Download three dimensional geometry

11
THREE DIMENSIONAL
GEOMETRY
INTRODUCTION
In class XI, we studied Analytical Geometry in two dimensions and introduced
3-dimensional geometry, and restricted our study to cartesian method only. Now we shall
use vector algebra to 3-dimensional geometry which makes its study simple and elegant.
In this chapter, we shall study the direction cosines and direction ratios of a line. We shall
also find :
(i) angle between two lines when their direction cosines/ratios are known
(ii) the vector and cartesian equations of a line in different forms
(iii) angle between two lines when their equations are known
(iv) shortest distance between two lines
(v) the vector and cartesian equation of a plane in different forms
(vi) angle between two planes
(vii) angle between a line and a plane
(viii) distance of a point from a plane.
11.1 DIRECTION COSINES AND DIRECTION RATIOS OF A LINE
In the previous chapter, we have studied the direction cosines and direction ratios of a
position vector.
We must bear in mind the difference between a position vector and a free vector. In
→
general, we may assume any given vector v = a iˆ + b ˆj + c kˆ to be a free vector, so the vector
→
→
v can be represented by the position vector OP of the point P (a, b, c). We can also represent
→
→
the vector v by any other line segment of the same length and with direction of OP .
In this section, we shall study the direction cosines and direction ratios of a line.
11.1.1 Direction cosines of a line
Let AB be a line in space. Through origin O, draw a
line QP parallel to AB as shown in fig. 11.1. If the ray
OP makes angles α, β and γ with the positive
directions of coordinate axes OX, OY and OZ
respectively, then the ray AB also makes angles α, β
and γ with the positive directions of the coordinate
axes OX, OY and OZ respectively. The angles α, β and
γ are called the direction angles of the ray AB and the
cosines of these angles i.e. cos α, cos β and cos γ are
called the direction cosines of the ray AB. The
Z
B
A
γ
β
O
Q
X
α
Fig. 11.1.
P
Y
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THREE DIMENSIONAL GEOMETRY
direction cosines are usually denoted by l, m and n i.e. l = cos α, m = cos β and n = cos γ. These
are written as < l, m, n >.
→
Notice that the direction cosines of the vector OP (studied in the previous chapter) are the
same as the direction cosines of the ray OP. It is due to the following reasons :
(i) a vector and a ray differ only that a vector has definite magnitude whereas a ray does
not have definite magnitude.
(ii) the measure of an angle is independent of magnitude of vector or ray by which it is
made up of.
When the ray OP makes angles α, β and γ with
the positive directions of the coordinate axes OX, OY
and OZ respectively, then the ray OQ makes angles
π – α, π – β and π – γ with the positive directions of
the coordinate axes OX, OY and OZ respectively.
Therefore, the direction cosines of the ray BA are
cos (π – α), cos (π – β) and cos (π – γ) i.e. – cos α,
– cos β and – cos γ i.e. – l, – m and – n.
Z
B
A
π– γ
P
O
Y
Thus, if the direction cosine of the ray AB are l, m
and n, then the direction cosines of the ray BA are
– l, – m and – n.
Q π–α
The direction cosines of either the ray AB or the
ray BA are taken as the direction cosines of the
line AB.
X
π– β
Fig. 11.2.
Thus, the direction cosines of a line AB are either < l, m, n > or < – l, – m, – n >, where
l = cos α, m = cos β and n = cos γ.
REMARKS
π π
,
with X-axis, Y-axis and Z-axis respectively, its direction
2 2
π
π
> i.e. < 1, 0, 0 >. Similarly, direction cosines of Y-axis are
cosines are < cos 0, cos , cos
2
2
Since X-axis makes angles 0,
< 0, 1, 0 > and that of Z-axis are < 0, 0, 1 >.
11.1.2 Relation between direction cosines
Let < l, m, n > be direction cosines of a given line L.
Through O (origin), draw OP parallel to the line L. From
P, draw PA, PB, PC perpendiculars to coordinate axes
and complete the ‘rectangular box’. Let coordinates of P
be (x, y, z), then
Z
L
C
P(x, y, z)
OA = x, OB = y and OC = z
Let
OP = r, then r2 = x2 + y2 + z2
…(i)
O
If the line L makes angles α, β, γ with the coordinate
axes, then l = cos α , m = cos β and n = cos γ.
In ∆OAP, ∠OAP = 90° ( Q AP ⊥ OA) and ∠AOP = α,
∴
cos α =
Similarly,
OA
x
⇒l=
⇒ x = lr.
OP
r
y = mr and z = nr.
Substituting these values of x, y, z in (i), we get
r2 = (lr)2 + (mr)2 + (nr)2 ⇒ 1 = l 2 + m 2 + n 2.
Hence, l2 + m2 + n2 = 1.
Corollary. The coordinates of P are (lr, mr, nr).
α
A
X
Fig. 11.3.
B
Y
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MATHEMATICS – XII
11.1.3 Direction ratios of a line
If < l, m, n > are direction cosines of a line L and λ is any non-zero real number, then < λ l, λ m, λ n >
