,
–1
3
,
3
1
and
3
1
,
3
1
,
–1
3
, respectively.
Let l, m, n be the direction cosines of the given line which makes angles α, β, γ, δ
with OE, AF, BG, CD, respectively. Then
cosα =
1
3
(l + m+ n); cos β =
1
3
(– l + m + n);
cosγ =
1
3
(l – m + n); cos δ =
1
3
(l + m – n) (Why?)
Squaring and adding, we get
cos
2
α + cos
2
β + cos
2
γ + cos
2
δ
=
1
3
[ (l + m + n )
2
+ (–l + m + n)
2
] + (l – m + n)
2
+ (l + m –n)
2
]
=
1
3
[ 4 (l
2
+ m
2
+ n
2
) ] =
3
4
(as l
2
+ m
2
+ n
2
= 1)
Example 27 Find the equation of the plane that contains the point (1, – 1, 2) and is
perpendicular to each of the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8.
Solution The equation of the plane containing the given point is
A (x – 1) + B(y + 1) + C (z – 2) = 0 ... (1)
Applying the condition of perpendicularly to the plane given in (1) with the planes
2x + 3y – 2z = 5 and x + 2y – 3z = 8, we have
2A + 3B – 2C = 0 and A + 2B – 3C = 0
Solving these equations, we find A = – 5C and B = 4C. Hence, the required
equation is
– 5C (x – 1) + 4 C (y + 1) + C(z – 2) = 0
i.e. 5x – 4y – z = 7
Example 28 Find the distance between the point P(6, 5, 9) and the plane determined
by the points A (3, – 1, 2), B (5, 2, 4) and C(– 1, – 1, 6).
Solution Let A, B, C be the three points in the plane. D is the foot of the perpendicular
drawn from a point P to the plane. PD is the required distance to be determined, which
is the projection of
AP
on
AB AC
×
.
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Hence, PD = the dot product of
AP
with the unit vector along
AB AC
×
.
So
AP
= 3
k ji
ˆ
7
ˆ
6
ˆ
++
and
AB AC
×
=
ˆ
ˆˆ
ˆ
ˆˆ
2 3 2 12 16 12
4 04
i jk
i jk
=−+
−
Unit vector along
AB AC
×
=
ˆ
ˆˆ
34 3
34
i jk
−+
Hence PD = (
)
ˆ
7
ˆ
6
ˆ
3
kji ++
.
ˆ
ˆˆ
34 3
34
i jk
−+
=
17
3 43
Alternatively, find the equation of the plane passing through A, B and C and then
compute the distance of the point P from the plane.
Example 29 Show that the lines
xad
−+
α−δ
=
ya zad
− −−
=
α α+δ
and
xbc
−+
β−γ
=
yb zbc
− −−
=
β β+γ
are coplanar.
Solution
Here x
1
= a – dx
2
= b – c
y
1
= a y
2
= b
z
1
= a + dz
2
= b + c
a
1
= α – δ a
2
= β – γ
b
1
= α b
2
= β
c
1
= α + δ c
2
= β + γ
Now consider the determinant
21 2 121
1 1 1
2 2 2
xxy yz z
abc
abc
−−−
=
bcadb abc a d
−− + − + − −
α−δ α α+δ
β−γ β β+γ
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THREE DIMENSIONAL GEOMETRY 497
Adding third column to the first column, we get
2
bababcad
− − +− −
α α α +δ
β β β+γ
= 0
Since the first and second columns are identical. Hence, the given two lines are
coplanar.
Example 30 Find the coordinates of the point where the line through the points
A (3, 4, 1) and B (5, 1, 6) crosses the XY-plane.
