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1 | 5718-5721 | λb
r
= r
a−
r
r
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
rr =
λ
r
ra +
b (1)
Remark If
ˆ
ˆ
ˆ
b
ai
bj
ck
= + +
r
, then a, b, c are direction ratios of the line and conversely,,
if a, b, c are direction ratios of a line, then
ˆ
ˆ
ˆ
=
+
+
br
ai
bj
ck will be the parallel to
the line Here, b should not be confused with |
r
b| |
1 | 5719-5722 | Hence, the vector equation of the line is given by
rr =
λ
r
ra +
b (1)
Remark If
ˆ
ˆ
ˆ
b
ai
bj
ck
= + +
r
, then a, b, c are direction ratios of the line and conversely,,
if a, b, c are direction ratios of a line, then
ˆ
ˆ
ˆ
=
+
+
br
ai
bj
ck will be the parallel to
the line Here, b should not be confused with |
r
b| Derivation of cartesian form from vector form
Let the coordinates of the given point A be (x1, y1, z1) and the direction ratios of
the line be a, b, c |
1 | 5720-5723 | (1)
Remark If
ˆ
ˆ
ˆ
b
ai
bj
ck
= + +
r
, then a, b, c are direction ratios of the line and conversely,,
if a, b, c are direction ratios of a line, then
ˆ
ˆ
ˆ
=
+
+
br
ai
bj
ck will be the parallel to
the line Here, b should not be confused with |
r
b| Derivation of cartesian form from vector 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) |
1 | 5721-5724 | Here, b should not be confused with |
r
b| Derivation of cartesian form from vector 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) Then
zk
jy
ix
r
ˆ
ˆ
ˆ
+
+
r=
;
k
z
j
y
xi
a
ˆ
ˆ
ˆ
1
1
1
+
+
=
r
and
ˆ
ˆ
ˆ
b
a i
b j
c k
=
+
+
r
Substituting these values in (1) and equating the coefficients of ˆ
ˆ
,i
j and kˆ , we get
x = x1 + λa; y = y1 + λ b; z = z1+ λc |
1 | 5722-5725 | Derivation of cartesian form from vector 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) Then
zk
jy
ix
r
ˆ
ˆ
ˆ
+
+
r=
;
k
z
j
y
xi
a
ˆ
ˆ
ˆ
1
1
1
+
+
=
r
and
ˆ
ˆ
ˆ
b
a i
b j
c k
=
+
+
r
Substituting these values in (1) and equating the coefficients of ˆ
ˆ
,i
j and kˆ , we get
x = x1 + λa; y = y1 + λ b; z = z1+ λc (2)
Fig 11 |
1 | 5723-5726 | Consider the coordinates of any point P be (x, y, z) Then
zk
jy
ix
r
ˆ
ˆ
ˆ
+
+
r=
;
k
z
j
y
xi
a
ˆ
ˆ
ˆ
1
1
1
+
+
=
r
and
ˆ
ˆ
ˆ
b
a i
b j
c k
=
+
+
r
Substituting these values in (1) and equating the coefficients of ˆ
ˆ
,i
j and kˆ , we get
x = x1 + λa; y = y1 + λ b; z = z1+ λc (2)
Fig 11 4
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
469
These are parametric equations of the line |
1 | 5724-5727 | Then
zk
jy
ix
r
ˆ
ˆ
ˆ
+
+
r=
;
k
z
j
y
xi
a
ˆ
ˆ
ˆ
1
1
1
+
+
=
r
and
ˆ
ˆ
ˆ
b
a i
b j
c k
=
+
+
r
Substituting these values in (1) and equating the coefficients of ˆ
ˆ
,i
j and kˆ , we get
x = x1 + λa; y = y1 + λ b; z = z1+ λc (2)
Fig 11 4
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
469
These are parametric equations of the line Eliminating the parameter λ from (2),
we get
1
x – x
a
=
1
1
y – y
=z – z
b
c |
1 | 5725-5728 | (2)
Fig 11 4
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
469
These are parametric equations of the line Eliminating the parameter λ from (2),
we get
1
x – x
a
=
1
1
y – y
=z – z
b
c (3)
This is the Cartesian equation of the line |
1 | 5726-5729 | 4
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
469
These are parametric equations of the line Eliminating the parameter λ from (2),
we get
1
x – x
a
=
1
1
y – y
=z – z
b
c (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
=
1
1
y – y
=z – z
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
ˆ
ˆ
ˆ
3
2
8
i
j
k
+
− |
1 | 5727-5730 | Eliminating the parameter λ from (2),
we get
1
x – x
a
=
1
1
y – y
=z – z
b
c (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
=
1
1
y – y
=z – z
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
ˆ
ˆ
ˆ
3
2
8
i
j
k
+
− Solution We have
ar =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
5
2
4
and
3
2
8
i
j
k
b
i
j
k
+
−
=
+
−
r
Therefore, the vector equation of the line is
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
5
2
4
( 3
2
8
)
i
j
k
i
j
k
+
−
+ λ
+
−
Now, rr is the position vector of any point P(x, y, z) on the line |
1 | 5728-5731 | (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
=
1
1
y – y
=z – z
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
ˆ
ˆ
ˆ
3
2
8
i
j
k
+
− Solution We have
ar =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
5
2
4
and
3
2
8
i
j
k
b
i
j
k
+
−
=
+
−