are called direction ratios or direction numbers (abbreviated d.n.) of the line L. Thus, if
< a, b, c > are direction ratios of a line L, then a = λ l, b = λ m, c = λ n for some non-zero
real number λ .
Since, λ is any non-zero real number, it follows that there are infinitely many sets of
direction ratios of a line.
To find (actual) direction cosines of a line from its direction ratios.
Let < a, b, c > be direction ratios of a line L and < l, m, n > be its (actual) direction cosines,
then
a = λ l, b = λ m, c = λ n for some real number λ (≠ 0)
⇒ l=
a
b
c
, m= , n=
λ
λ
λ
…(i)
As < l, m, n > are direction cosines of the line L, we have
l 2 + m2 + n2 = 1 ⇒
a2
b2
c2
=1
+
+
λ2
λ2
λ2
⇒ λ2 = a2 + b2 + c2 ⇒ λ = ±
∴ l=±
a
a2 + b2 + c 2
, m=±
(using (i))
a2 + b2 + c 2
b
a2 + b2 + c 2
, n=±
c
a2 + b2 + c 2
where signs to be taken all positive or all negative.
REMARKS
1. Here any one or two of the numbers a, b, c could be zero. If a = 0, then so is l.
2. While l 2 + m 2 + n 2 = 1 but a 2 + b 2 + c 2 ≠ 1, in general.
11.1.4 Direction cosines of a line joining two points
Let P (x1, y1, z1) and Q (x2, y2, z2) be two given points.
Since one and only one line passes through two given
points, we can find the direction cosines of the line PQ.
Z
cos γ =
Q
L
O
Y
Then LQ = NQ – NL = NQ – MP = z2 – z1.
X
From right angled triangle PQL, we get
γ
P
From P and Q, draw PM and QN perpendiculars to
the XY-plane. From P, draw PL perpendicular to NQ as
shown in fig. 11.4.
In ∆ PLQ, ∠PQL = γ.
γ
γ
Let l, m and n be the direction cosines of the line PQ
and let it make angle α, β and γ with the positive
directions of the coordinate axes OX, OY and OZ
respectively.
M
N
Fig. 11.4.
z − z1
LQ
= 2
.
PQ
PQ
Similarly, by drawing perpendiculars to YZ and ZX planes from P and Q, we get
cos α =
x2 − x1
y − y1
and cos β = 2
.
PQ
PQ
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THREE DIMENSIONAL GEOMETRY
Hence, the direction cosines of the line joining the points P (x1, y1, z1) and Q (x2, y2, z2) are
<
x 2 − x 1 y2 − y1 z2 − z1
,
,
>
PQ
PQ
PQ
where
PQ =
( x2 − x1 ) 2 + ( y2 − y1 ) 2 + ( z2 − z1 ) 2 .
Corollary. Direction ratios of the line joining the points P (x1, y1, z1) and Q (x2, y2, z2) are
< x2 – x1, y2 – y1, z2 – z1 >.
11.1.5 Angle between two lines
The concept of an angle between two coplanar
(intersecting or parallel) lines is already known to the
reader. If two lines are non-coplanar (skew), then an
angle between them is defined as follows :
Let L1, L2 be two non-coplanar (skew) lines. Through any
point, say A, draw two straight lines L′1, L′2 parallel to L1, L2
respectively (shown in fig. 11.5), then an angle between L′1 and
L′2 is called an angle between the given lines L1 and L2 .
In fact, there are two angles, θ and π – θ, between two
lines (intersecting or non-intersecting) as shown in
fig. 11.5. If the lines are parallel, then angle between them
is taken as zero.
L1
L2
L′1
π–θ
θ
L′2
A
Fig. 11.5.
REMARK
Usually, we draw lines through origin parallel to given lines ; it follows that the angles
between a given line and coordinate axes are the angles which the line drawn through origin
and parallel to the given line makes with the axes.
11.1.6 To find angle between two lines whose direction cosines are given
Let L1, L2 be two lines with direction cosines
< l1, m1, n1 >, < l2, m2, n2 > respectively.
P(l1, m1, n1)
Let θ be an angle between given lines L1, L2, then θ
is also the angle between OP, OQ.
PQ2
=
OP2
+
OQ2
O
Z
θ
L2
Q
(l2, m2, n2)
θ
( Q x = lr, y = mr, z = nr).
From ∆OPQ, by cosine formula, we get
L1
Y
Through O (origin), draw lines OP, OQ parallel to
L1, L2 respectively. Then direction cosines of OP, OQ
are < l1, m1, n1 >, < l2, m2, n2 > respectively. Cut off
OP = 1 = OQ, then the coordinates of P, Q are
(l1, m1, n1), (l2, m2, n2) respectively
X
Fig. 11.6.
– 2 . OP . OQ cos θ
⇒ (l2 – l1)2 + (m2 – m1)2 + (n2 – n1)2 = 12 + 12 – 2 . 1 . 1 . cos θ
⇒ (l22 + m22 + n22 ) + (l12 + m12 + n12 ) – 2 (l1l2 + m1m2 + n1n2) = 1 + 1 – 2 cos θ
⇒ 1 + 1 – 2 (l1l2 + m1m2 + n1n2) = 1 + 1 – 2 cos θ
⇒ – 2 (l1l2 + m1m2 + n1n2) = – 2 cos θ
⇒ cos θ = l1l2 + m1m2 + n1n2 .
Corollary 1. Formula for sin θ
sin2 θ = 1 – cos2 θ = 1 – (l1 l2 + m1 m2 + n1 n2)2
= (l12 + m12 + n12 ) (l22 + m22 + n22 ) – (l1l2 + m1m2 + n1n2)2
(Q l12 + m12 + n12 = 1, l22 + m22 + n22 = 1)
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MATHEMATICS – XII
= ( l12 m22 + l22 m12 – 2 l1l2m1m2) + ( m12 n22 + m22 n12 – 2m1m2n1n2)
+ ( n12 l22 + n22 l12 – 2n1n2l1l2)
= (l1m2 – l2m1)2 + (m1n2 – m2n1)2 + (n1l2 – n2l1)2
= Σ (l1m2 – l2m1)2
⇒
Σ (l1 m 2 − l2 m1 ) 2
sin θ =
(0 ≤ θ < π ⇒ sin θ ≥ 0)
Corollary 2. Condition of perpendicularity.