Solution The vector equation of the line through the points A and B is
r
=
ˆ ˆ
ˆˆ ˆˆ
3 4 [ (5 3) (1 4 ) ( 6 1) ]
i jk i jk
+ + +λ − + − + −
i.e.
r
=
ˆ ˆ
ˆˆ ˆˆ
3 4 (2 3 5 )
i jk i j k
+ + +λ − +
... (1)
Let P be the point where the line AB crosses the XY-plane. Then the position
vector of the point P is of the form
j yi x
ˆˆ
+
.
This point must satisfy the equation (1). (Why ?)
i.e.
ˆˆ
xi y j
+
=
ˆ
ˆ ˆ
(32) (43) (15)
i j k
+λ+−λ++λ
Equating the like coefficients of
ˆ
ˆˆ
, and
ij k
, we have
x = 3 + 2 λ
y = 4 – 3 λ
0 = 1 + 5 λ
Solving the above equations, we get
x =
13 23
and
55
y =
Hence, the coordinates of the required point are
0,
5
23
,
5
13
.
Miscellaneous Exercise on Chapter 11
1. Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the
line determined by the points (3, 5, – 1), (4, 3, – 1).
2. If l
1
, m
1
, n
1
and l
2
, m
2
, n
2
are the direction cosines of two mutually perpendicular
lines, show that the direction cosines of the line perpendicular to both of these
are
12 21 12 2112 21
, ,
m n m n nl n l lm l m
− − −
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3. Find the angle between the lines whose direction ratios are a, b, c and
b – c, c – a, a – b.
4. Find the equation of a line parallel to x-axis and passing through the origin.
5. If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and
(2, 9, 2) respectively, then find the angle between the lines AB and CD.
6. If the lines
1 23 1 1 6
and
32 2 3 1 5
x y z xy z
k k
− −− −−−
==
==
− −
are perpendicular,
find the value of k.
7. Find the vector equation of the line passing through (1, 2, 3) and perpendicular to
the plane
09 )
ˆ
5
ˆ
2
ˆ
(.
= +− + kji r
.
8. Find the equation of the plane passing through (a, b, c) and parallel to the plane
ˆ
ˆˆ
( ) 2.
ri jk
⋅++ =
9. Find the shortest distance between lines
ˆ ˆ
ˆˆ ˆ ˆ
622 (22)
ri j k i j k
= + + +λ − +
and
ˆ ˆ
ˆ ˆˆ
4 (3 2 2 )
r ik i j k
=− − +µ − −
.
10. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,1)
crosses the YZ-plane.
11. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1)
crosses the ZX-plane.
12. Find the coordinates of the point where the line through (3, – 4, – 5) and
(2, – 3, 1) crosses the plane 2x + y + z = 7.
13. Find the equation of the plane passing through the point (– 1, 3, 2) and perpendicular
to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
14. If the points (1, 1, p) and (– 3, 0, 1) be equidistant from the plane
ˆ
ˆˆ
(3 4 12 ) 13 0,
⋅ + − +=
ri j k
then find the value of p.
15. Find the equation of the plane passing through the line of intersection of the
planes
ˆ
ˆˆ
( )1
ri jk
⋅ ++ =
and
ˆ
ˆˆ
(2 3 ) 4 0
r i jk
⋅ + − +=
and parallel to x-axis.
16. If O be the origin and the coordinates of P be (1, 2, – 3), then find the equation of
the plane passing through P and perpendicular to OP.
17. Find the equation of the plane which contains the line of intersection of the planes
ˆ
ˆˆ
( 2 3) 4 0
ri j k
⋅ + + −=
,
ˆ
ˆˆ
(2 ) 5 0
r i jk
⋅ +− +=
and which is perpendicular to the
plane
ˆ
ˆˆ
(5 3 6 ) 8 0
ri jk
⋅ + − +=
.
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THREE DIMENSIONAL GEOMETRY 499
18. Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the
line
ˆ ˆ
ˆˆ ˆˆ
2 2 (3 4 2 )
r ij k i j k
= − + +λ + +
and the plane
ˆ
ˆˆ
( )5
ri j k
⋅ −+ =
.