r
Therefore, the vector equation of the line is
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
5
2
4
( 3
2
8
)
i
j
k
i
j
k
+
−
+ λ
+
−
Now, rr is the position vector of any point P(x, y, z) on the line Therefore,
ˆ
ˆ
ˆ
xi
y j
z k
+
+
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
5
2
4
( 3
2
8
)
i
j
k
i
j
k
+
−
+ λ
+
−
=
$
$
(5
3 )
(2
2 )
( 4
8 )
i
j
k
+ λ
+
+ λ
+ −
− λ
$
Eliminating λ , we get
5
3
x −
=
2
4
2
8
y
z
−
+
=
−
which is the equation of the line in Cartesian form |
1 | 5729-5732 | �Note If l, m, n are the direction cosines of the line, the equation of the line is
1
x – x
l
=
1
1
y – y
=z – z
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
ˆ
ˆ
ˆ
3
2
8
i
j
k
+
− Solution We have
ar =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
5
2
4
and
3
2
8
i
j
k
b
i
j
k
+
−
=
+
−
r
Therefore, the vector equation of the line is
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
5
2
4
( 3
2
8
)
i
j
k
i
j
k
+
−
+ λ
+
−
Now, rr is the position vector of any point P(x, y, z) on the line Therefore,
ˆ
ˆ
ˆ
xi
y j
z k
+
+
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
5
2
4
( 3
2
8
)
i
j
k
i
j
k
+
−
+ λ
+
−
=
$
$
(5
3 )
(2
2 )
( 4
8 )
i
j
k
+ λ
+
+ λ
+ −
− λ
$
Eliminating λ , we get
5
3
x −
=
2
4
2
8
y
z
−
+
=
−
which is the equation of the line in Cartesian form 11 |
1 | 5730-5733 | Solution We have
ar =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
5
2
4
and
3
2
8
i
j
k
b
i
j
k
+
−
=
+
−
r
Therefore, the vector equation of the line is
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
5
2
4
( 3
2
8
)
i
j
k
i
j
k
+
−
+ λ
+
−
Now, rr is the position vector of any point P(x, y, z) on the line Therefore,
ˆ
ˆ
ˆ
xi
y j
z k
+
+
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
5
2
4
( 3
2
8
)
i
j
k
i
j
k
+
−
+ λ
+
−
=
$
$
(5
3 )
(2
2 )
( 4
8 )
i
j
k
+ λ
+
+ λ
+ −
− λ
$
Eliminating λ , we get
5
3
x −
=
2
4
2
8
y
z
−
+
=
−
which is the equation of the line in Cartesian form 11 3 |
1 | 5731-5734 | Therefore,
ˆ
ˆ
ˆ
xi
y j
z k
+
+
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
5
2
4
( 3
2
8
)
i
j
k
i
j
k
+
−
+ λ
+
−
=
$
$
(5
3 )
(2
2 )
( 4
8 )
i
j
k
+ λ
+
+ λ
+ −
− λ
$
Eliminating λ , we get
5
3
x −
=
2
4
2
8
y
z
−
+
=
−
which is the equation of the line in Cartesian form 11 3 2 Equation of a line passing through two given points
Let ar and b
r
be the position vectors of two
points A (x1, y1, z1) and B(x 2, y 2, z2),
respectively that are lying on a line (Fig 11 |
1 | 5732-5735 | 11 3 2 Equation of a line passing through two given points
Let ar and b
r
be the position vectors of two
points A (x1, y1, z1) and B(x 2, y 2, z2),
respectively that are lying on a line (Fig 11 5) |
1 | 5733-5736 | 3 2 Equation of a line passing through two given points
Let ar and b
r
be the position vectors of two
points A (x1, y1, z1) and B(x 2, y 2, z2),
respectively that are lying on a line (Fig 11 5) Let rr be t he position vector of an
arbitrary point P(x, y, z), then P is a point on
the line if and only if AP
r
a
= −
uuur
r
r and
AB
b
a
=
−
uuur
r
r are collinear vectors |
1 | 5734-5737 | 2 Equation of a line passing through two given points
Let ar and b
r
be the position vectors of two
points A (x1, y1, z1) and B(x 2, y 2, z2),
respectively that are lying on a line (Fig 11 5) Let rr be t he position vector of an
arbitrary point P(x, y, z), then P is a point on
the line if and only if AP
r
a
= −
uuur
r
r and
AB
b
a
=
−
uuur
r
r are collinear vectors Therefore,
P is on the line if and only if
(
)
r
a
b
a
−=λ −
r
r
r
r
Fig 11 |
1 | 5735-5738 | 5) Let rr be t he position vector of an
arbitrary point P(x, y, z), then P is a point on
the line if and only if AP
r
a
= −
uuur
r
r and
AB
b
a
=
−
uuur
r
r are collinear vectors Therefore,
P is on the line if and only if
(
)
r
a
b
a
−=λ −
r
r
r
r
Fig 11 5
© NCERT
not to be republished
MATHEMATI CS
470
or
(
)
r
a
b
a
=
+ λ
−
r
r
r
r , λ ∈ R |
1 | 5736-5739 | Let rr be t he position vector of an
arbitrary point P(x, y, z), then P is a point on
the line if and only if AP
r
a
= −
uuur
r
r and
AB
b
a
=
−
uuur
r
r are collinear vectors Therefore,
P is on the line if and only if
(
)
r
a
b
a
−=λ −
r
r
r
r
Fig 11 5
© NCERT
not to be republished
MATHEMATI CS
470
or
(
)
r
a
b
a
=
+ λ
−
r
r
r
r , λ ∈ R (1)
This is the vector equation of the line |
1 | 5737-5740 | Therefore,