If the lines are perpendicular, then angle between them is 90° i.e. θ = 90°
∴ cos θ = cos 90° = 0
⇒ l1l2 + m1m2 + n1n2 = 0.
Corollary 3. Condition of parallelism.
If the given lines are parallel, then they make equal angles with X-axis ⇒ α1 = α2
⇒ cos α1 = cos α2 ⇒ l1 = l2 .
Similarly, m1 = m2, n1 = n2 .
l1 = l2, m1 = m2, n1 = n2 .
Hence,
11.1.7 To find angle between two lines whose direction ratios are given
Let < a1, b1, c1 > and < a2, b2, c2 > be the direction ratios of the given lines, then their direction
cosines are
<±
a1
Σ
a12
and < ±
b1
,±
Σ
a2
a12
b2
, ±
Σ a22
, ±
Σ a22
c1
> where signs to be taken all positive or all negative
Σ a12
, ±
c2
Σ a22
> where signs to be taken all positive or all negative.
If θ is an angle between the given lines, then
a1a2 + b1 b2 + c 1 c 2
cos θ = ±
a12
+ b12 + c 12
(cos θ = l1 l2 + m1 m2 + n1 n2)
a22 + b22 + c 22
Σ ( a1 b2 – a2 b1 ) 2
and sin θ =
a12
+ b12
+
c 12
a22
+ b22
+
(sin θ =
c 22
Σ (l1 m2 − l2 m1 )2
)
Corollary 1. Condition of perpendicularity.
If the lines are perpendicular, then θ = 90°,
∴ cos θ = cos 90° = 0
⇒ ± (a1a2 + b1b2 + c1c2) = 0
⇒ a1 a2 + b1 b2 + c1 c2 = 0.
Corollary 2. Condition of parallelism.
If lines are parallel, then θ = 0
∴
sin θ = sin 0 = 0 ⇒ sin2 θ = 0
⇒ (a1b2 – a2b1)2 + (b1c2 – b2c1)2 + (c1a2 – c2a1)2 = 0
⇒ a1b2 – a2b1 = 0, b1c2 – b2c1 = 0, c1a2 – c2a1 = 0
( ∵ if A, B are reals, then A2 + B2 = 0 ⇒ A = 0 and B = 0)
⇒
a1
a2
=
b1
b2
,
b1
b2
=
c1
c2
,
c1
c2
=
a1
a2
⇒
a1
a2
=
b1
b2
=
c1
c2
.
REMARK
If the lines are non-perpendicular, then the acute angle θ between the lines is given by
cos θ = | l1l2 + m1m2 + n1n2 | or cos θ =
| a1 a2 + b1b2 + c1c 2 |
a12
+ b12 + c12
a22 + b22 + c 22
.
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THREE DIMENSIONAL GEOMETRY
ILLUSTRATIVE EXAMPLES
Example 1. If a line makes angles 90°, 135° and 45° with the positive directions of X, Y and Z axes, then
find its direction cosines.
(NCERT)
Solution. Let the direction cosines of the line be l, m and n. Then
l = cos 90° = 0, m = cos 135° = –
1
2
1
and n = cos 45° =
Hence, direction cosines of the line are < 0, –
1
2
,
1
2
2
.
>.
Example 2. If a line makes angles α, β and γ with the coordinate axes, then prove that
sin2α + sin2 β + sin2 γ = 2.
Solution. Let < l, m, n > be the direction cosines of the given line, then
l = cos α, m = cos β and n = cos γ.
We know that l 2 + m2 + n2 = 1
⇒ cos2 α + cos2 β + cos2 γ = 1
⇒ (1 – sin2 α) + (1 – sin2 β) + (1 – sin2 γ ) = 1
⇒ – sin2 α – sin2 β – sin2 γ = – 2
⇒ sin2 α + sin2 β + sin2 γ = 2.
Example 3. Find the direction cosines of the line that makes equal angles with the coordinate axes.
(C.B.S.E. 2011)
Solution. As the line makes equal angles with coordinate axes,
α = β = γ ⇒ cos α = cos β = cos γ
⇒ l = m = n, where < l, m, n > are direction cosines of the line.
We know that l 2 + m2 + n2 = 1
1
1
⇒l=±
.
3
3
⇒ l2 + l2 + l2 = 1 ⇒ 3l2 = 1 ⇒ l2 =
The direction cosines of the line are
<
1
,
3
1
,
3
1
1
1
1
, −
, −
> or < −
>
3
3
3
3
Example 4. A straight line makes angles 60° and 45° with the positive directions of X-axis and Y-axis
respectively. What angle does it make with the Z-axis?
Solution. Let γ be the angle which the given line makes with Z-axis. Since this line makes angles
of 60° and 45° with the X-axis and Y-axis respectively, we have
1 2
1 2
+ cos2 γ = 1
cos2 60° + cos2 45° + cos2 γ = 1 ⇒   + 
 2
⇒ cos2 γ = 1 –
 2
1 1
1
1
–
=
⇒ cos γ = ±
⇒ γ = 60° or 120°.
4 2
4
2
Therefore, the acute angle which the given line makes with the positive direction of Z-axis
is 60°.
Example 5. If the direction cosines of a line are –
9
6
2
,
,–
, then what are its direction ratios?
11 11
11
Solution. Obviously < – 9, 6, – 2 > is one set of direction ratios. All sets of direction ratios are
given by < – 9λ, 6λ, – 2λ > where λ is any non-zero real number.
Example 6. If direction ratios of a line are < – 18, 12, – 4 >, then find its direction cosines.
(NCERT)
Solution. Given direction ratios of a line are < – 18, 12, – 4 >.
Dividing by
( − 18)2 + 122 + ( − 4 )2 =
the direction cosines of the line are < –
484 = 22,
9
6
2
,
, –
>.
11 11
11
882
MATHEMATICS – XII
Example 7. Find the direction cosines of the line passing through the two points (– 2, 4, – 5)
and (1, 2, 3).
(C.B.S.E. 2009)
Solution. Direction ratios of the line joining the points P (– 2, 4, – 5) and Q (1, 2, 3) are
< 1 – (– 2), 2 – 4, 3 – (– 5) >
|< x2 – x1, y2 – y1, z2 – z1 >
i.e. < 3, – 2, 8 >
3 2 + ( − 2)2 + 8 2 =
Dividing by
9 + 4 + 64 =
3
the direction cosines of the line are <
77
,–
Example 8. The direction cosines of the two lines are <
between them.
2
77
77 ,
,
8
77
>.
2
1
2
3 2 6
, – , – > and < , ,
>. Find the angle
3
3
3
7 7 7
Solution. Let θ be the angle between the two lines, then
 2  3

