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THREE D IMENSIONAL G EOMETRY
11
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THREE DIMENSIONAL GEOMETRY
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Chapter
463
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v The moving power of mathematical invention is not
reasoning but imagination. – A. DEMORGAN v
11.1 Introduction
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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. T he 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
Leonhard Euler
planes, a line and a plane, shortest distance between two
(1707-1783)
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 andDirection Ratios of a Line
From Chapter 10, recall that if a directed line L passing through the origin makes
angles a, b and g with x, y and z-axes, respectively, called direction angles, then cosine
of these angles, namely, cos a, cos b and cos g are called direction cosines of the
directed line L.
If we reverse the direction of L,then the direction angles arereplaced by their supplements,
i.e.,
,
and
. Thus, the signs of the direction cosines are reversed.
p-a p-b p-g
* Fo r va ri ou s ac ti vi ti es i n th re e di me ns io na l ge omet ry, on e ma y re fe r to t he B oo k
“A Hand Book for designing Mathematics Laboratory in Schools”, NCERT, 2005
M ATHE MATI CS
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Fig 11.1
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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. T hese 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.
Nowtake 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 = ll, b=lm and c = ln, for any nonzero l Î R.
ANote
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
T herefore
But
T herefore
or
l
m
=
=
a
b
l = ak, m =
2
l + m 2 + n2 =
2
k (a 2 + b 2 + c2) =
n
= k (say), k being a constant.
c
bk, n = ck
1
1
k= ±
1
a + b 2 + c2
2
... (1)
THREE D IMENSIONAL G EOMETRY
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Hence, from (1), the d.c.’s of the line are
a
a 2 + b 2 + c2
,m = ±
b
a 2 + b 2 + c2
c
,n = ±
a 2 + b 2 + c2
d
l =±
he
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 ¹ 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.
Z
bl
S
R
x
a=OA
=
. T his gives x = lr..
OP r
P (x, y , z)
O
a
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Let OP = r. T hen cos
is
11.2.1 Relation between the direction cosines of a line
Consider a line RSwith direction cosines l, m, n. T hrough
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).
Similarly,
T hus
But
Hence
y = mr and z = nr
x + y + z2 = r2 (l2 + m 2 + n 2)
x2 + y2 + z2 = r 2
l2 + m 2 + n2 = 1
2
2
Y
P
a
A
X
r
O
a
x
A
Fig 11.2
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(x1, y1, z1) and Q(x2, y2, z2)
as follows (Fig 11.3 (a)).
Fig 11.3
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M ATHE MATI CS
cosg =
NQ
z -z
= 2 1
PQ
PQ
Similarly
cosa =
x2 - x1
y -y
and cos b = 2 1
PQ
PQ
is
T herefore,
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Let l, m, n be the direction cosines of the line PQ and let it makes angles a, b and g
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, ÐPQN=
g (Fig 11.3 (b).
Hence, the direction cosines of the line segment joining the points P(x1, y1, z1) and
Q(x2, y2, z2) are
bl
x2 - x1 y2 - y1 z 2 - z1
,
,
PQ
PQ
PQ
-x ) +( y -y ) +(z -z )
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whe re
( x2
PQ =
2
2
2
1
2
1
2
1
ANote
T he direction ratios of the line segment joining P(x1, y1, z1) and Q(x2, y2, z2)
may be taken as
x2 – x1, y2 – y1, z2 – z1 or x1 – x2, y1 – y2, z1 – z2
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. T hen l = cos 90 0 = 0, m = cos 60 0 =
1
,
2
3
.
2
Example 2 If a line has direction ratios 2, – 1, – 2, determine its direction cosines.
n = cos 30 0 =
Solution Direction cosines are
2
2 + (-1) + (-2)
2
2
2
,
-1
2 + (-1) + (-2 )
2
2
2
,
-2
2 + ( - 1) + ( -2) 2
2
2
2 -1 -2
, ,
3 3 3
Example 3 Find the direction cosines of the line passing through the two points
(– 2, 4, – 5) and (1, 2, 3).
or
THREE D IMENSIONAL G EOMETRY
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Solution We know the direction cosines of the line passing through two points
P(x1, y1, z1) and Q(x2, y2, z2) are given by
( x2 - x1 ) 2 + ( y2 - y1 ) 2 + ( z2 - z1 )
PQ =
2
he
whe re
Here P is (– 2, 4, – 5) and Q is (1, 2, 3).
,
-2
,
8
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77
77
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Example 4 Find the direction cosines of x, y and z-axis.
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So
PQ = (1 - ( -2)) 2 + (2 - 4) 2 + (3 - ( -5)) 2 =
T hus, the direction cosines of the line joining two points is
3
d
x2 - x1 y2 - y1 z 2 - z1
,
,
PQ
PQ
PQ
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Solution T he x-axis makes angles 0°, 90° and 90° respectively with x, y and z-axis.
T herefore, 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.
T he 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 ofAB and BC are proportional, hence, AB is parallel
to BC. But po int B is common to both AB and BC. T herefore , A, B, C are
collinear points.
EXERCIS E 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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11.3 Equation of a Line in S pace
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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.
r
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11.3.1 Equation of a line through a given point and parallel to a given vector b
r
Let a be the position vector of the given point
A wit h r esp ect to th e o rigin O o f t he
rectangular coordinate system. Let l be the
line which passes through the point A and is
r
r
parallel to a given vector b . Let r be the
position vector of an arbitrary point P on the
line (Fig 11.4).
r
uuur
T hen AP is parallel to the vector b , i.e.,
r
uuur
Fig 11.4
AP = lb , where l is some real number..
uuur
uuur uuur
AP = OP – OA
But
r
r r
i.e.
lb = r a
Conversely, for each value of the parameter l, this equation gives the position
vector of a point P on the line. Hence, the vector equation of the line is given by
r
r
r
r = a+λ b
... (1)
r
Remark If b aiˆ bjˆ ckˆ , then a, b, c are direction ratios of the line and conversely,,
r
if a, b, c are direction ratios of a line, then b = aiˆ + bjˆ + ckˆ will be the parallel to
r
the line. Here, b should not be confused with | b |.
-
= ++
Derivation of carte sian form from ve ctor form
Let the coordinates of the given point A be (x1, y1, z1) and the direction ratios of
the line be a, b, c. Consider the coordinates of any point P be (x, y, z). T hen
r
r
r = x iˆ + yˆj + zkˆ ; a = x1 ˆi + y 1 ˆj + z1 kˆ
and
r
b = a iˆ + b ˆj + c kˆ
Substituting these values in (1) and equating the coefficients of iˆ , jˆ and k̂ , we get
x = x1 + la; y = y1 + l b; z = z1+ lc
... (2)
THREE D IMENSIONAL G EOMETRY
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T hese are parametric equations of the line. Eliminating the parameter l from (2),
we get
x – x1
y – y1 z – z1
=
=
a
b
c
T his is the Cartesian equation of the line.
d
If l, m, n are the direction cosines of the line, the equation of the line is
he
ANote
... (3)
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x – x1
y – y1 z – z 1
=
=
l
m
n
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 3 ˆi + 2 ˆj - 8 kˆ .
Solution We have
r
r
a = 5 ˆi + 2 ˆj - 4 kˆ and b = 3 iˆ + 2 ˆj - 8 kˆ
T herefore, the vector equation of the line is
r
r = 5 iˆ + 2 jˆ - 4 kˆ + l ( 3 iˆ + 2 jˆ - 8 kˆ )
r
Now, r is the position vector of any point P(x, y, z) on the line.
x iˆ + y ˆj + z kˆ = 5 ˆi + 2 ˆj - 4 kˆ + l ( 3 iˆ + 2 ˆj - 8 kˆ)
T herefore,
Eliminating l , we get
= (5 + 3l) $i + (2 + 2 l) $j + ( - 4 - 8l ) k$
x -5
y- 2 z + 4
=
=
3
2
-8
which is the equation of the line in Cartesian form.