P is on the line if and only if
(
)
r
a
b
a
−=λ −
r
r
r
r
Fig 11 5
© NCERT
not to be republished
MATHEMATI CS
470
or
(
)
r
a
b
a
=
+ λ
−
r
r
r
r , λ ∈ R (1)
This is the vector equation of the line Derivation of cartesian form from vector form
We have
1
1
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
,
r
xi
y j
z k a
x i
y j
z k
= + +
= + +
r
r
and
2
2
2 ˆ
ˆ
ˆ
,
b
x i
y j
z k
= + +
r
Substituting these values in (1), we get
$
$
$
$
$
$
1
1
1
2
1
2
1
2
1
[(
)
(
)
(
) ]
xi
y j
z k
x i
y j
z k
x
x i
y
y
j
z
z k
+
+
=
+
+
+ λ
−
+
−
+
−
$
$
$
Equating the like coefficients of
jk
i
ˆ
,ˆ
,ˆ
, we get
x = x1 + λ (x2 – x1); y = y1 + λ (y2 – y1); z = z1 + λ (z2 – z1)
On eliminating λ, we obtain
1
1
1
2
1
2
1
2
1
x
x
y
y
z
z
x
x
y
y
z
z
−
−
−
=
=
−
−
−
which is the equation of the line in Cartesian form |
1 | 5738-5741 | 5
© NCERT
not to be republished
MATHEMATI CS
470
or
(
)
r
a
b
a
=
+ λ
−
r
r
r
r , λ ∈ R (1)
This is the vector equation of the line Derivation of cartesian form from vector form
We have
1
1
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
,
r
xi
y j
z k a
x i
y j
z k
= + +
= + +
r
r
and
2
2
2 ˆ
ˆ
ˆ
,
b
x i
y j
z k
= + +
r
Substituting these values in (1), we get
$
$
$
$
$
$
1
1
1
2
1
2
1
2
1
[(
)
(
)
(
) ]
xi
y j
z k
x i
y j
z k
x
x i
y
y
j
z
z k
+
+
=
+
+
+ λ
−
+
−
+
−
$
$
$
Equating the like coefficients of
jk
i
ˆ
,ˆ
,ˆ
, we get
x = x1 + λ (x2 – x1); y = y1 + λ (y2 – y1); z = z1 + λ (z2 – z1)
On eliminating λ, we obtain
1
1
1
2
1
2
1
2
1
x
x
y
y
z
z
x
x
y
y
z
z
−
−
−
=
=
−
−
−
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) |
1 | 5739-5742 | (1)
This is the vector equation of the line Derivation of cartesian form from vector form
We have
1
1
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
,
r
xi
y j
z k a
x i
y j
z k
= + +
= + +
r
r
and
2
2
2 ˆ
ˆ
ˆ
,
b
x i
y j
z k
= + +
r
Substituting these values in (1), we get
$
$
$
$
$
$
1
1
1
2
1
2
1
2
1
[(
)
(
)
(
) ]
xi
y j
z k
x i
y j
z k
x
x i
y
y
j
z
z k
+
+
=
+
+
+ λ
−
+
−
+
−
$
$
$
Equating the like coefficients of
jk
i
ˆ
,ˆ
,ˆ
, we get
x = x1 + λ (x2 – x1); y = y1 + λ (y2 – y1); z = z1 + λ (z2 – z1)
On eliminating λ, we obtain
1
1
1
2
1
2
1
2
1
x
x
y
y
z
z
x
x
y
y
z
z
−
−
−
=
=
−
−
−
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 ar and b
r
be the position vectors of the point A(– 1, 0, 2) and B(3, 4, 6) |
1 | 5740-5743 | Derivation of cartesian form from vector form
We have
1
1
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
,
r
xi
y j
z k a
x i
y j
z k
= + +
= + +
r
r
and
2
2
2 ˆ
ˆ
ˆ
,
b
x i
y j
z k
= + +
r
Substituting these values in (1), we get
$
$
$
$
$
$
1
1
1
2
1
2
1
2
1
[(
)
(
)
(
) ]
xi
y j
z k
x i
y j
z k
x
x i
y
y
j
z
z k
+
+
=
+
+
+ λ
−
+
−
+
−
$
$
$
Equating the like coefficients of
jk
i
ˆ
,ˆ
,ˆ
, we get
x = x1 + λ (x2 – x1); y = y1 + λ (y2 – y1); z = z1 + λ (z2 – z1)
On eliminating λ, we obtain
1
1
1
2
1
2
1
2
1
x
x
y
y
z
z
x
x
y
y
z
z
−
−
−
=
=
−
−
−
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 ar and b
r
be the position vectors of the point A(– 1, 0, 2) and B(3, 4, 6) Then
ˆ
ˆ
2
a
i
=−+k
r
and
ˆ
ˆ
ˆ
3
4
6
b
i
j
k
= + +
r
Therefore
ˆ
ˆ
ˆ
4
4
4
b
a
i
j
k
−
=
+
+
r
r
Let rr be the position vector of any point on the line |
1 | 5741-5744 | Example 7 Find the vector equation for the line passing through the points (–1, 0, 2)
and (3, 4, 6) Solution Let ar and b
r
be the position vectors of the point A(– 1, 0, 2) and B(3, 4, 6) Then
ˆ
ˆ
2
a
i
=−+k
r
and
ˆ
ˆ
ˆ
3
4
6
b
i
j
k
= + +
r
Therefore
ˆ
ˆ
ˆ
4
4
4
b
a
i
j
k
−
=
+
+
r
r
Let rr be the position vector of any point on the line Then the vector equation of
the line is
ˆ
ˆ
ˆ
ˆ
ˆ
2
(4
4
4 )
r
i
k
i
j
=−+ +λ + +k
r
Example 8 The Cartesian equation of a line is
3
5
6
2
4
2
x
y
z
+ − +
=
=
Find the vector equation for the line |
1 | 5742-5745 | Solution Let ar and b
r
be the position vectors of the point A(– 1, 0, 2) and B(3, 4, 6) Then
ˆ
ˆ
2
a
i
=−+k
r
and
ˆ
ˆ
ˆ
3
4
6
b
i
j
k
= + +
r
Therefore
ˆ
ˆ
ˆ
4
4
4
b
a
i
j
k
−
=
+
+
r
r
Let rr be the position vector of any point on the line Then the vector equation of
the line is
ˆ
ˆ
ˆ
ˆ
ˆ
2
(4
4
4 )
r
i
k
i
j
=−+ +λ + +k
r
Example 8 The Cartesian equation of a line is
3
5
6
2
4
2
x
y
z
+ − +
=
=
Find the vector equation for the line Solution Comparing the given equation with the standard form
1
1
1
x
x
y
y
z
z
a
b
c
−
−
−