1  2

2  6
cos θ =     +  −    +  −   
 3 7
 3 7
 3 7
=
⇒
|cos θ = l1l2 + m1m2 + n1n2
6
2
12
8
−
−
=–
21
21
21
21

θ = cos–1  −
8
.
21 
In fact, there are two angles between the given lines. The acute angle θ between these lines
is given by
cos θ = −
8
8
 8
=
⇒ θ = cos–1   .
21
21
21
Example 9. Find the angle between the lines whose direction numbers are < 1, 2, 2 > and < 3, 2, – 6 >.
Solution. The direction numbers of the given lines are < 1, 2, 2 > and < 3, 2, – 6 >.
Hence, the (acute) angle θ between the lines is given by
cos θ =
=
⇒
|1 . 3 + 2 . 2 + 2 .(– 6 )|
12 + 2 2 + 2 2
3 2 + 2 2 + (– 6 ) 2

Q cos θ =



a12 + b12 + c12 ⋅ a22 + b22 + c 22 
| a1 a2 + b1b2 + c1c 2 |
| 3 + 4 – 12 | | – 5 |
5
=
=
3.7
21
21
5
θ = cos–1   .
 21 
Example 10. Find the angle between the lines whose direction ratios are < a, b, c > and
< b – c, c – a, a – b >.
(NCERT)
Solution. Let θ be the angle between the two lines, then
cos θ =
⇒
a( b − c ) + b ( c − a ) + c ( a − b )
=0
Σa 2 Σ ( b − c ) 2
θ = 90°.
Example 11. Show that the line joining the points A (7, 8, 6) and B (9, 11, 7) is perpendicular to the line
joining the points C (9, 1, – 6) and D (7, 2, – 5).
Solution. Direction numbers of the line AB are < 9 – 7, 11 – 8, 7 – 6 > i.e. < 2, 3, 1 >
and direction numbers of the line CD are
< 7 – 9, 2 – 1, – 5 + 6 > i.e. < – 2, 1, 1 >
∴
a1 a2 + b1 b2 + c1 c2 = 2 . (– 2) + 3 . 1 + 1 . 1 = 0
⇒ the line AB is perpendicular to the line CD.
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THREE DIMENSIONAL GEOMETRY
Example 12. For what values of p and q will the line joining points A (3, 2, 5) and B (p, 5, 0) be parallel
to the line joining points C (1, 3, q) and D (6, 4, – 1).
Solution. Direction ratios of the line AB are
< p – 3, 5 – 2, 0 – 5 > i.e. < p – 3, 3, – 5 >
and direction ratios of the line CD are
< 6 – 1, 4 – 3, – 1 – q > i.e. < 5, 1, – 1 – q >.
Now, the line AB will be parallel to the line CD if
p–3
=
5
⇒
a1
–5
3
=
1
–1 – q
a2
=
b1
b2
=
c1
c2
p – 3 = 15 and 3 (1 + q) = 5
⇒
p = 18 and 3 q = 2 i.e. q =
Hence,
p = 18 and q =
2
.
3
2
.
3
Example 13. Using direction numbers, show that the points A (– 2, 4, 7), B (3, – 6, – 8) and C (1, – 2, – 2) are
collinear.
Solution. Direction numbers of the line AB are
< 3 + 2, – 6 – 4, – 8 – 7 > i.e. < 5, – 10, – 15 > i.e. < 1, – 2, – 3 >
and direction numbers of the line AC are
< 1 + 2, – 2 – 4, – 2 – 7 > i.e. < 3, – 6, – 9 > i.e. < 1, – 2, – 3 >.
Clearly, direction numbers of the two lines AB and AC are proportional, therefore, these lines
are parallel.
But the lines AB and AC have a common point A and hence the points A, B and C are
collinear.
Example 14. If A (8, 2, 0), B (4, 6, – 7), C (– 3, 1, 2) and D (– 9, – 2, 4) are four given points, then find
angle between AB and CD.
Solution. Direction ratios of the line AB are
< 4 – 8, 6 – 2, – 7 – 0 > i.e. < – 4, 4, – 7 > i.e. < 4, – 4, 7 >
and direction ratios of the line CD are
< – 9 + 3, – 2 – 1, 4 – 2 > i.e. < – 6, – 3, 2 > i.e. < 6, 3, – 2 >.
Let θ be the (acute) angle between the lines, then
cos θ =
⇒
| 4 . 6 + (– 4 ). 3 + 7 .(– 2 )|
42
+
(– 4 ) 2
+
72
62
+
32
+
(– 2 ) 2
=
|– 2 |
2
=
9.7
63
2
θ = cos–1   .
 63 
Example 15. Find the direction cosines of the line which is perpendicular to the lines with direction
cosines proportional to < 1, – 2, – 2 > and < 0, 2, 1 >.
Solution. Direction ratios of the given lines are < 1, – 2, – 2 > and < 0, 2, 1 >.
Let < a, b, c > be direction ratios of the line which is perpendicular to the given lines, then
a . 1 + b . (– 2) + c . (– 2) = 0 ⇒ a – 2b – 2c = 0
and a . 0 + b . 2 + c . 1 = 0 ⇒ 2b + c = 0
From (ii), we get c = – 2b.
Substituting this value of c in (i), we get
a – 2b – 2 (– 2b) = 0 ⇒ a + 2b = 0 ⇒ a = – 2b.
…(i)
…(ii)
884
MATHEMATICS – XII
∴ Direction ratios of the line are < – 2b, b, – 2b > i.e. < 2, – 1, 2 >.
2 2 + (– 1)2 + 2 2 = 3,
Dividing by
the direction cosines of the line are <
2
1 2
, – ,
>.
3
3 3
Example 16. If < l1, m1, n1 > and < l2, m2, n2 > are direction numbers of two lines, then show that
direction numbers of a line which is perpendicular to both the lines are
< m1n2 – m2n1, n1l2 – n2l1, l1m2 – l2m1 >.
(NCERT)
Solution. Let < a, b, c > be direction number of a line which is perpendicular to the two lines with
direction numbers < l1, m1, n1 > and < l2, m2, n2 >, therefore, we have
l1a + m1b + n1c = 0
…(i)
l2a + m2b + n2c = 0
…(ii)
Solving (i) and (ii) by cross-multiplication, we get
a
b
c
=
=
.
m1n 2 − m2n1
n1l2 − n 2l1
l1m2 − l2m1
Hence, direction numbers of the line are
< m1n2 – m2n1, n1l2 – n2l1, l1m2 – l2m1 >.
Example 17. Find the angle between the two lines whose direction cosines are given by the equations
l + m + n = 0, l 2 + m2 – n2 = 0.
(NCERT Examplar Problems)
Solution. The equations giving the direction cosines of the two lines are
l+m+n=0
and
2
l +
m2
–
n2
…(i)
=0
…(ii)
From (i), l = – m – n
…(iii)
Putting this value of l in (ii), we get
[– (m + n)]2 + m2 – n2 = 0 ⇒ 2m2 + 2mn = 0
⇒
m (m + n) = 0
either m = 0
or
m = –n
⇒
⇒
l = 0, from (iii)
l = – n, from (iii)
∴ Direction numbers of the two lines are
< – n, 0, n > and < 0, – n, n > i.e. < – 1, 0, 1 > and < 0, – 1, 1 >.
If θ is the acute angle between the lines, then
cos θ =
⇒
θ =
| – 1 . 0 + 0 .(– 1) + 1 . 1|
(–1) 2 + 0 2 + 12
0 2 + (–1) 2 + 12
=
1
1
=
2
2 2
π
.
3
Example 18. Find the direction cosines of the four diagonals and the angle between two diagonals of a
cube.
Z
Solution. Take one vertex, say O, of the cube as origin and
the three mutually perpendicular edges OA, OB and
C(0, 0, a)
N(0, a, a)
OC as coordinate axes. Let the length of the side of the
L
cube be a units. Then the coordinates of the vertices are
P(a, a, a)
(a, 0, a)
shown in fig. 11.7.
The cube has four diagonals OP, MC, AN and BL. The
direction numbers of OP are < a – 0, a – 0, a – 0 >
i.e. < 1, 1, 1 > and hence direction cosines are
<
1
,
3
1
,
3
1
>.
3
B(0, a, 0)
O
A(a, 0, 0)
X
Fig. 11.7.
Y
M(a, a, 0)
885
THREE DIMENSIONAL GEOMETRY
Direction numbers of MC are < a – 0, a – 0, 0 – a > i.e. < 1, 1, – 1 > and hence its direction
cosines are <
1
,
3
1
1
, –
>.
3
3
Similarly, direction cosines of AN are < –
<
1
1
, –
,
3
3
1
,
3
1
,
3
1
> and that of BL are
3
1
>.
3
Consider any two diagonals, say OP and MC, let θ be the (acute) angle between these
diagonals, then
1
1
1
1
1  1 
1
⋅
+
⋅
+
⋅ –
=