11.3.2 Equation of a line passing through two given points
r
r
Let a and b be the position vectors of two
po in ts A ( x1, y1, z1 ) an d B( x 2, y 2, z 2) ,
respectively that are lying on a line (Fig 11.5).
r
Let r be t he positio n vector of an
arbitrary point P(x, y, z), then P is a point on
uuur r r
th e lin e if a nd on ly if AP r a an d
=-
uuur r r
AB = b - a are collinear vectors. T herefore,
P is on the line if and only if
r r
r r
r a
(b a )
- =l -
Fig 11.5
M ATHE MATI CS
r r
r r
or
r = a + l ( b - a ) , l Î R.
T his is the vector equation of the line.
Derivation of carte sian form from ve ctor form
We have
=+ + = + +
... (1)
= + +
is
Equating the like coefficients of ˆi , ˆj , kˆ , we get
he
r
r
r
r x iˆ y jˆ z kˆ , a x1iˆ y1 ˆj z1 kˆ and b x2 iˆ y2 jˆ z 2 kˆ,
Substituting these values in (1), we get
x $i + y $j + z k$ = x1 $i + y1 $j + z1 k$ + l [( x2 - x1 ) $i + ( y2 - y1 ) $j + ( z2 - z1 ) k$ ]
d
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x = x1 + l (x2 – x1); y = y1 + l (y2 – y1); z = z1 + l (z2 – z1)
On eliminating l, we obtain
x - x1
y - y1 z - z1
=
=
x2 - x1 y2 - y1 z 2 - z1
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).
r
r
Solution Let a and b be the position vectors of the point A(– 1, 0, 2) and B(3, 4, 6).
r
Then
a
iˆ 2 kˆ
r
and
b 3 iˆ 4 ˆj 6 kˆ
r r
T herefore
b - a = 4 iˆ + 4 ˆj + 4 kˆ
r
Let r be the position vector of any point on the line. Then the vector equation of
the line is
r
r
iˆ 2 kˆ (4 iˆ 4 ˆj 4 kˆ )
=-+
=+ +
=-+ +l + +
Example 8 T he Cartesian equation of a line is
x 3 y 5 z 6
2
4
2
Find the vector equation for the line.
+= - = +
Solution Comparing the given equation with the standard form
x - x1 y - y1 z - z1
=
=
a
b
c
We observe that
x1 = – 3, y1 = 5, z1 = – 6; a = 2, b = 4, c = 2.
THREE D IMENSIONAL G EOMETRY
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T hus, the required line passes through the point (– 3, 5, – 6) and is parallel to the
r
vector 2 iˆ + 4 ˆj + 2 kˆ . Let r be the position vector of any point on the line, then the
vector equation of the line is given by
r
r ( 3 ˆi 5 ˆj 6 kˆ ) + l (2 iˆ + 4 ˆj + 2 kˆ )
d
=- + -
is
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Fig 11.6
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Let L1 and L2 be two lines passing through the origin
and with direction ratios a 1, b 1, c1 and a 2, b 2, c2,
respectively. Let P be a point on L1 and Q be a point
on L2. Consider the directed lines OP and OQ as
given in Fig 11.6. Let q be the acute angle between
OP an d OQ. No w recall that the directed line
segments OP and OQ are vectors with components
a 1, b 1, c1 and a 2, b 2, c2, respectively. T herefore, the
angle q between them is given by
he
11.4 Angle between Two Lines
cos q =
a1 a2 + b1 b2 + c1 c2
... (1)
a12 + b12 + c12 a 22 + b 22 + c22
T he angle between the lines in terms of sin q is given by
- q
sin q = 1 cos 2
1-
=
(a12 + b12 + c12 )( a22 + b22 + c22 )- ( a1a 2 + b1b2 + c1c2 )2
(a12 + b12 + c12 ) ( a22 + b22 + c22 )
=
ANote
( a1a 2 + b1b2 + c1 c2 ) 2
=
( a12 + b12 + c12 )(a 22 + b22 + c22 )
( a1 b2
-a
2
+(b c -b c ) +(c a -c
+b +c a +b +c
b1 ) 2
a12
2
1
2
1
2
2
1
2
1
2
2
1
2
2
2
2
2
2
a1 ) 2
... (2)
In case the lines L1 and L2 do not pass through the origin, we may take
lines L¢1 and L¢2 which are parallel to L1 and L2 respectively and pass through
the origin.
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M ATHE MATI CS
If instead of direction ratios for the lines L1 and L2, direction cosines, namely,
l1, m 1, n 1 for L1 and l2, m 2, n2 for L2 are given, then (1) and (2) takes the following form:
(l1 m2 - l2 m1 )2 - (m1 n 2 - m2 n1 )2 + (n1 l2 - n 2 l1 )2
Two lines with direction ratios a 1, b 1, c1 and a 2, b 2, c2 are
(i) perpendicular i.e. if q = 90° by (1)
a 1a 2 + b 1b 2 + c1c2 = 0
(ii) parallel i.e. if q = 0 by (2)
... (4)
he
sin q =
... (3)
is
and
(as l12 + m12 + n12 =1 = l22 + m22 + n 22 )
d
cos q = |l1 l2 + m 1m 2 + n 1n 2 |
+m
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a1
b1 c1
=
a2 = b2 c2
Now, we find the angle between two lines when their equations are given. If q is
acute the angle between the lines
r
ur
r
r
r
b2
b1 and r = a2
r = a1
r r
b1 × b 2
then
cosq = r r
b1 b 2
In Cartesian form, if q is the angle between the lines
and
x - x1
y - y1 z - z1
=
=
a1
b1
c1
... (1)
x - x2
y - y2 z - z 2
=
=
a2
b2
c2
... (2)
where, a 1 ,b 1, c1 and a 2, b 2, c2 are the direction ratios of the lines (1) and (2), respectively,
then
cos q =
a1 a2 + b1 b2 + c1 c2
a12 + b12 + c12
a22 + b22 + c22
Example 9 Find the angle between the pair of lines given by
r
r = 3 ˆi + 2 ˆj - 4 kˆ + l (iˆ + 2 ˆj + 2 kˆ )
r
and
r = 5 iˆ - 2 jˆ + m (3 ˆi + 2 ˆj + 6 kˆ)
THREE D IMENSIONAL G EOMETRY
473
he
d
r
r
Solution Here b1 = iˆ + 2 jˆ + 2 kˆ and b2 = 3 iˆ + 2 jˆ + 6 kˆ
T he angle q between the two lines is given by
r r
b1 × b2
(iˆ + 2 ˆj + 2kˆ) × (3 iˆ + 2 ˆj + 6 kˆ)
cos q = r r =
1 + 4 + 4 9 + 4 + 36
b1 b2
3 + 4 + 12 19
=
3´ 7
21
æ 19 ö
Hence
q = cos–1 ç ÷
è 21 ø
Example 10 Find the angle between the pair of lines
x+3
y -1 z + 3
=
=
3
5
4
x +1
y-4 z-5
=
and
=
1
1
2
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=
Solution T he direction ratios of the first line are 3, 5, 4 and the direction ratios of the
second line are 1, 1, 2. If q is the angle between them, then
cos q =
3.1 + 5.1 + 4.2
32 + 52 + 42
12 + 12 + 2 2
=
16
50 6
=
16
5 2 6
=
8 3
15
æ8 3 ö
Hence, the required angle is cos–1 çç
÷÷ .
è 15 ø
11.5 S hortest 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
Fig 11.7
x, y and z-axes respectively Fig 11.7.
474
M ATHE MATI CS
he
d
T he 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. T hese
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.
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11.5.1 Distance between two skew lines
We now determine the shortest distance between two skew lines in the following way:
Let l1 and l2 be two skew lines with equations (Fig. 11.8)
r
r
r
r = a1 + l b1
... (1)
r
r
r
r = a2 + m b2
and
... (2)
r
r
Take any point S on l1 with position vector a1 and T on l2, with position vector a 2.