19. Find the vector equation of the line passing through (1, 2, 3) and parallel to the
planes
ˆ
ˆˆ
( 2) 5
ri j k
⋅ −+ =
and
ˆ
ˆˆ
(3 ) 6
r i jk
⋅ ++ =
.
20. Find the vector equation of the line passing through the point (1, 2, – 4) and
perpendicular to the two lines:
7
10
16
19
3
8
−
=
−
+
=
−
zy x
and
15
3
x
−
=
29 5
85
yz
−−
=
−
.
21. Prove that if a plane has the intercepts a, b, c and is at a distance of p units from
the origin, then
2
222
11 1 1
pc b a
=+ +
.
Choose the correct answer in Exercises 22 and 23.
22. Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is
(A) 2 units (B) 4 units (C) 8 units (D)
2
29
units
23. The planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are
(A) Perpendicular (B) Parallel
(C) intersect y-axis (D) passes through
5
0,0,
4
Summary
Direction cosines of a line are the cosines of the angles made by the line
with the positive directions of the coordinate axes.
If l, m, n are the direction cosines of a line, then l
2
+ m
2
+ n
2
= 1.
Direction cosines of a line joining two points P(x
1
, y
1
, z
1
) and Q(x
2
, y
2
, z
2
) are
2 1 2 12 1
,,
PQ PQ PQ
x xy yz z
−−−
where PQ =
()
2
12
2
12
2
12
)( ) ( z z y yx x − +− +−
Direction ratios of a line are the numbers which are proportional to the
direction cosines of a line.
If l, m, n are the direction cosines and a, b, c are the direction ratios of a line
2019-20
MATHEMATICS
500
then
l =
222
cba
a
++
; m =
222
cba
b
++
; n =
222
cba
c
++
Skew lines are lines in space which are neither parallel nor intersecting.
They lie in different planes.
Angle between skew lines is the angle between two intersecting lines
drawn from any point (preferably through the origin) parallel to each of the
skew lines.
If l
1
, m
1
, n
1
and l
2
, m
2
, n
2
are the direction cosines of two lines; and θ is the
acute angle between the two lines; then
cosθ = | l
1
l
2
+ m
1
m
2
+ n
1
n
2
|
If a
1
, b
1
, c
1
and a
2
, b
2
, c
2
are the direction ratios of two lines and θ is the
acute angle between the two lines; then
cosθ =
1 2 12 12
2 22 2 2 2
111222
aa bb cc
abcabc
++
++ ++
Vector equation of a line that passes through the given point whose position
vector is
a
and parallel to a given vector
b
is
ra b
= +λ
.
Equation of a line through a point (x
1
, y
1
, z
1
) and having direction cosines l, m, n is
1 11
xx y y z z
l mn
−−−
==
The vector equation of a line which passes through two points whose position
vectors are
a
and
b
is
()
ra ba
= +λ −
.
Cartesian equation of a line that passes through two points (x
1
, y
1
, z
1
) and
(x
2
, y
2
, z
2
) is
1 1 1
21 2 1 21
xx yy zz
xx yyzz
− −−
==
− −−
.
If θ is the acute angle between
11
ra b
= +λ
and
22
ra b
= +λ
, then
12
12
cos
| || |
bb
bb
⋅
θ=
If
1
1
1
1
1
1
n
zz
m
yy
l
xx
−
=
−
=
−
and
2
2
2
2
2
2
n
zz
m
yy
l
xx
−
=
−
=
−
are the equations of two lines, then the acute angle between the two lines is
given by cos θ = |l
1
l
2
+ m
1
m
2
+ n
1
n
2
|.
2019-20
THREE DIMENSIONAL GEOMETRY 501
Shortest distance between two skew lines is the line segment perpendicular
to both the lines.