=
=
We observe that
x1 = – 3, y1 = 5, z1 = – 6; a = 2, b = 4, c = 2 |
1 | 5743-5746 | Then
ˆ
ˆ
2
a
i
=−+k
r
and
ˆ
ˆ
ˆ
3
4
6
b
i
j
k
= + +
r
Therefore
ˆ
ˆ
ˆ
4
4
4
b
a
i
j
k
−
=
+
+
r
r
Let rr be the position vector of any point on the line Then the vector equation of
the line is
ˆ
ˆ
ˆ
ˆ
ˆ
2
(4
4
4 )
r
i
k
i
j
=−+ +λ + +k
r
Example 8 The Cartesian equation of a line is
3
5
6
2
4
2
x
y
z
+ − +
=
=
Find the vector equation for the line Solution Comparing the given equation with the standard form
1
1
1
x
x
y
y
z
z
a
b
c
−
−
−
=
=
We observe that
x1 = – 3, y1 = 5, z1 = – 6; a = 2, b = 4, c = 2 © NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
471
Thus, the required line passes through the point (– 3, 5, – 6) and is parallel to the
vector
ˆ
ˆ
ˆ
2
4
2
i
j
k
+
+ |
1 | 5744-5747 | Then the vector equation of
the line is
ˆ
ˆ
ˆ
ˆ
ˆ
2
(4
4
4 )
r
i
k
i
j
=−+ +λ + +k
r
Example 8 The Cartesian equation of a line is
3
5
6
2
4
2
x
y
z
+ − +
=
=
Find the vector equation for the line Solution Comparing the given equation with the standard form
1
1
1
x
x
y
y
z
z
a
b
c
−
−
−
=
=
We observe that
x1 = – 3, y1 = 5, z1 = – 6; a = 2, b = 4, c = 2 © NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
471
Thus, the required line passes through the point (– 3, 5, – 6) and is parallel to the
vector
ˆ
ˆ
ˆ
2
4
2
i
j
k
+
+ Let rr be the position vector of any point on the line, then the
vector equation of the line is given by
ˆ
ˆ
ˆ
(
3
5
6 )
r
i
j
r=− + −k
+ λ
ˆ
ˆ
ˆ
(2
4
2 )
i
j
k
+
+
11 |
1 | 5745-5748 | Solution Comparing the given equation with the standard form
1
1
1
x
x
y
y
z
z
a
b
c
−
−
−
=
=
We observe that
x1 = – 3, y1 = 5, z1 = – 6; a = 2, b = 4, c = 2 © NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
471
Thus, the required line passes through the point (– 3, 5, – 6) and is parallel to the
vector
ˆ
ˆ
ˆ
2
4
2
i
j
k
+
+ Let rr be the position vector of any point on the line, then the
vector equation of the line is given by
ˆ
ˆ
ˆ
(
3
5
6 )
r
i
j
r=− + −k
+ λ
ˆ
ˆ
ˆ
(2
4
2 )
i
j
k
+
+
11 4 Angle between Two Lines
Let L1 and L2 be two lines passing through the origin
and with direction ratios a1, b1, c1 and a2, b2, c2,
respectively |
1 | 5746-5749 | © NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
471
Thus, the required line passes through the point (– 3, 5, – 6) and is parallel to the
vector
ˆ
ˆ
ˆ
2
4
2
i
j
k
+
+ Let rr be the position vector of any point on the line, then the
vector equation of the line is given by
ˆ
ˆ
ˆ
(
3
5
6 )
r
i
j
r=− + −k
+ λ
ˆ
ˆ
ˆ
(2
4
2 )
i
j
k
+
+
11 4 Angle between Two Lines
Let L1 and L2 be two lines passing through the origin
and with direction ratios a1, b1, c1 and a2, b2, c2,
respectively Let P be a point on L1 and Q be a point
on L2 |
1 | 5747-5750 | Let rr be the position vector of any point on the line, then the
vector equation of the line is given by
ˆ
ˆ
ˆ
(
3
5
6 )
r
i
j
r=− + −k
+ λ
ˆ
ˆ
ˆ
(2
4
2 )
i
j
k
+
+
11 4 Angle between Two Lines
Let L1 and L2 be two lines passing through the origin
and with direction ratios a1, b1, c1 and a2, b2, 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 |
1 | 5748-5751 | 4 Angle between Two Lines
Let L1 and L2 be two lines passing through the origin
and with direction ratios a1, b1, c1 and a2, b2, 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 |
1 | 5749-5752 | 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 θ be the acute angle between
OP and OQ |
1 | 5750-5753 | 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
a1, b1, c1 and a2, b2, c2, respectively |
1 | 5751-5754 | 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
a1, b1, c1 and a2, b2, c2, respectively Therefore, the
angle θ between them is given by
cos θ =
1 2
1 2
1 2
2
2
2
2
2
2
1
1
1
2
2
2
a a
b b
c c
a
b
c
a
b
c
+
+
+
+
+
+ |
1 | 5752-5755 | Let θ be the acute angle between
OP and OQ Now recall that the directed line
segments OP and OQ are vectors with components
a1, b1, c1 and a2, b2, c2, respectively Therefore, the
angle θ between them is given by
cos θ =
1 2
1 2
1 2
2
2
2
2
2
2
1
1
1
2
2
2
a a
b b
c c
a
b
c
a
b
c
+
+
+
+
+
+ (1)
The angle between the lines in terms of sin θ is given by
sin θ =
2
1
cos
− θ
=
(
)(
)
2
1 2
1 2
1 2
2
2
2
2
2
2
1
1
1
2
2
2
(
)
1
a a
bb
c c
a
b
c
a
b
c
+
+
−
+
+
+
+
=
(
)(
) (
)
(
) (
)
2
2
2
2
2
2
2
1
1
1
2
2
2
1 2
1 2
1 2
2
2
2
2
2
2