T hen the magnitude of the shortest distance vector
T
will be equal to that of the projection of ST along the
Q
direction of the line of shortest distance (See 10.6.2).
l2
uuur
If PQ is the shortest distance vector between
r
l1 and l2 , then it being perpendicular to both b1 and
l1
r
uuur
S
P
b2 , the unit vector n̂ along PQ would therefore be
r
r
Fig 11.8
b1 ´ b2
r
r
... (3)
n̂ =
| b1 ´ b2 |
uuur
Then
PQ = d n̂
where, d is the magnitude of the shortest distance vector. Let q be the angle between
uur
uuur
ST and PQ . T hen
PQ = ST |cos q |
uuur uur
PQ × ST
But
cos q = uuuur uur
| PQ | | ST |
r
r
d nˆ × (a 2 - a1 )
uur
r
r
=
(since ST = a2 - a1 )
d ST
r
r
r
r
(b1 ´ b2 ) × (a 2 - a1)
r r
=
[From (3)]
ST b1 ´ b2
THREE D IMENSIONAL G EOMETRY
he
x - x2
y - y2
z - z2
=
=
a2
b2
c2
is
l2 :
bl
x - x1
y - y1
z - z1
=
=
a1
b1
c1
x2 - x1
a1
y2 - y1
b1
z 2 - z1
c1
a2
b2
c2
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and
l1 :
d
Hence, the required shortest distance is
d = PQ = ST |cos q |
r
r
r
r
( b1 ´ b2 ) . ( a 2 - a 1 )
r
r
or
d =
| b1 ´ b2 |
Carte sian form
T he shortest distance between the lines
is
475
( b1 c2 - b2 c1 ) 2 + ( c1 a 2 - c2 a1 ) 2 + ( a1 b2 - a 2 b1 ) 2
11.5.2 Distance between parallel lines
If two lines l1 and l2 are parallel, then they are coplanar. Let the lines be given by
r
r
r
... (1)
r = a1 + l b
r
r
r
and
… (2)
r = a2 + m b
where, ar1 is the position vector of a point S on l1 and
r
a 2 is the position vector of a point T on l2 Fig 11.9.
As l1, l2 are coplanar, if the foot of the perpendicular
from T on the line l1 is P, then the distance between the
lines l1 and l2 = |T P |.
r
uur
Let q be the angle between the vectors ST and b .
Then
r uur
r uur
b ´ ST = ( | b || ST | sin q) nˆ
Fig 11.9
where n̂ is the unit vector perpendicular to the plane of the lines l1 and l2.
uur r
r
But
ST = a2 a1
-
... (3)
M ATHE MATI CS
T herefore, from (3), we get
r
r
r r
b ´ ( a2 - a1 ) = | b | PT nˆ
(since PT = ST sin q)
r r
r
r
| b ´ ( a 2 - a1 ) | = | b | PT 1
i.e.,
(as | n̂ | = 1)
Hence, the distance between the given parallel lines is
r r
r
uuur
b ´ ( a 2 - a1 )
|
PT
|
=
r
d =
|b|
he
×
d
476
bl
is
Example 11 Find the shortest distance between the lines l1 and l2 whose vector
equations are
r
... (1)
r = ˆi + ˆj + l (2 ˆi - jˆ + kˆ )
r
and
... (2)
r = 2 iˆ + ˆj - kˆ + m (3 iˆ - 5 ˆj + 2 kˆ )
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r
r
r
r
r
Solution Comparing (1) and (2) with rr = a1 + l b1 and r = a2 + m b 2 respectively,,
r
r
we get
a1 = ˆi + ˆj , b1 = 2 iˆ - ˆj + kˆ
r
r
a2 = 2 î + ĵ – kˆ and b2 = 3 î – 5 ĵ + 2 kˆ
r
r
T herefore
a2 - a1 = î - kˆ
r r
and
b1 ´ b2 = ( 2 iˆ - ˆj + kˆ ) ´ ( 3 ˆi - 5 ˆj + 2 kˆ )
=
iˆ jˆ kˆ
2 -1 1
= 3 iˆ - ˆj - 7 kˆ
3 -5 2
r r
| b1 ´ b2 | = 9 +1 + 49 = 59
Hence, the shortest distance between the given lines is given by
r
r
r
r
( b1 ´ b2 ) . ( a 2 - a1 ) = | 3 - 0 + 7 | = 10
r
r
d =
59
59
| b1 ´ b2 |
So
Example 12 Find the distance between the lines l1 and l2 given by
r
r = ˆi + 2 ˆj - 4 kˆ + l ( 2 iˆ + 3 jˆ + 6 kˆ )
and
r
r = 3 iˆ + 3 jˆ - 5 kˆ + m ( 2 iˆ + 3 ˆj + 6 kˆ )
THREE D IMENSIONAL G EOMETRY
477
d
Solution T he two lines are parallel (Why? ) We have
r
r
r
a1 = ˆi + 2 ˆj - 4 kˆ , a2 = 3 ˆi + 3 ˆj - 5 kˆ and b = 2 iˆ + 3 ˆj + 6 kˆ
T herefore, the distance between the lines is given by
he
iˆ ˆj kˆ
2 3 6
2 1 -1
r r r
b ´ ( a 2 - a1 )
r
d=
=
|b |
| - 9 iˆ + 14 ˆj - 4 kˆ |
=
49
293
49
=
293
7
EXERCIS E 11.2
bl
=
1. Showthat the three lines with direction cosines
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is
4 + 9 + 36
12 -3 -4
4 12 3
3 -4 12
,
,
;
,
, ;
,
,
are mutually perpendicular..
13 13 13 13 13 13 13 13 13
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 3 iˆ + 2 ˆj - 2 kˆ .
5. Find the equation of the line in vector and in cartesian form that passes through
the point with position vector 2 iˆ - j + 4 kˆ and is in the direction iˆ + 2 jˆ - kˆ .
6. Find the cartesian equation of the line which passes through the point (– 2, 4, – 5)
and parallel to the line given by
x +3 y- 4 z +8
.
=
=
3
5
6
x -5 y+ 4 z -6
. Write its vector form.
=
=
3
7
2
8. Find the vector and the cartesian equations of the lines that passes through the
origin and (5, – 2, 3).
7. T he cartesian equation of a line is
478
M ATHE MATI CS
9. Find the vector and the cartesian equations of the line that passes through the
points (3, – 2, – 5), (3, – 2, 6).
he
+= - =-
is
- = -= +
-
d
10. Find the angle between the following pairs of lines:
r
(i) r = 2 iˆ - 5 ˆj + kˆ + l (3 ˆi + 2 ˆj + 6 kˆ ) and
r
r = 7 iˆ - 6 kˆ + m ( iˆ + 2 ˆj + 2 kˆ )
r
(ii) r = 3 iˆ + jˆ - 2 kˆ + l (iˆ - jˆ - 2 kˆ ) and
r
r = 2 iˆ - jˆ - 56 kˆ + m (3 iˆ - 5 ˆj - 4 kˆ )
11. Find the angle between the following pair of lines:
x 2 y 1 z 3
x 2 y 4 z 5
and
2
5
3
1
8
4
x y z
x - 5 y - 2 z -3
= = and
=
=
(ii)
2 2 1
4
1
8
1 - x 7 y - 14 z - 3
12. Find the values of p so that the lines
=
=
3
2p
2
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(i)
and
7 - 7x y - 5 6 - z
=
=
are at right angles.
3p
1
5
13. Show that the lines
x -5 y+ 2 z
x y z
=
= and = = are perpendicular to
7
-5
1
1 2 3
each other.