Shortest distance between
11
ra b
= +λ
and
22
ra b
= +µ
is
12 2 1
12
( )( – )
||
bb a a
bb
×⋅
×
Shortest distance between the lines:
1 1 1
1 1 1
x x y y zz
abc
−−−
==
and
2 2
2 2
xx yy
ab
−−
=
=
2
2
zz
c
−
is
2 1 2 12 1
111
222
2
2 2
12 21 1 2 2 1 12 21
( )( )( )
x xy yz z
abc
abc
bc b c ca c a ab a b
−−−
− +− +−
Distance between parallel lines
1
ra b
= +λ
and
2
ra b
= +µ
is
21
()
||
baa
b
×−
In the vector form, equation of a plane which is at a distance d from the
origin, and
n
ˆ
is the unit vector normal to the plane through the origin is
ˆ
rn d
⋅=
.
Equation of a plane which is at a distance of d from the origin and the direction
cosines of the normal to the plane as l, m, n is lx + my + nz = d.
The equation of a plane through a point whose position vector is
a
and
perpendicular to the vector
N
is
( ).N 0
ra
−=
.
Equation of a plane perpendicular to a given line with direction ratios A, B, C
and passing through a given point (x
1
, y
1
, z
1
) is
A (x – x
1
) + B (y – y
1
) + C (z – z
1
) = 0
Equation of a plane passing through three non collinear points (x
1
, y
1
, z
1
),
2019-20
MATHEMATICS
502
(x
2
, y
2
, z
2
) and (x
3
, y
3
, z
3
) is
1 3 13 13
121212
111
z z yy xx
z z yy xx
z zy yx x
−−−
−−−
−−−
= 0
Vector equation of a plane that contains three non collinear points having
position vectors
ba
,
and
c
is
().[()()]0
ra ba ca
− −× − =
Equation of a plane that cuts the coordinates axes at (a, 0, 0), (0, b, 0) and
(0, 0, c) is
1
=++
c
z
b
y
a
x
Vector equation of a plane that passes through the intersection of
planes
11 22
and
rn d rn d
⋅= ⋅ =
is
1 21 2
()
rn n d d
⋅ +λ = +λ
, where λ is any
nonzero constant.
Cartesian equation of a plane that passes through the intersection of two
given planes A
1
x + B
1
y + C
1
z + D
1
= 0 and A
2
x + B
2
y + C
2
z + D
2
= 0
is (A
1
x + B
1
y + C
1
z + D
1
) + λ(A
2
x + B
2
y + C
2
z + D
2
) = 0.
Two lines
11
ra b
= +λ
and
22
ra b
= +µ
are coplanar if
2 1 12
( )( )
aabb
−⋅×
= 0
In the cartesian form two lines =
xx
a
yy
b
zz
c
and
xx
a
−
=
−
=
−−
1
1
1
1
1
1
2
2
2 2
2 2
––
yy zz
bC
==
are coplanar if
222
111
1 2 1212
c ba
c ba
z z yy xx − −−
= 0.
In the vector form, if θ is the angle between the two planes,
11
rn d
⋅=
and
22
rn d
⋅=
, then θ = cos
–1
12
12
||
| || |
nn
nn
⋅
.
The angle φ between the line
ra b
= +λ
and the plane
ˆ
rn d
⋅=
is
2019-20
THREE DIMENSIONAL GEOMETRY 503
ˆ
sin
ˆ
| || |
bn
bn
⋅
φ=
The angle θ between the planes A
1
x + B
1
y + C
1
z + D
1
= 0 and
A
2
x + B
2
y + C
2
z + D
2
= 0 is given by
cos
θ =
1 2 12 12
222 222
111 222
AA BB CC
ABC ABC
++
++ ++
The distance of a point whose position vector is
a
from the plane
ˆ
rn d
⋅=
is
ˆ
||
d an
−⋅
The distance from a point (x
1
, y
1
, z
1
) to the plane Ax + By + Cz + D = 0 is
1 11
222
A BCD
ABC
x yz
+++
++
.
—
—
2019-20