1
1
1
2
2
2
a
b
c
a
b
c
a a
bb
c c
a
b
c
a
b
c
+
+
+
+
−
+
+
+
+
+
+
=
2
2
2
1
2
2
1
1
2
2
1
1
2
2
1
2
2
2
2
2
2
1
1
1
2
2
2
(
)
(
)
(
)
−
+
−
+
−
+ +
+ +
a b
a b
b c
b c
c a
c a
a
b
c
a
b
c |
1 | 5753-5756 | Now recall that the directed line
segments OP and OQ are vectors with components
a1, b1, c1 and a2, b2, c2, respectively Therefore, the
angle θ between them is given by
cos θ =
1 2
1 2
1 2
2
2
2
2
2
2
1
1
1
2
2
2
a a
b b
c c
a
b
c
a
b
c
+
+
+
+
+
+ (1)
The angle between the lines in terms of sin θ is given by
sin θ =
2
1
cos
− θ
=
(
)(
)
2
1 2
1 2
1 2
2
2
2
2
2
2
1
1
1
2
2
2
(
)
1
a a
bb
c c
a
b
c
a
b
c
+
+
−
+
+
+
+
=
(
)(
) (
)
(
) (
)
2
2
2
2
2
2
2
1
1
1
2
2
2
1 2
1 2
1 2
2
2
2
2
2
2
1
1
1
2
2
2
a
b
c
a
b
c
a a
bb
c c
a
b
c
a
b
c
+
+
+
+
−
+
+
+
+
+
+
=
2
2
2
1
2
2
1
1
2
2
1
1
2
2
1
2
2
2
2
2
2
1
1
1
2
2
2
(
)
(
)
(
)
−
+
−
+
−
+ +
+ +
a b
a b
b c
b c
c a
c a
a
b
c
a
b
c (2)
�Note In case the lines L1 and L2 do not pass through the origin, we may take
lines
1
′L andL2
′ which are parallel to L1 and L2 respectively and pass through
the origin |
1 | 5754-5757 | Therefore, the
angle θ between them is given by
cos θ =
1 2
1 2
1 2
2
2
2
2
2
2
1
1
1
2
2
2
a a
b b
c c
a
b
c
a
b
c
+
+
+
+
+
+ (1)
The angle between the lines in terms of sin θ is given by
sin θ =
2
1
cos
− θ
=
(
)(
)
2
1 2
1 2
1 2
2
2
2
2
2
2
1
1
1
2
2
2
(
)
1
a a
bb
c c
a
b
c
a
b
c
+
+
−
+
+
+
+
=
(
)(
) (
)
(
) (
)
2
2
2
2
2
2
2
1
1
1
2
2
2
1 2
1 2
1 2
2
2
2
2
2
2
1
1
1
2
2
2
a
b
c
a
b
c
a a
bb
c c
a
b
c
a
b
c
+
+
+
+
−
+
+
+
+
+
+
=
2
2
2
1
2
2
1
1
2
2
1
1
2
2
1
2
2
2
2
2
2
1
1
1
2
2
2
(
)
(
)
(
)
−
+
−
+
−
+ +
+ +
a b
a b
b c
b c
c a
c a
a
b
c
a
b
c (2)
�Note In case the lines L1 and L2 do not pass through the origin, we may take
lines
1
′L andL2
′ which are parallel to L1 and L2 respectively and pass through
the origin Fig 11 |
1 | 5755-5758 | (1)
The angle between the lines in terms of sin θ is given by
sin θ =
2
1
cos
− θ
=
(
)(
)
2
1 2
1 2
1 2
2
2
2
2
2
2
1
1
1
2
2
2
(
)
1
a a
bb
c c
a
b
c
a
b
c
+
+
−
+
+
+
+
=
(
)(
) (
)
(
) (
)
2
2
2
2
2
2
2
1
1
1
2
2
2
1 2
1 2
1 2
2
2
2
2
2
2
1
1
1
2
2
2
a
b
c
a
b
c
a a
bb
c c
a
b
c
a
b
c
+
+
+
+
−
+
+
+
+
+
+
=
2
2
2
1
2
2
1
1
2
2
1
1
2
2
1
2
2
2
2
2
2
1
1
1
2
2
2
(
)
(
)
(
)
−
+
−
+
−
+ +
+ +
a b
a b
b c
b c
c a
c a
a
b
c
a
b
c (2)
�Note In case the lines L1 and L2 do not pass through the origin, we may take
lines
1
′L andL2
′ which are parallel to L1 and L2 respectively and pass through
the origin Fig 11 6
© NCERT
not to be republished
MATHEMATI CS
472
If instead of direction ratios for the lines L1 and L2, direction cosines, namely,
l1, m 1, n1 for L1 and l2, m 2, n2 for L2 are given, then (1) and (2) takes the following form:
cos θ = |l1 l2 + m 1m 2 + n1n2| (as
2
2
2
1
1
1
1
l
m
n
+
+
=
2
2
2
2
2
2
l
m
n
=
+
+
) |
1 | 5756-5759 | (2)
�Note In case the lines L1 and L2 do not pass through the origin, we may take
lines
1
′L andL2
′ which are parallel to L1 and L2 respectively and pass through
the origin Fig 11 6
© NCERT
not to be republished
MATHEMATI CS
472
If instead of direction ratios for the lines L1 and L2, direction cosines, namely,
l1, m 1, n1 for L1 and l2, m 2, n2 for L2 are given, then (1) and (2) takes the following form:
cos θ = |l1 l2 + m 1m 2 + n1n2| (as
2
2
2
1
1
1
1
l
m
n
+
+
=
2
2
2
2
2
2
l
m
n
=
+
+
) (3)
and
sin θ =
(
)
2
2
2
1
2
2
1
1
2
2
1
1 2
2 1
(
)
(
)
l m
l m
m n
m n
n l
n l
−
−
−
+
− |
1 | 5757-5760 | Fig 11 6
© NCERT
not to be republished
MATHEMATI CS
472
If instead of direction ratios for the lines L1 and L2, direction cosines, namely,
l1, m 1, n1 for L1 and l2, m 2, n2 for L2 are given, then (1) and (2) takes the following form:
cos θ = |l1 l2 + m 1m 2 + n1n2| (as
2
2
2
1
1
1
1
l
m
n
+
+
=
2
2
2
2
2
2
l
m
n
=
+
+
) (3)
and
sin θ =
(
)
2
2
2
1
2
2
1
1
2
2
1
1 2
2 1
(
)
(
)
l m
l m
m n
m n
n l
n l
−
−
−
+
− (4)
Two lines with direction ratios a1, b1, c1 and a2, b2, c2 are
(i)
perpendicular i |
1 | 5758-5761 | 6
© NCERT
not to be republished
MATHEMATI CS
472
If instead of direction ratios for the lines L1 and L2, direction cosines, namely,
l1, m 1, n1 for L1 and l2, m 2, n2 for L2 are given, then (1) and (2) takes the following form:
cos θ = |l1 l2 + m 1m 2 + n1n2| (as
2
2
2
1
1
1
1
l
m
n
+
+
=
2
2
2
2
2
2
l
m
n
=
+
+
) (3)
and
sin θ =
(
)
2
2
2
1
2
2
1
1
2
2
1
1 2
2 1
(
)
(
)
l m