14. Find the shortest distance between the lines
r
r = ( iˆ + 2 jˆ + kˆ ) + l ( iˆ - ˆj + kˆ ) and
r
r = 2 iˆ - jˆ - kˆ + m (2 iˆ + jˆ + 2 kˆ )
15. Find the shortest distance between the lines
x +1 y + 1 z + 1
x -3 y-5 z -7
=
=
=
=
and
7
-6
1
1
-2
1
16. Find the shortest distance between the lines whose vector equations are
r
r = ( iˆ + 2 jˆ + 3 kˆ ) + l ( iˆ - 3 ˆj + 2 kˆ )
r
and r = 4 iˆ + 5 ˆj + 6 kˆ + m (2 iˆ + 3 ˆj + kˆ )
17. Find the shortest distance between the lines whose vector equations are
r
r = (1 - t ) ˆi + ( t - 2) ˆj + (3 - 2 t) kˆ and
r
r = ( s + 1) iˆ + (2 s - 1) jˆ - (2 s + 1) kˆ
THREE D IMENSIONAL G EOMETRY
479
11.6 Plane
he
d
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
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Consider a plane whose perpendicular distance from the origin is d (d ¹ 0). Fig 11.10.
uuur
If ON is the normal from the origin to the plane, and n̂ is the unit normal vector
uuur
uuur
Z
along ON . T hen ON = d n̂ . Let P be any
uuur
po int on th e p lan e. T her efo re, NP is
uuur
perpendicular to ON .
uuur uuur
T herefore, NP × ON = 0
... (1)
P( x, y,z )
r
Let r be the position vector of the point P,,
r
uuur r
uuur uuur uuur
then NP = r - d nˆ (as ON NP OP )
d N
+ =
O
T herefore, (1) becomes
r
Ù
Ù
(r - d n ) × d n = 0
or
r
Ù
Ù
(r - d n ) × n = 0
or
r Ù
Ù Ù
r ×n - d n× n = 0
Y
X
(d ¹ 0)
r Ù
Ù Ù
r×n = d
i.e.,
(as n × n = 1)
T his is the vector form of the equation of the plane.
Carte sian form
Fig 11.10
… (2)
Equation (2) gives the vector equation of a plane, where n̂ is the unit vector normal to
the plane. Let P(x, y, z) be any point on the plane. T hen
uuur
r
OP = r x iˆ y ˆj z kˆ
= + +
Let l, m, n be the direction cosines of n̂ . Then
n̂ = l iˆ
+m ˆj +n kˆ
480
M ATHE MATI CS
T herefore, (2) gives
A
he
d
( x iˆ + y ˆj + z kˆ ) × (l ˆi + m ˆj + n kˆ ) = d
i.e.,
lx + my + nz = d
... (3)
T his is the cartesian equation of the plane in the normal form.
r
Note Equation (3) shows that if r × ( a iˆ + b jˆ + c kˆ ) = 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
is
- +
6
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from the origin and its normal vector from the origin is 2 iˆ 3 ˆj 4 kˆ .
r
Solution Let n = 2 iˆ - 3 jˆ + 4 kˆ . T hen
r
n
2 iˆ - 3 ˆj + 4 kˆ
2 iˆ - 3 ˆj + 4 kˆ
nˆ = r =
=
| n|
4 + 9 + 16
29
Hence, the required equation of the plane is
r æ 2 ˆ -3 ˆ
4 ˆö
6
r ×ç
i+
j+
k÷ =
29
29 ø
29
è 29
Example 14 Find the direction cosines of the unit vector perpendicular to the plane
r
r × (6 iˆ - 3 jˆ - 2 kˆ ) + 1 = 0 passing through the origin.
Solution T he given equation can be written as
r
r × ( - 6 ˆi + 3 ˆj + 2 kˆ ) = 1
+ + =7
... (1)
No w
| - 6 iˆ + 3 ˆj + 2 kˆ | = 36 9 4
T herefore, dividing both sides of (1) by 7, we get
r æ 6
3 ˆ 2 ˆö
1
r × ç - iˆ +
j+
k÷ =
7
7 ø
7
è 7
r
× =d .
which is the equation of the plane in the form r nˆ
T his shows that nˆ = -
6 ˆ 3 ˆ 2 ˆ
i + j+
k is a unit vector perpendicular to the
7
7
7
plane through the origin. Hence, the direction cosines of n̂ are - 6 , 3 , 2 .
7
7 7
THREE D IMENSIONAL G EOMETRY
481
Example 15 Find the distance of the plane 2x – 3y + 4z – 6 = 0 from the origin.
2 + (- 3) + 4
2
2
2
,
-3
2 + (- 3) + 4
2
2
4
,
2
2
2 + (- 3) + 4
2
2
, i.e.,
2
29
,
-3
29
,
4
29
he
2
d
Solution Since the direction ratios of the normal to the plane are 2, –3, 4; the direction
cosines of it are
Hence, dividing the equation 2x – 3y + 4z – 6 = 0 i.e., 2x – 3y + 4z = 6 throughout by
29 , we get
x +
-3
4
y +
6
z =
is
2
bl
29
29
29
29
T his 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
6
29
.
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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
planeis(x1, y1, z1) (Fig 11.11).
Z
T hen, the direction ratios of the line OP are
x1, y1, z1.
P (x1 , y 1, z1 )
Writing the equation of the plane in the normal
form, we have
2
29
x-
3
29
y+
4
z=
6
29
O
29
2 , -3
4
,
ar e th e dire ct io n X
29 29 29
cosines of the OP.
wh er e,
Fig 11.11
Since d.c.’s and direction ratios of a line are proportional, we have
x1
y1
=
=
2
-3
29
29
i.e.,
x1 =
2k
29
z1
=k
4
29
, y1 =
-3k
29
, z1 =
4k
29
Y
482
M ATHE MATI CS
Substituting these in the equation of the plane, we get k =
6
29
.
d
æ 12 -18 , 24 ö
Hence, the foot of the perpendicular is ç ,
÷.
è 29 29 29 ø
ANote
is
bl
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(x1, y1, z1), only one such plane exists (see
Fig 11.12).
he
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).
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Let a plane pass through a point A with ur
position
Fig 11.12
r
vector a and perpendicular to the vector N .
r
Let r be the position vector of any point P(x, y, z) in the plane. (Fig 11.13).
T hen the point P lies in the plane if and only if
uuur
ur
uuur ur
AP is perpendicular to N . i.e., AP . N = 0. But
uuur r r
r r r
AP = r - a . Therefore, ( r - a ) × N = 0
… (1)
T his is the vector equation of the plane.
Carte sian form
Let the given point A be (x1, y1, z1), P be (x, y, z)
ur
and direction ratios of N are A, B and C. T hen,
No w
Fig 11.13
= + +
r
r
r
a = x1 iˆ + y1 ˆj + z1 kˆ, r = x iˆ + y jˆ + z kˆ and N A iˆ B ˆj C kˆ
r r r
( r – a ) N= 0
×
So
é( x - x1 ) iˆ + ( y - y1 ) jˆ + ( z - z1 ) kˆ ù × (A iˆ + B jˆ + C kˆ ) = 0
ë
û
i.e.
A (x – x 1) + B (y – y1) + C (z – z1) = 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.
THREE D IMENSIONAL G EOMETRY
483
r
Solution We have the position vector of point (5, 2, – 4) as a = 5 iˆ + 2 jˆ - 4kˆ and the
r
r
normal vector N perpendicular to the plane as N =2 ˆi + 3 ˆj kˆ
r r r
T herefore, the vector equation of the plane is given by ( r a ) .N 0
r
or
... (1)
[ r - (5 iˆ + 2 ˆj - 4 kˆ )] × (2 iˆ + 3 ˆj - kˆ ) = 0
T ransforming (1) into Cartesian form, we have
-+ -- +=
bl
is
or
2( x 5) 3( y 2) 1( z 4) 0
i.e.
2x + 3y – z = 20
which is the cartesian equation of the plane.
he
[( x – 5) iˆ + ( y - 2) ˆj + ( z + 4) kˆ ] × (2 iˆ + 3 ˆj - kˆ ) = 0
d
- =
11.6.3 Equation of a plane passing through three non collinear points
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r r
Let R, S and T be three non collinear points on the plane with position vectors a , b and
r
c respectively (Fig 11.14).