3
3
3
3
3
3
3
1


θ = cos–1
.
 3
cos θ =
⇒
Example 19. Find the coordinates of the foot of the perpendicular drawn from the point A (1, 8, 4) to the
line joining the points B (0, – 1, 3) and C (2, – 3, – 1).
(C.B.S.E. 2005, 04)
Solution. Let N be the foot of perpendicular drawn from A
A(1, 8, 4)
to the line BC, and let N divide segment BC in the ratio
λ : 1, then coordinates of the point N are
− 3 λ − 1 − λ + 3
 2λ
,
,

.
λ +1
λ+1
λ+1 
∴ Direction ratios of the line AN are
− 3λ − 1
−λ + 3
2λ
<
− 1,
− 8,
− 4 >
λ + 1
λ + 1
λ + 1
i.e. <
λ
B
(0, –1, 3)
:
1
N
C
(2, –3, –1)
Fig. 11.8.
λ − 1 − 11λ − 9 − 5λ − 1
,
,
> i.e. < λ – 1, – 11 λ – 9, – 5 λ – 1 >.
λ + 1
λ + 1
λ + 1
Direction ratios of the line BC are < 2 – 0, – 3 – (– 1), – 1 – 3 >
i.e. < 2, – 2, – 4 > i.e. < 1, – 1, – 2 >.
Since the line AN is perpendicular to the line BC, we have
(λ – 1) (1) + (– 11 λ – 9) (– 1) + (– 5 λ – 1) (– 2) = 0
⇒ 22 λ + 10 = 0 ⇒ λ = –
5
.
11
∴ The coordinates of the point N are
15
5
 − 10
−1
+3 
11
11
11
 i.e.

,
,

 5
5
5

−
1
1
1
+
−
+
−
+

 11
11
11
 − 10 , 4 , 38  i.e.  − 5 , 2 , 19  .
 6 6 6
 3 3 3
Example 20. Prove that the lines whose direction cosines are given by al + bm + cn = 0
and fmn + gnl + hlm = 0 are
(1) perpendicular if
f
f
f
+ + = 0 and
a b c
(2) parallel if a2f 2 + b2g2 + c2h2 – 2(bcgh + cahf + abfg) = 0.
Solution. (1) Direction cosines of the lines are given by
al + bm + cn = 0
and fmn + gnl + hlm = 0
al + bm
c
From (i), n = –
…(i)
…(ii)
…(iii)
Substituting this value of n in (ii), we get
al + bm 
(fm + gl)  –
+ hlm = 0