l m
m n
m n
n l
n l
−
−
−
+
− (4)
Two lines with direction ratios a1, b1, c1 and a2, b2, c2 are
(i)
perpendicular i e |
1 | 5759-5762 | (3)
and
sin θ =
(
)
2
2
2
1
2
2
1
1
2
2
1
1 2
2 1
(
)
(
)
l m
l m
m n
m n
n l
n l
−
−
−
+
− (4)
Two lines with direction ratios a1, b1, c1 and a2, b2, c2 are
(i)
perpendicular i e if θ = 90° by (1)
a1a2 + b1b2 + c1c2 = 0
(ii)
parallel i |
1 | 5760-5763 | (4)
Two lines with direction ratios a1, b1, c1 and a2, b2, c2 are
(i)
perpendicular i e if θ = 90° by (1)
a1a2 + b1b2 + c1c2 = 0
(ii)
parallel i e |
1 | 5761-5764 | e if θ = 90° by (1)
a1a2 + b1b2 + c1c2 = 0
(ii)
parallel i e if θ = 0 by (2)
1
2
a
a =
1
1
2
2
b
c
b
c
=
Now, we find the angle between two lines when their equations are given |
1 | 5762-5765 | if θ = 90° by (1)
a1a2 + b1b2 + c1c2 = 0
(ii)
parallel i e if θ = 0 by (2)
1
2
a
a =
1
1
2
2
b
c
b
c
=
Now, we find the angle between two lines when their equations are given If θ is
acute the angle between the lines
rr =
1
1
a
b
+λ
ur
and rr =
2
2
a
b
+µ
r
r
then
cosθ =
1
2
1
2
b
b
b
b
⋅
r
r
r
r
In Cartesian form, if θ is the angle between the lines
1
1
x
x
a
−
=
1
1
1
1
y
y
z
z
b
c
−
−
= |
1 | 5763-5766 | e if θ = 0 by (2)
1
2
a
a =
1
1
2
2
b
c
b
c
=
Now, we find the angle between two lines when their equations are given If θ is
acute the angle between the lines
rr =
1
1
a
b
+λ
ur
and rr =
2
2
a
b
+µ
r
r
then
cosθ =
1
2
1
2
b
b
b
b
⋅
r
r
r
r
In Cartesian form, if θ is the angle between the lines
1
1
x
x
a
−
=
1
1
1
1
y
y
z
z
b
c
−
−
= (1)
and
2
2
x
x
a
−
=
2
2
2
2
y
y
z
z
b
c
−
−
= |
1 | 5764-5767 | if θ = 0 by (2)
1
2
a
a =
1
1
2
2
b
c
b
c
=
Now, we find the angle between two lines when their equations are given If θ is
acute the angle between the lines
rr =
1
1
a
b
+λ
ur
and rr =
2
2
a
b
+µ
r
r
then
cosθ =
1
2
1
2
b
b
b
b
⋅
r
r
r
r
In Cartesian form, if θ is the angle between the lines
1
1
x
x
a
−
=
1
1
1
1
y
y
z
z
b
c
−
−
= (1)
and
2
2
x
x
a
−
=
2
2
2
2
y
y
z
z
b
c
−
−
= (2)
where, a1, b1, c1 and a2, b2, c2 are the direction ratios of the lines (1) and (2), respectively,
then
cos θ =
1 2
1 2
1 2
2
2
2
2
2
2
1
1
1
2
2
2
a a
b b
c c
a
b
c
a
b
c
+
+
+
+
+
+
Example 9 Find the angle between the pair of lines given by
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
2
4
(
2
2 )
i
j
k
i
j
k
+
−
+ λ
+
+
and
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
5
2
(3
2
6 )
i
j
i
j
k
−
+ µ
+
+
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
473
Solution Here 1b
r
=
ˆ
ˆ
2ˆ
2
i
j
k
+
+
and
2b
r
=
ˆ
ˆ
ˆ
3
2
6
i
j
k
+
+
The angle θ between the two lines is given by
cos θ =
1
2
1
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
2
2 ) (3
2
6 )
1
4
4
9
4
36
b b
i
j
k
i
j
k
b b
⋅
+
+
⋅
+
+
=
+ +
+
+
r
r
r
r
=
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 +
=
1
3
5
4
y
z
−
+
=
and
1
1
x +
=
4
5
1
2
y
z
−
−
=
Solution The direction ratios of the first line are 3, 5, 4 and the direction ratios of the
second line are 1, 1, 2 |
1 | 5765-5768 | If θ is
acute the angle between the lines
rr =
1
1
a
b
+λ
ur
and rr =
2
2
a
b
+µ
r
r
then
cosθ =
1
2
1
2
b
b
b
b
⋅
r
r
r
r
In Cartesian form, if θ is the angle between the lines
1
1
x
x
a
−
=
1
1
1
1
y
y
z
z
b
c
−
−
= (1)
and
2
2
x
x
a
−
=
2
2
2
2
y
y
z
z
b
c
−
−
= (2)
where, a1, b1, c1 and a2, b2, c2 are the direction ratios of the lines (1) and (2), respectively,
then
cos θ =
1 2
1 2
1 2
2
2
2
2
2
2
1
1
1
2
2
2
a a
b b
c c
a
b
c
a
b
c
+
+
+
+
+
+
Example 9 Find the angle between the pair of lines given by
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
2
4
(
2
2 )
i
j
k
i
j
k
+
−
+ λ
+
+
and
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
5
2
(3
2
6 )
i
j
i
j
k
−
+ µ
+
+
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
473
Solution Here 1b
r
=
ˆ
ˆ
2ˆ
2
i
j
k
+
+
and
2b
r
=
ˆ
ˆ
ˆ
3
2
6
i
j
k
+
+
The angle θ between the two lines is given by
cos θ =
1
2
1
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
2
2 ) (3
2
6 )
1
4
4
9
4
36
b b
i
j
k
i
j
k
b b
⋅
+
+
⋅
+
+
=
+ +
+
+
r
r
r
r
=
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 +
=
1
3
5
4
y
z
−
+
=
and
1
1
x +
=
4
5
1
2
y
z
−
−
=
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 =
2
2
2
2
2
2
3 |
1 | 5766-5769 | (1)
and
2
2
x
x
a
−
=
2
2
2
2
y
y
z
z
b
c
−
−
= (2)
where, a1, b1, c1 and a2, b2, c2 are the direction ratios of the lines (1) and (2), respectively,