Z
(RS X RT)
R
P
O
r a
b
S
c
T
Y
X
Fig 11.14
uuur
uuur
uuur
T he vectors RS and RT are in the given plane. Therefore, the vector RS
uuur
´RT
r
is perpendicular to the plane containing points R, S and T. Let r be the position vector
of any point P in the plane. T herefore, the equation of the plane passing through R and
uuur uuur
perpendicular to the vector RS RT is
r r uuur uuur
( r - a ) × (RS´ RT) = 0
r r r r r r
( r – a ).[( b – a )×(c – a )] = 0
or
… (1)
´
484
M ATHE MATI CS
Note Why was it necessary to say that the three points
A
had to be non collinear? If the three points were on the same
R
line, then there will be many planes that will contain them
(Fig 11.15).
he
S
T
d
T his is the equation of the plane in vector form passing through three noncollinear
points.
is
T hese planes will resemble the pages of a book where the
line containing the points R, Sand T are members in the binding
of the book.
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Fig 11.15
Carte sian form
Let (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) 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
r
vector r . T hen
uuur
RP = (x – x1) î + (y – y1) ĵ + (z – z1) k̂
uuur
RS = (x2 – x1) î + (y2 – y1) ĵ + (z2 – z1) k̂
uuur
RT = (x3 – x1) î + (y3 – y1) ĵ + (z3 – z1) k̂
Substituting these values in equation (1) of the vector form and expressing it in the
form of a determinant, we have
x - x1
x2 - x1
y - y1
y2 - y1
z - z1
z 2 - z1 = 0
x3 - x1
y3 - y1
z 3 - z1
which is the equation of the plane in Cartesian form passing through three non collinear
points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3).
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).
r
r
r
Solution Let a 2 iˆ 5 jˆ 3 kˆ , b = - 2 iˆ - 3 ˆj + 5 kˆ , c 5 iˆ 3 ˆj 3 kˆ
r
r
T hen the vector equation of the plane passing through ar , b and c and is
given by
r r uuur uuur
( r - a ) × (RS´ RT) = 0 (Why?)
r r r r
r r
or
( r - a ) × [(b - a ) ´(c - a )] = 0
r
i.e.
[ r (2 iˆ 5 jˆ 3 kˆ )] [( 4 iˆ 8 jˆ 8 kˆ ) (3 iˆ 2 jˆ )] 0
=+-
=+ -
- + - ×-- + ´ - =
THREE D IMENSIONAL G EOMETRY
485
-D
a
-D
Bb + D = 0 or B =
b
-D
Cc + D = 0 or C =
is
Aa + D = 0 or A =
bl
T herefore
he
d
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 ¹ 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.
Fig 11.16
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c
Substituting these values in the equation (1) of the
plane and simplifying, we get
x y z
+ + =1
a b c
which is the required equation of the plane in the intercept form.
... (1)
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
x y z
+ + =1
... (1)
a b c
He re
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
x y z
+ + = 1 or 6x + 4y + 3z = 12.
2 3 4
11.6.5 Plane passing thro ugh the intersection
of two given planes
Let p1 an d p 2 be two p lan es with equat io ns
×
×
r
r
r nˆ1 = d 1 and r nˆ 2 = d2 respectively. T he position
vector of any point on the line of intersection must
satisfy both the equations (Fig 11.17).
Fig 11.17
486
M ATHE MATI CS
is
he
d
r
If t is the position vector of a point on the line, then
r
r
t × nˆ1 = d 1 and t × nˆ 2 = d 2
T herefore, for all real values of l, we have
r
t × ( nˆ1 +lnˆ2 ) = d1 +ld 2
r
Since t is arbitrary, it satisfies for any point on the line.
r r
r
Hence, the equation r × ( n1 + ln2 ) = d1 +l d 2 represents a plane p3 which is such
r
that if any vector r satisfies both the equations p1 and p2, it also satisfies the equation
p3 i.e., any plane passing through the intersection of the planes
r r
r r
r × n1 = d1 and r × n2 = d 2
r r
r
has the equation
r × ( n1 + ln2 ) = d 1 + ld 2
... (1)
bl
Carte sian form
In Cartesian system, let
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r
n1 = A1 iˆ + B 2 ˆj + C1 kˆ
r
n2 = A2 iˆ + B 2 ˆj + C 2 kˆ
r
r = x iˆ + y ˆj + z kˆ
and
T hen (1) becomes
x (A1 + lA2) + y (B1 + lB2) + z (C1 + lC2) = d 1 + ld 2
or
(A1 x + B 1 y + C 1z – d 1) + l(A2 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 l.
Example 20 Find the vector equation of the plane passing through the intersection of
r
r
the planes r × ( iˆ + jˆ + kˆ ) = 6 and r × (2iˆ + 3 ˆj + 4 kˆ ) = - 5, and the point (1, 1, 1).
r
r
Solution Here, n1 = iˆ + jˆ + kˆ and n2 = 2 iˆ + 3 ˆj + 4 kˆ;
and
d 1 = 6 and d 2 = –5
r
r
r
Hence, using the relation r × ( n1 + l n2 ) = d1 + l d 2 , we get
r
r ×[ iˆ + jˆ + kˆ +l (2 iˆ + 3 ˆj + 4 kˆ )] = 6 - 5 l
or
r
r ×[(1 + 2 l) iˆ + (1 + 3 l) jˆ + (1 + 4 l)kˆ ] = 6 - 5 l
where, l is some real number.
… (1)
THREE D IMENSIONAL G EOMETRY
Taking
487
=+ +
r
r x iˆ y jˆ z kˆ , we get
(1 + 2l ) x + (1 + 3l) y + (1 + 4l) z = 6 – 5l
or
(x + y + z – 6 ) + l (2x + 3y + 4 z + 5) = 0
... (2)
he
or
d
( x iˆ + y ˆj + z kˆ ) ×[(1 + 2l ) iˆ + (1 + 3l ) ˆj + (1+ 4l )kˆ ]= 6 - 5l
Given that the plane passes through the point (1,1,1), it must satisfy (2), i.e.
3
14
Putting the values of l in (1), we get
or
l=
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r éæ 3 ö æ 9 ö æ 6 ö ù
15
r êç 1 + ÷ iˆ + ç1 + ÷ ˆj + ç1 + ÷ kˆ ú = 6ëè 7 ø è 14 ø è 7 ø û
14
is
(1 + 1 + 1 – 6) + l (2 + 3 + 4 + 5) = 0
r æ 10 23 13 ö 69
r ç iˆ + jˆ + kˆ ÷ =
14
7 ø 14
è 7
r
or
r × (20 iˆ + 23 ˆj + 26 kˆ ) = 69
which is the required vector equation of the plane.
or
11.7 Coplanarity of Two Lines
Let the given lines be
r
r
r
r = a1 +l b1
r
r
r
and
r = a2 + mb2
... (1)
... (2)
r
T he line (1) passes through the point, say A, with position vector a1 and is parallel
r
r
to b1 . T he line (2) passes through the point, say B with position vector a2 and is parallel
r
to b2 .
uuur
r r
Thus,
AB = a2 a1
uuur
r r
T he given lines are coplanar if and only if AB is perpendicular to b1 ´b2 .
uuur r r
r r
r r
i.e.
AB.( b1 ´ b2 ) = 0 or ( a2 - a1 ) × (b1 ´ b2 ) = 0
-
Carte sian form
Let (x1, y1, z1) and (x2, y2, z2) be the coordinates of the points A and B respectively.