⇒
c

agl 2 + (af + bg – ch) lm + bf m2 = 0
…(iv)
886
MATHEMATICS – XII
Note that both l and m cannot be zero, for, if l = m = 0, then from (iii), we get n = 0, which
is wrong as l 2 + m2 + n2 = 1. So, without any loss of generality, we may take
m ≠ 0. Dividing both sides of (iv) by m2, we get
l 2
l
+ (af + bg – ch)   + bf = 0
 m
 m
ag 
The above equation is a quadratic in
…(v)
l
l
, giving two values of
. If < l1, m1, n1 > and
m
m
< l2, m2, n2 > are the direction cosines of the two lines given by the equations (i) and (ii), then
l1
m1
l2
,
are roots of (v).
m2
∴
Product of roots =
l1l2
⇒
=
m1m2
l1l2
⇒
f /a
=
ll
mm
f /a
⇒ 12 = 1 2
g/b
f /a
g/b
m1m2
g/b
=
f
a
⇒
l
l
bf
bf
⇒ 1 ⋅ 2 =
ag
m1 m2
ag
n1n2
(by symmetry of result)
h/c
l1l2 = λ , m1m2 = λ
g
h
, n1n2 = λ for some non-zero real number λ.
b
c
Now the lines whose direction cosines are given by (i) and (ii) are at right angles if
l1l2 + m1m2 + n1n2 = 0
if λ
i.e.
f
g
h
f
g
h
= 0.
+ +
+ λ + λ = 0 i.e. if
a
b
c
a
b
c
(2) The lines will be parallel if l1 = l2, m1 = m2, n1 = n2
i.e. if
l1
l
= 2 i.e. if (v) and hence (iv) has equal roots
m1
m2
i.e. if discriminant of (iv) is zero
i.e. if (af + bg – ch)2 – 4agbf = 0
i.e. if a2f 2 + b2g2 + c2h2 – 2(bcgh + cahf + abfg) = 0.
EXERCISE 11.1
Very short answer type questions (1 to 12) :
1. If a line makes angles 90°, 60° and 30° with the positive directions of x, y and z-axes
respectively, find its direction cosines.
(NCERT)
2. Can the numbers
1
,
2
Hint. l 2 + m2 + n2 ≠ 1
1
,
2
1
be the direction cosines of a line?
2
3. Can a line have direction angles 45°, 60°, 120°?
4. Prove that 1, 1, 1 cannot be direction cosines of a straight line.
5.
(i) Write the direction cosines of z-axis.
(C.B.S.E. 2012)
(ii) Write the direction cosines of a line equally inclined to the three coordinate axes.
(C.B.S.E. 2009)
6. If the direction cosines of a line are <
1 1 1
, ,
>, then find value(s) of c.
c c c
7. If a line makes angles α, β and γ with the coordinate axes, then find the value of
cos 2 α + cos 2 β + cos 2 γ.
8. A line makes an angle of
π
with each of X-axis and Y-axis. What angle does it make with
4
Z-axis?
9. A line in the XY-plane makes an angle of 30° with the X-axis, find the direction cosines
of the line.
887
THREE DIMENSIONAL GEOMETRY
10. If the direction cosines of a line are
2
1
2
, – , – , then find its direction ratios.
3
3
3
(i) If a line has direction ratios 2, – 1, – 2, determine its direction cosines. (C.B.S.E. 2012)
11.
(ii) If direction numbers of a line are < 2, – 6, 3 >, find its direction cosines.
(i) Write the direction cosines of the line joining the points (1, 0, 0) and (0, 1, 1).
(C.B.S.E. 2011)
(ii) Find the direction cosines of the line passing through the points P (2, 3, 5) and
Q (– 1, 2, 4).
(NCERT Examplar Problems)
13. 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) and (4, 3, – 1).
(NCERT)
12.
14. Show that the line through the points (1, – 1, 2) and (3, 4, – 2) is perpendicular to the line
through the points (0, 3, 2) and (3, 5, 6).
(NCERT)
15. Show that the line through the points (4, 7, 8) and (2, 3, 4) is parallel to the line through
(NCERT)
the points (– 1, – 2, 1) and (1, 2, 5).
16. For what value of p will the line through (4, 1, 2) and (5, p, 0) be perpendicular to the line
through (2, 1, 1) and (3, 3, – 1).
17. Find the values of p and q so that the line joining the points (7, p, 2) and (q, – 2, 5) may
be parallel to the line joining the points (2, – 3, 5) and (– 6, – 15, 11).
18. Show that the points A (2, 3, – 4), B (1, – 2, 3) and C (3, 8, – 11) are collinear. (NCERT)
19. Show that the points (2, 3, 4), (– 1, – 2, 1) and (5, 8, 7) are collinear.
(NCERT)
20. If the points (– 1, 3, 2), (– 4, 2, – 2) and (5, 5, λ) are collinear, then find the value of λ.
21. Find the direction cosines of the sides of the triangle whose vertices are (3, 5, – 4),
(– 1, 1, 2) and (– 5, – 5, – 2).
(NCERT)
22. Find the direction ratios and the direction cosines of the line joining the points (5, 4, – 3)
and (3, – 2, 6), the angle between the line and X-axis taken acute.
23. Find the angle between the lines whose direction cosines are
<
3
1
,
,
2
4
3
> and < –
4
3
1
,
,
2
4
3
>.
4
24. Find the angle between the lines whose direction numbers are
< 1, 2, 1 >, < 2, – 3, 4 >.