then
cos θ =
1 2
1 2
1 2
2
2
2
2
2
2
1
1
1
2
2
2
a a
b b
c c
a
b
c
a
b
c
+
+
+
+
+
+
Example 9 Find the angle between the pair of lines given by
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
2
4
(
2
2 )
i
j
k
i
j
k
+
−
+ λ
+
+
and
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
5
2
(3
2
6 )
i
j
i
j
k
−
+ µ
+
+
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
473
Solution Here 1b
r
=
ˆ
ˆ
2ˆ
2
i
j
k
+
+
and
2b
r
=
ˆ
ˆ
ˆ
3
2
6
i
j
k
+
+
The angle θ between the two lines is given by
cos θ =
1
2
1
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
2
2 ) (3
2
6 )
1
4
4
9
4
36
b b
i
j
k
i
j
k
b b
⋅
+
+
⋅
+
+
=
+ +
+
+
r
r
r
r
=
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 +
=
1
3
5
4
y
z
−
+
=
and
1
1
x +
=
4
5
1
2
y
z
−
−
=
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 =
2
2
2
2
2
2
3 1
5 |
1 | 5767-5770 | (2)
where, a1, b1, c1 and a2, b2, c2 are the direction ratios of the lines (1) and (2), respectively,
then
cos θ =
1 2
1 2
1 2
2
2
2
2
2
2
1
1
1
2
2
2
a a
b b
c c
a
b
c
a
b
c
+
+
+
+
+
+
Example 9 Find the angle between the pair of lines given by
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
2
4
(
2
2 )
i
j
k
i
j
k
+
−
+ λ
+
+
and
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
5
2
(3
2
6 )
i
j
i
j
k
−
+ µ
+
+
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
473
Solution Here 1b
r
=
ˆ
ˆ
2ˆ
2
i
j
k
+
+
and
2b
r
=
ˆ
ˆ
ˆ
3
2
6
i
j
k
+
+
The angle θ between the two lines is given by
cos θ =
1
2
1
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(
2
2 ) (3
2
6 )
1
4
4
9
4
36
b b
i
j
k
i
j
k
b b
⋅
+
+
⋅
+
+
=
+ +
+
+
r
r
r
r
=
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 +
=
1
3
5
4
y
z
−
+
=
and
1
1
x +
=
4
5
1
2
y
z
−
−
=
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 =
2
2
2
2
2
2
3 1
5 1
4 |
1 | 5768-5771 | If θ is the angle between them, then
cos θ =
2
2
2
2
2
2
3 1
5 1
4 2
16
16
8 3
15
50
6
5 2
6
3
5
4
1
1
2
+
+
=
=
=
+
+
+
+
Hence, the required angle is cos–1 8
3
15
|
1 | 5769-5772 | 1
5 1
4 2
16
16
8 3
15
50
6
5 2
6
3
5
4
1
1
2
+
+
=
=
=
+
+
+
+
Hence, the required angle is cos–1 8
3
15
11 |
1 | 5770-5773 | 1
4 2
16
16
8 3
15
50
6
5 2
6
3
5
4
1
1
2
+
+
=
=
=
+
+
+
+
Hence, the required angle is cos–1 8
3
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 |
1 | 5771-5774 | 2
16
16
8 3
15
50
6
5 2
6
3
5
4
1
1
2
+
+
=
=
=
+
+
+
+
Hence, the required angle is cos–1 8
3
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 |
1 | 5772-5775 | 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 |
1 | 5773-5776 | 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 |
1 | 5774-5777 | 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 |
1 | 5775-5778 | 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 |
1 | 5776-5779 | 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 |
1 | 5777-5780 | 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 |
1 | 5778-5781 | 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 |
1 | 5779-5782 | 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
© NCERT
not to be republished
MATHEMATI CS
474
l2
S
T
Q
P
l1
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 |
1 | 5780-5783 | 7 Fig 11 7
© NCERT
not to be republished
MATHEMATI CS
474
l2
S
T
Q
P
l1
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 |
1 | 5781-5784 | Fig 11 7
© NCERT
not to be republished
MATHEMATI CS
474
l2
S
T
Q
P
l1
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 |
1 | 5782-5785 | 7
© NCERT
not to be republished
MATHEMATI CS
474
l2
S
T
Q
P
l1
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 |
1 | 5783-5786 | 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 |
1 | 5784-5787 | 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 | 5785-5788 | 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 l1 and l2 be two skew lines with equations (Fig |
1 | 5786-5789 | 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 |
1 | 5787-5790 | 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)
rr =
1
1
a
b
+ λ
r
r |
1 | 5788-5791 | 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)
rr =
1
1
a
b
+ λ
r
r (1)
and
rr =
2
2
a
b
+ µ
r
r |
1 | 5789-5792 | 11 8)
rr =
1
1
a
b
+ λ
r
r (1)
and
rr =
2
2
a
b
+ µ
r
r (2)
Take any point S on l1 with position vector
1ar
and T on l2, with position vector
ar |
1 | 5790-5793 | 8)
rr =