M ATHE MATI CS
r
r
Let a 1, b 1, c1 and a 2, b 2, c2 be the direction ratios of b1 and b2 , respectively. T hen
uuur
AB ( x2 x1 ) iˆ ( y2 y1 ) ˆj (z 2 z1 )kˆ
r
r
b1 a1 iˆ b1 ˆj c1 kˆ and b2 a 2 iˆ b2 ˆj c2 kˆ
= - + - + = + +
= uuu+
+
r r r
d
488
-
-
a2
b2
c2
=
is
-
he
T he given lines are coplanar if and only if AB× ( b1 ´b2 ) = 0 . In the cartesian form,
it can be expressed as
x2 x1 y2 y1 z2 z1
... (4)
a1
b1
c1
0
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Example 21 Showthat the lines
x +1 y - 2 z - 5
x +3 y -1 z - 5
=
=
and
=
=
are coplanar..
–1
2
5
–3
1
5
Solution Here, x1 = – 3, y1 = 1, z1 = 5, a 1 = – 3, b 1 = 1, c1 = 5
x2 = – 1, y2 = 2, z2 = 5, a 2 = –1, b 2 = 2, c2 = 5
Now, consider the determinant
x2
-x
1
y2
-y
1
z2
-z
1
a1
b1
c1
a2
b2
c2
2 1 0
3 1 5
=-1
=0
2 5
T herefore, lines are coplanar.
11.8 Angle between Two Planes
De finition 2 T he angle between two planes is defined as the angle between their
normals (Fig 11.18 (a)). Observe that if q is an angle between the two planes, then so
is 180 – q (Fig 11.18 (b)). We shall take the acute angle as the angles between
two planes.
Fig 11.18
THREE D IMENSIONAL G EOMETRY
489
r
r
If n1 and n2 are normals to the planes and q be the angle between the planes
r r
r r
r n1 = d 1 and r . n 2 = d 2 .
T hen q is the angle between the normals to the planes drawn from some common
point.
r r
n1 × n2
We have,
cos q = r
r
| n1 | | n2 |
r r
Note T he planes are perpendicular to each other if n . n
A
r
r
1
= 0 and parallel if
is
n1 is parallel to n2 .
2
he
d
×
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Carte sian form Let q be the angle between the planes,
A1 x + B1 y + C1z + D1 = 0 and A2x + B2 y + C2 z + D2 = 0
T he direction ratios of the normal to the planes are A1, B1, C1 and A2, B2, C2
respectively.
T herefore, cos q =
A1 A2 + B1 B2 + C1 C2
A12
+ B 12 + C12
A 22 + B 22 + C22
ANote
1. If th e p lan es are at right angles, t hen q = 9 0 o a nd so cos q = 0 .
Hence, cos q = A1A2 + B1B2 + C1C2 = 0.
2. If the planes are parallel, then
A1
A2
=BB =CC .
1
1
2
2
Example 22 Find the angle between the two planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7
using vector method.
Solution T he angle between two planes is the angle between their normals. From the
equation of the planes, the normal vectors are
ur
ur
N1 = 2 iˆ + ˆj - 2 kˆ and N 2 = 3 iˆ - 6 ˆj - 2 kˆ
ur ur
N1 × N2
(2 iˆ + jˆ - 2 kˆ ) × (3 iˆ - 6 jˆ - 2 kˆ )
æ4 ö
ur
=
T herefore
cos q = ur
= ç ÷
| N1 | |N 2 |
4 + 1 + 4 9 + 36 + 4
è 21 ø
Hence
æ4 ö
q = cos – 1 ç ÷
è 21 ø
490
M ATHE MATI CS
Example 23 Find the angle between the two planes 3x – 6y + 2z = 7 and 2x + 2y – 2z =5.
5
=
7´2 3
=
7 3
5 3
21
æ5 3ö
q = cos-1 çç
÷÷
è 21 ø
bl
T herefore,
-10
is
=
(32 + (- 6)2 + (-2)2 ) (2 2 + 22 + (-2)2 )
he
3 ´ 2 + ( -6) (2) + (2) (-2)
cos q =
d
Solution Comparing the given equations of the planes with the equations
A1 x + B1 y + C1 z + D1 = 0 and A2 x + B2 y + C2 z + D2 = 0
We get
A1 = 3, B1 = – 6, C1 = 2
A2 = 2, B2 = 2, C2 = – 2
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11.9 Distance of a Point from a Plane
Ve ctor form
r
Consider a point P with positio n vector a and a plane p 1 whose equation is
r
r × nˆ = d (Fig 11.19).
Z
Z
Q
P
a
N’
O
X
(a)
p2
p1
p2
d
p1
d
N
O
Y
a
P
N’
Y
N
X
(b)
Fig 11.19
Consider a plane p2 through P parallel to the plane p1. T he unit vector normal to
r
r
p2 is n̂ . Hence, its equation is ( r - a ) × nˆ = 0
i.e.,
r
r
r × nˆ = a × nˆ
r
T hus, the distance ON¢ of this plane from the origin is | a × nˆ | . T herefore, the distance
PQ from the plane p1 is (Fig. 11.21 (a))
i.e.,
r
×
ON – ON¢ = |d – a nˆ |
THREE D IMENSIONAL G EOMETRY
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)).
ANote
× =d , where urN is normal
r ur
| a ×N -d | .
to the plane, then the perpendicular distance is
ur
r ur
r N
d
If theequationof theplanep2 is in the form
| N|
× =d is || Nurd ||
r ur
T he length of the perpendicular from origin O to the plane r N
is
2.
he
1.
bl
r
(since a = 0).
Carte sian form
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r
Let P(x1, y1, z1) be the given point with position vector a and
Ax + By + Cz = D
be the Cartesian equation of the given plane. T hen
r
a = x1 iˆ + y1 jˆ + z1 kˆ
ur
N = A iˆ + B ˆj + C kˆ
Hence, from Note 1, the perpendicular from P to the plane is
( x1 iˆ + y1 jˆ + z1 kˆ ) × ( A iˆ + B jˆ + C kˆ ) - D
A2 + B2 + C 2
=
A x1
+B y +C z -D
A +B +C
1
2
2
1
2
Example 24 Find the distance of a point (2, 5, – 3) from the plane
r
r × ( 6 iˆ - 3 ˆj + 2 kˆ ) = 4
ur
r
Solution Here, a 2 iˆ 5 ˆj 3 kˆ , N 6 ˆi 3 ˆj 2 kˆ and d = 4.
= + -
= - +
T herefore, the distance of the point (2, 5, – 3) from the given plane is
| (2 iˆ + 5 ˆj - 3 kˆ ) × (6 ˆi - 3 ˆj + 2 kˆ ) - 4|
| 12 - 15 - 6 - 4 | 13
=
=
ˆ
ˆ
ˆ
| 6 i -3 j + 2 k |
7
36 + 9 + 4
492
M ATHE MATI CS
11.10 Angle between a Line and a Plane
he
d
De finition 3 T he angle between a line and a plane is
the complement of the angle between the line and
normal to the plane (Fig 11.20).
Ve ctor form If th e equation of the line is
r
r r
r = a + l b a nd the equation of the pla ne is
r r
r × n = d . T hen the angle q between the line and the
normal to the plane is
r r
b ×n
cos q = r r
| b | ×| n |
is
Fig 11.20
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and so the angle f between the line and the plane is given by 90 – q, i.e.,
sin (90 – q) = cos q
r r
b×n
–1 b × n
i.e.
sin f = r r or f = sin
|b | | n |
b n
Example 25 Find the angle between the line
x+1
y
z-3
=
=
2
3
6
and the plane 10 x + 2y – 11 z = 3.
Solution Let q be the angle between the line and the normal to the plane. Converting the
given equations into vector form, we have
r
r = ( – iˆ + 3 kˆ ) + l ( 2 iˆ + 3 ˆj + 6 kˆ )
r
and
r × ( 10 ˆi + 2 ˆj - 11 kˆ ) = 3
r
r
He re
b = 2 iˆ + 3 ˆj + 6 kˆ and n = 10 iˆ + 2 ˆj - 11 kˆ
sin f =
=
(2 iˆ + 3 ˆj + 6 kˆ ) × (10 ˆi + 2 ˆj - 11 kˆ )
2 2 + 32 + 6 2
10 2 + 22 + 112
- 40
-8
8
æ 8 ö
=
=
or f = sin -1 ç ÷
7 ´ 15
21
21
è 21 ø
THREE D IMENSIONAL G EOMETRY
493
EXERCIS E 11.3
he
d
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 3 iˆ + 5 ˆj - 6 kˆ.