25. Find the angle between the lines whose direction ratios are
< 1, 1, 2 > and < 3 – 1, –
3 – 1, 4 >.
26. Show that the lines with direction cosines
<
12
3
4
4 12 3
3
4 12
,–
,–
>, <
> and <
,–
,
> are mutually perpendicular.
,
,
13
13
13
13 13 13
13
13 13
27. If A, B are the points (2, 1, – 2), (3, – 4, 5), then find the angle that OA makes with OB,
where O is the origin.
28. If A, B, C are the points (1, 4, 2), (– 2, 1, 2), (2, – 3, 4), then find the angles of the triangle
ABC. What type of triangle is it?
29. If the coordinates of the points A, B, C and D are (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and
(2, 9, 2) respectively, then find the angle between the lines AB and CD.
(NCERT)
30. Show that the points A (4, 7, 8), B (2, 3, 4), C (– 1, – 2, 1) and D (1, 2, 5), taken in order,
are the vertices of a parallelogram.
31. Show that the points A (4, 2, – 6), B (5, – 3, 1), C (12, 4, 5) and D (11, 9, – 2), taken in order,
are the vertices of a rectangle.
32. Find the direction cosines of a line which is perpendicular to the two lines formed by
joining A (2, 3, – 4) to B (– 3, 3, – 2) and C (– 1, 4, 2) to D (3, 5, 1).
888
MATHEMATICS – XII
33.
(i) Show that the straight lines whose direction cosines are given by 2l + 2m – n = 0 and
mn + nl + lm = 0 are at right angles.
(NCERT Examplar Problems)
(ii) Find the angle between the lines whose direction cosines are given by the equations
3l + m + 5n = 0 and 6mn – 2nl + 5 lm = 0.
(NCERT Examplar Problems)
34. If a line L makes angles α, β, γ, δ with the four diagonals of a cube, then prove that
cos2 α + cos2 β + cos2 γ + cos2 δ =
4
.
3
(C.B.S.E. 2007)
Hint. Find direction cosines of four diagonals of a cube ; these are
<
1
1
1
1
1
1
1
1
1
1
1
1
,–
,
>, <
,
, –
>.
,
,
>, < –
,
,
>, <
3
3
3
3
3
3
3
3
3
3
3
3
Let < l, m, n > be direction cosines of line L, then
cos α =
1
(l + m + n), cos β =
3
1
( − l + m + n) etc. Square and add these.
3
35. Find the point in which the join of A (– 9, 4, 5) and B (11, 0, – 1) is met by the
perpendicular from the origin.
36. Find the coordinates of the foot of perpendicular drawn from the point A(1, 2, 1) on the
line joining the points B(1, 4, 6) and C(5, 4, 4).
37. Prove that the lines whose direction cosines are given by
ul2 + vm2 + wn2 = 0 and al + bm + cn = 0 are
(1) perpendicular if u(b2 + c2) + v(c2 + a2) + w(a2 + b2) = 0 and
(2) parallel if
a2 b 2 c 2
+
+
= 0.
u
v
w
38. The fig. 11.9 shows a rectangular box of size a,
b and c units along x, y and z-axes respectively.
Prove that the angles between the four
Z
C
 ± a2 ± b2 ± c 2 
diagonals are given by cos–1  2
2
2 
 a +b +c 
N
L
P
where all signs cannot be +ve or –ve.
O
Hint. The rectangular box has four diagonals
OP, MC, AN and BL. Number of pairs
of diagonals is 4C2 i.e. 6. So there are six
angles between the pairs of diagonals.
B
Y
M
A
X
Fig. 11.9.
11.2 STRAIGHT LINE (IN SPACE)
We know that in space a straight line is uniquely determined if
(i) it passes through a given point and has a given direction or
(ii) it passes through two given points.
We shall obtain the vector equation and cartesian equation of a line (in space).
→
Z
position vector a and let it be parallel to the given
>
11.2.1 Equation of a straight line passing through a given point and parallel to a given vector
Let the line pass through the given point A with
→
b
→
>
vector b (as shown in fig. 11.10).
P
→
>
>
→
a
>
>→
r
O
>
Y
>
Let P with position vector r be any point, then P lies
→
→
on the line iff AP is parallel to the vector b i.e. iff
→
→
AP = λ b , where λ is a real number (called parameter)
→
→
→
i.e. iff OP − OA = λ b
A
X
Fig. 11.10.
964
MATHEMATICS – XII
ANSWERS
EXERCISE 11.1
1. < 0,
1
3
,
>
2 2
2. No
6. ±
1
1
1
1
1
1
,
,
> or < −
,−
,−
>
3
3
3
3
3
3
(ii) <
5. (i) < 0, 0, 1 >
3
3. Yes
7. – 1
10. < 2λ, – λ, – 2λ >, λ ≠ 0
12. (i) < −
2
,
17
22. < 2, 6, – 9 >; <
(ii) <
20. 10
2
3
,−
,−
17
17
3
>, < −
17
2
6
9
,
,–
>
11 11
11
3 1
, , 0>
2 2
9. <
11. (i) <
1
1
1
,
,
>
3
3
3
17. p = 4, q = 3
2
21. < −
,−
17
π
2
8.
2
1
2
,– ,–
>
3
3
3
3
1
1
11
,
2
>, <
17
23. 60°
11
,
11
4
,
42
(ii) <
>
2
6 3
,– ,
>
7
7 7
3
16. –
2
5
,−
42
24. 90°
1
>
42
25. 60°
 4 2
27. cos–1 
 15 
 2
 1 
28. A = cos–1 
, B = 90°, C = cos–1 
; right-angled triangle