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3. Find the Cartesian equation of the following planes:
r
r
(a) r × ( iˆ + jˆ - kˆ ) = 2
(b) r × (2 iˆ + 3 ˆj - 4 kˆ ) = 1
r
(c) r ×[( s - 2 t) iˆ + (3 - t ) ˆj + (2 s + t ) kˆ ] = 15
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
iˆ ˆj kˆ .
+-
(b) that passes through the point (1,4, 6) and the normal vector to the plane is
ˆi 2 ˆj kˆ .
-+
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 e quatio n of t he pla ne thro ugh th e inte rsection 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
r
r
planes r .( 2 iˆ + 2 jˆ - 3 kˆ ) = 7 , r .( 2 iˆ + 5 ˆj + 3 kˆ ) = 9 and through the point
(2, 1, 3).
11. Find the equation of the plane thro ugh the line of interse ction of the
planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane
x – y + z = 0.
494
M ATHE MATI CS
12. Find the angle between the planes whose vector equations are
r
r
r × (2 iˆ + 2 jˆ - 3 kˆ ) = 5 and r × (3 iˆ - 3 jˆ + 5 kˆ ) = 3.
d
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.
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
is
(b) 2x + y + 3z – 2 = 0
he
(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0
Point
Plane
bl
14. In the following cases, find the distance of each of the given points from the
corresponding given plane.
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
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(a) (0, 0, 0)
Miscellaneous Examples
Example 26 A line makes angles a, b, g and d with the diagonals of a cube, prove that
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)
Z
T he four diagonals are OE, AF, BG and CD.
T he direction cosines of the diagonal OE which
C(0, 0, a)
is the line joining two points O and E are
F(0, a, a)
cos2 a + cos2 b + cos2 g + cos2 d =
+a +a
a 0
a2
i.e.,
2
1
3
,
1
3
2
,
,
+a +a
a 0
a2
1
3
2
2
,
+a +a
(a, 0, a) G
a 0
a2
2
2
O
X
Y
B(0, a, 0)
D(a, a, 0)
Fig 11.21
THREE D IMENSIONAL G EOMETRY
Similarly, the direction cosines of AF, BG and CD are
–1
1
1
1
cosg =
1
3
;
1
3
,
–1
3
1
d
(l + m+ n); cos b =
(l – m + n); cos d =
3
Squaring and adding, we get
cos2a + cos2 b + cos2 g + cos2 d
1
3
1
3
he
1
,
3
,
(– l + m + n);
(l + m – n)
is
cosa =
,
1
(Why?)
bl
and
3
,
, respectively..
3
3
3
3
Let l, m, n be the direction cosines of the given line which makes angles a, b, g, d
with OE, AF, BG, CD, respectively. T hen
3
,
–1
495
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1
[ (l + m + n ) 2 + (–l + m + n) 2 ] + (l – m + n) 2 + (l + m –n) 2]
3
1
4
= [ 4 (l2 + m 2 + n 2 ) ] =
(as l2 + m 2 + n 2 = 1)
3
3
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 T he 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
uuur
uuur uuur
is the projection of AP on AB AC .
´
M ATHE MATI CS
uuur
uuur
Hence, PD = the dot product of AP with the unit vector along AB
uuur
So
AP = 3 ˆi + 6 ˆj + 7 kˆ
uuur
Unit vector along AB
´
-4
uuur
´AC
=
ˆj kˆ
3 2 = 12 iˆ - 16 ˆj + 12 kˆ
he
and
iˆ
2
uuur
AC =
0 4
3 iˆ - 4 ˆj + 3 kˆ
34
is
uuur
AB
uuur
.
´AC
d
496
bl
3 iˆ - 4 ˆj + 3 kˆ
PD = ( 3 iˆ + 6 jˆ + 7 kˆ ) .
34
Hence
3 34
17
Alte rnative ly, find the equation of the plane passing through A, B and C and then
compute the distance of the point P from the plane.
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=
Example 29 Showthat the lines
x- a+d
y-a z-a-d
=
=
a-d
a
a+d
x-b+c
y-b
z-b-c
=
=
are coplanar..
b- g
b
b+g
and
Solution
He re
x1 = a – d
y1 = a
z1 = a + d
a1 = a – d
b1 = a
c1 = a + d
Nowconsider the determinant
x2 - x1
a1
y2 - y1
b1
z 2 - z1
c1
a2
b2
c2
x2
y2
z2
a2
b2
c2
=
=
=
=
=
=
=
b–c
b
b+c
b–g
b
b+g
b -c - a + d
a -d
b-g
b - a b +c - a - d
a
a+d
b
b+g
THREE D IMENSIONAL G EOMETRY
497
Adding third column to the first column, we get
b - a b +c - a - d
a
a +d
=0
b
b+g
d
b-a
a
2
b
he
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.
r
... (1)
r = 3 iˆ + 4 ˆj + kˆ + l ( 2 iˆ - 3 ˆj + 5 kˆ )
Let P be the point where the line AB crosses the XY-plane. T hen the position
bl
i.e.
is
Solution T he vector equation of the line through the points A and B is
r
r = 3 iˆ + 4 ˆj + kˆ + l [ (5 - 3) iˆ + (1 - 4) ˆj + ( 6 - 1) kˆ ]
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vector of the point P is of the form x iˆ + y ˆj .
T his point must satisfy the equation (1). (Why ?)
i.e.
x iˆ + y ˆj = (3 + 2 l ) iˆ + ( 4 - 3 l) ˆj + ( 1 + 5 l ) kˆ
Equating the like coefficients of ˆi , ˆj and kˆ , we have
x= 3+2l
y=4–3l
0= 1 +5 l
Solving the above equations, we get
x=
13
23
and y =
5
5
æ 13 23 ö
, 0÷ .
Hence, the coordinates of the required point are ç ,
5
è5
ø
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 l1, m 1, n 1 and l2, 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 m1 n 2
-m
2
n1 , n1 l2
-n
2 l1 ,
l1 m2
-l
2
m1
498
M ATHE MATI CS
3. Find the angle between the lines whose direction ratios are a, b, c a nd
b – c, c – a, a – b.
4. Find the equation of a line parallel to x-axis and passing through the origin.
d
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.
he
x-1
y - 2 z -3
x -1 y - 1 z - 6
are perpendicular,,
=
=
and
=
=
-3
2k
2
3k
1
-5
find the value of k.
7. Find the vector equation of the line passing through (1, 2, 3) and perpendicular to
r
the plane r . ( ˆi + 2 ˆj - 5 kˆ ) + 9 = 0 .
8. Find the equation of the plane passing through (a, b, c) and parallel to the plane
r
r ( iˆ jˆ kˆ ) 2.
bl
× ++ =
is
6. If the lines
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r
9. Find the shortest distance between lines r = 6 iˆ + 2 jˆ + 2 kˆ + l ( iˆ - 2 jˆ + 2 kˆ )
r
and r = - 4 iˆ - kˆ + m (3 iˆ - 2 ˆj - 2 kˆ ) .
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 poin ts ( 1, 1 , p) and (– 3 , 0, 1) be e quidist ant from the pla ne
r
r (3 iˆ 4 jˆ 12 kˆ ) 13 0, then find the value of p.
× + - +=
15. Find the equation of the plane passing through the line of intersection of the
r
r
planes r × ( iˆ + jˆ + kˆ ) = 1 and r × (2 iˆ + 3 jˆ - kˆ ) + 4 = 0 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
r
r
r × ( iˆ + 2 jˆ + 3 kˆ) - 4 = 0 , r × (2 iˆ + jˆ - kˆ) + 5 = 0 and which is perpendicular to the
r
plane r × (5 iˆ + 3 ˆj - 6 kˆ ) + 8 = 0 .
THREE D IMENSIONAL G EOMETRY
499
18. Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the
r
r
line r = 2 iˆ - jˆ + 2 kˆ + l (3 iˆ + 4 jˆ + 2 kˆ ) and the plane r × ( iˆ - jˆ + kˆ ) = 5 .
d
19. Find the vector equation of the line passing through (1, 2, 3) and parallel to the
r
r
planes r × ( iˆ - jˆ + 2 kˆ ) = 5 and r × (3 iˆ + ˆj + kˆ ) = 6 .
he
20. Find the vector equation of the line passing through the point (1, 2, – 4) and
perpendicular to the two lines:
1
1
1
1
+ 2 + 2 = 2.
2
a
b
c
p
bl
the origin, then
is
x - 8 y + 19 z -10
x - 15
y - 29 z - 5 .
=
=
=
and
=
3
- 16
7
3
8
-5
21. Prove that if a plane has the intercepts a, b, c and is at a distance of p units from
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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
units
29
23. T he planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are
(A) Perpendicular
(B) Parallel
5ö
æ
(D) passes through ç 0, 0, ÷
è
4ø
(C) intersect y-axis
Summary
®
®
®
Dire ction cosine s 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 l2 + m 2 + n 2 = 1.
Direction cosines of a line joining two points P(x1, y1, z1) and Q(x2, y2, z2) are
x2
-x , y -y , z -z
PQ
1
2
PQ
where PQ =
®
®
1
2
1
PQ
( x 2 - x1 ) 2 + ( y 2 - y1 ) 2 + ( z 2 - z1 )2
Dire ction 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
500
M ATHE MATI CS
then
®
b
2
2
2
;n=
c
2
he
®
2
is
®
;m =
a +b +c
a +b + c
a + b 2 + c2
Ske w line s are lines in space which are neither parallel nor intersecting.
T hey lie in different planes.
Angle be twe e n ske w line s is the angle between two intersecting lines
drawn from any point (preferably through the origin) parallel to each of the
skew lines.
If l1, m 1, n 1 and l2, m 2, n 2 are the direction cosines of two lines; and q is the
acute angle between the two lines; then
cosq = |l1l2 + m 1m 2 + n 1n 2 |
If a 1, b 1, c1 and a 2, b 2, c2 are the direction ratios of two lines and q is the
acute angle between the two lines; then
2
a1 a 2 + b1 b2 + c1 c2
+ b12 + c12
a22 + b22 + c22
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cosq =
bl
®
a
2
®
®
®
®
d
l=
a12
Vector equation of a line that passes through the given point whose position
r
r
r
r r
vector is a and parallel to a given vector b is r = a + l b .
Equation of a line through a point (x1, y1,z1) and having direction cosines l, m, n is
x - x1 y - y1 z - z1
=
=
l
m
n
T he vector equation of a line which passes through two points whose position
r
r r
r
r r
vectors are a and b is r = a + l (b - a ) .
Cartesian equation of a line that passes through two points (x1, y1, z1) and
®
x - x1
y - y1
z - z1 .
=
=
x2 - x1
y2 - y1 z 2 - z1
r
r
r r
r r
If q is th e acute angle be twee n r = a1 + l b1 an d r = a 2 + l b2 , th en
r r
b1 × b2
cos q = r
r
| b1 | | b2 |
®
If
(x2, y2, z2) is
x - x1
y - y1
z - z1
x - x2
y - y2
z - z2
=
=
=
=
and
l
m
n2
l1
m1
n1
2
2
are the equations of two lines, then the acute angle between the two lines is
given by cos q = |l1 l2 + m 1 m 2 + n 1 n 2 |.
THREE D IMENSIONAL G EOMETRY
®
Shortest distance between two skew lines is the line segment perpendicular
to both the lines.
r
r
r r
r r
Shortest distance between r a1
b2 is
b1 and r a2
r r
r
r
( b1 ´ b2 ) × (a 2 – a1 )
r r
| b1 ´ b2 |
®
Shortest distance between the lines:
= +m
he
= +l
is
x - x1
y - y1
z - z1
=
=
and
a1
b1
c1
y2 - y1
z 2 - z1
a1
b1
c1
a2
b2
c2
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x2 - x1
bl
x - x2 y - y2
z - z2
=
=
is
a2
b2
c2
(b1c2 - b2 c1 )2 + (c1a 2 - c2 a1 )2 + (a1b2 - a 2 b1 )2
®
®
®
®
®
®
d
®
501
= +l
r
r r
r
Distance between parallel lines r a1
b and r
r r r
b ´ ( a 2 - a1 )
r
|b|
=ar +mbr is
2
In the vector form, equation of a plane which is at a distance d from the
origin, and n̂ is the unit vector normal to the plane through the origin is
r
r × nˆ = 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.
r
T he equation of a plane through a point whose position vector is a and
ur
r r ur
perpendicular to the vector N is ( r - a ) . N = 0 .
Equation of a plane perpendicular to a given line with direction ratiosA, B, C
and passing through a given point (x1, y1, z1) is
A (x – x1) + B (y – y1) + C (z – z1 ) = 0
Equation of a plane passing through three non collinear points (x1, y1, z1),
502
M ATHE MATI CS
(x2, y2, z2) and (x3, y3, z3) is
y - y1
y2 - y1
z - z1
z 2 - z1
x3 - x1
y3 - y1
z 3 - z1
=0
d
x - x1
x2 - x1
Vector equation of a plane that contains three non collinear points having
r r
r r
r
r r
r r
position vectors a , b and c is ( r - a ) . [ (b - a ) ´ ( c - a ) ] = 0
®
Equation of a plane that cuts the coordinates axes at (a, 0, 0), (0, b, 0) and
(0, 0, c) is
Vec tor e quation of a plane t hat p asses thro ugh t he in terse ction of
r r
r r
r r
r
planes r × n1 = d1 and r × n2 = d 2 is r × ( n1 + l n 2 ) = d1 + l d 2 , where l is any
nonzero constant.
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®
bl
x
y z
+ + =1
a b c
is
he
®
®
Cartesian equation of a plane that passes through the intersection of two
given planes A1 x + B1 y + C1 z + D1 = 0 and A2 x + B2 y + C2 z + D2 = 0
®
is (A1 x + B1 y + C1 z + D1) + l(A2 x + B2 y + C2 z + D2) = 0.
r
r
r
r
r
r
Two lines r = a1 + l b1 and r = a2 + m b2 are coplanar if
r r
r r
( a2 - a1 ) × (b1 ´ b2 ) = 0
®
In the cartesian form above lines passing through the points A(x 1, y 1, z1) and
B ( x 2 , y 2 , z2 )
y – y2 z – z 2
=
=
are coplanar if
b2
C2
®
®
x2 - x1
a1
y2 - y1
b1
z 2 - z1
c1
a2
b2
c2
= 0.
r r
In the vector form, if q is the angle between the two planes, r × n1 = d1 and
r r
| n1 × n2 |
r r
–1
r × n 2 = d 2 , then q = cos r r .
| n1 || n2 |
r
r r
r
T he angle f between the line r = a + l b and the plane r × nˆ = d is
THREE D IMENSIONAL G EOMETRY
r
b × nˆ
sin f = r
|b | | nˆ |
T he angle q between the planes A1x + B1y + C1z + D1 = 0 and
d
®
503
cos q =
he
A2 x + B2 y + C2 z + D2 = 0 is given by
A1 A 2 + B1 B 2 + C1 C 2
A21 + B21 + C12
A 22 + B 22 + C22
r
r
T he distance of a point whose position vector is a from the plane r × nˆ = d is
r
| d - a × nˆ |
®
T he distance from a point (x1, y1, z1) to the plane Ax + By + Cz + D = 0 is
Ax1 + By1 + Cz1 + D
.
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A 2 + B2 + C2
bl
is
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