 3 
 3
29. 0°
32. <
35. (1, 2, 2)
36. (3, 4, 5)
2
38
,–
3
38
,
5
38
 1
33. (ii) cos–1  
 6
>
EXERCISE 11.2
1.
→
(i) r = iˆ + 2 ˆj + 3 kˆ + λ ( 3 iˆ + 2 ˆj − 2 kˆ ) , where λ is a non-zero real number
→
(ii) r = iˆ − 2 ˆj + 3 kˆ + λ ( 3 iˆ − 2 jˆ + 6 kˆ ) , where λ is a non-zero real number.
2.
y+1
x−2
z−4
=
=
1
2
−1
→
3. (i) r = 5 iˆ − 4 ˆj + 6 kˆ + λ ( 3 iˆ + 7 ˆj + 2 kˆ )
(ii)
→
r = 5 iˆ − 4 ˆj + 6 kˆ + λ ( 3 iˆ + 7 ˆj − 2 kˆ )
→
(iii) r = − 3 iˆ + 5 ˆj − 6 kˆ + λ ( 2 iˆ + 4 ˆj + 2 kˆ )
4.
y−1
x−2
z+4
=
=
2
−1
−1
5.
x+2 y−4 z+5
=
=
3
−5
6
→
6.
7. r = λ( 5 iˆ – 2 ˆj + 3 kˆ ) ;
8. (i)
< 1, – 2, 4 >
9. < 2, 3, – 1 >
→
→
r = − iˆ + 2 kˆ + λ( 4 iˆ + 4 ˆj + 4 kˆ )
y
x
z
=
=
−2
5
3
(ii) < 1, – 7, 2 >
(iii) < –
10.
11. r = 5 iˆ + 2 ˆj − 4 kˆ + λ( 3 iˆ + 2 ˆj − 8 kˆ ) ;
2 3 6
, , >
7 7 7
(iv) <
→
r = λ iˆ
y − 2
x − 5
z + 4
=
=
3
2
−8
→
y+1
x−2
z−4
=
=
12. r = 2 iˆ − ˆj + 4 kˆ + λ (iˆ + ˆj − 2 kˆ ) ;
1
1
−2
3
,
55
4
,
55
6
>
55