1
1
a
b
+ λ
r
r (1)
and
rr =
2
2
a
b
+ µ
r
r (2)
Take any point S on l1 with position vector
1ar
and T on l2, with position vector
ar 2
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 |
1 | 5791-5794 | (1)
and
rr =
2
2
a
b
+ µ
r
r (2)
Take any point S on l1 with position vector
1ar
and T on l2, with position vector
ar 2
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 |
1 | 5792-5795 | (2)
Take any point S on l1 with position vector
1ar
and T on l2, with position vector
ar 2
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) |
1 | 5793-5796 | 2
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
uuur
is the shortest distance vector between
l1 and l2 , then it being perpendicular to both 1b
r
and
2b
r
, the unit vector nˆ along PQ
uuur
would therefore be
ˆn =
1
2
1
2
|
|
b
b
b
b
×
×
r
r
r
r |
1 | 5794-5797 | 6 2) If PQ
uuur
is the shortest distance vector between
l1 and l2 , then it being perpendicular to both 1b
r
and
2b
r
, the unit vector nˆ along PQ
uuur
would therefore be
ˆn =
1
2
1
2
|
|
b
b
b
b
×
×
r
r
r
r (3)
Then
PQ
uuur
= d nˆ
where, d is the magnitude of the shortest distance vector |
1 | 5795-5798 | 2) If PQ
uuur
is the shortest distance vector between
l1 and l2 , then it being perpendicular to both 1b
r
and
2b
r
, the unit vector nˆ along PQ
uuur
would therefore be
ˆn =
1
2
1
2
|
|
b
b
b
b
×
×
r
r
r
r (3)
Then
PQ
uuur
= d nˆ
where, d is the magnitude of the shortest distance vector Let θ be the angle between
ST
uur
and PQ
uuur |
1 | 5796-5799 | If PQ
uuur
is the shortest distance vector between
l1 and l2 , then it being perpendicular to both 1b
r
and
2b
r
, the unit vector nˆ along PQ
uuur
would therefore be
ˆn =
1
2
1
2
|
|
b
b
b
b
×
×
r
r
r
r (3)
Then
PQ
uuur
= d nˆ
where, d is the magnitude of the shortest distance vector Let θ be the angle between
ST
uur
and PQ
uuur Then
PQ = ST |cos θ|
But
cos θ =
PQ ST
| PQ | | ST |
⋅
uuur uur
uuuur
uur
=
2
1
ˆ (
)
ST
d n
a
a
⋅d
−
r
r
(since
2
1
ST
)
a
a
=
−
uur
r
r
=
1
2
2
1
1
2
(
) (
)
ST
b
b
a
a
b
b
×
⋅
−
×
r
r
r
r
r
r
[From (3)]
Fig 11 |
1 | 5797-5800 | (3)
Then
PQ
uuur
= d nˆ
where, d is the magnitude of the shortest distance vector Let θ be the angle between
ST
uur
and PQ
uuur Then
PQ = ST |cos θ|
But
cos θ =
PQ ST
| PQ | | ST |
⋅
uuur uur
uuuur
uur
=
2
1
ˆ (
)
ST
d n
a
a
⋅d
−
r
r
(since
2
1
ST
)
a
a
=
−
uur
r
r
=
1
2
2
1
1
2
(
) (
)
ST
b
b
a
a
b
b
×
⋅
−
×
r
r
r
r
r
r
[From (3)]
Fig 11 8
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
475
Hence, the required shortest distance is
d = PQ = ST |cos θ|
or
d =
1
2
2
1
1
2
(
) |
1 | 5798-5801 | Let θ be the angle between
ST
uur
and PQ
uuur Then
PQ = ST |cos θ|
But
cos θ =
PQ ST
| PQ | | ST |
⋅
uuur uur
uuuur
uur
=
2
1
ˆ (
)
ST
d n
a
a
⋅d
−
r
r
(since
2
1
ST
)
a
a
=
−
uur
r
r
=
1
2
2
1
1
2
(
) (
)
ST
b
b
a
a
b
b
×
⋅
−
×
r
r
r
r
r
r
[From (3)]
Fig 11 8
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
475
Hence, the required shortest distance is
d = PQ = ST |cos θ|
or
d =
1
2
2
1
1
2
(
) (
)
|
|
b
b
a
a
b
b
×
−
×
r
r
r
r
r
r
Cartesian form
The shortest distance between the lines
l1 :
1
1
x
x
−a
=
1
1
1
1
y
y
z
z
b
c
−
−
=
and
l2 :
2
2
x
x
−a
=
2
2
2
2
y
y
z
z
b
c
−
−
=
is
2
1
2
1
2
1
1
1
1
2
2
2
2
2
2
1 2
2 1
1
2
2 1
1 2
2 1
(
)
(
)
(
)
x
x
y
y
z
z
a
b
c
a
b
c
b c
b c
c a
c a
a b
a b
−
−
−
−
+
−
+
−
11 |
1 | 5799-5802 | Then
PQ = ST |cos θ|
But
cos θ =
PQ ST
| PQ | | ST |
⋅
uuur uur
uuuur
uur
=
2
1
ˆ (
)
ST
d n
a
a
⋅d
−
r
r
(since
2
1
ST
)
a
a
=
−
uur
r
r
=
1
2
2
1
1
2
(
) (
)
ST
b
b
a
a
b
b
×
⋅
−
×
r
r
r
r
r
r
[From (3)]
Fig 11 8
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
475
Hence, the required shortest distance is
d = PQ = ST |cos θ|
or
d =
1
2
2
1
1
2
(
) (
)
|
|
b
b
a
a
b
b
×
−
×
r
r
r
r
r
r
Cartesian form
The shortest distance between the lines
l1 :
1
1
x
x
−a
=
1
1
1
1
y
y
z
z
b
c
−
−
=
and
l2 :
2
2
x
x
−a
=
2
2
2
2
y
y
z
z
b
c
−
−
=
is
2
1
2
1
2
1
1
1
1
2
2
2
2
2
2
1 2
2 1
1
2
2 1
1 2
2 1
(
)
(
)
(
)
x
x
y
y
z
z
a
b
c
a
b
c
b c
b c
c a
c a
a b
a b
−
−
−
−
+
−
+
−
11 5 |
1 | 5800-5803 | 8
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
475
Hence, the required shortest distance is
d = PQ = ST |cos θ|
or
d =
1
2
2
1
1
2
(
) (
)
|
|
b
b
a
a
b
b
×
−
×
r
r
r
r
r
r
Cartesian form
The shortest distance between the lines
l1 :
1
1
x
x
−a
=
1
1
1
1
y
y
z
z
b
c
−
−
=
and
l2 :
2
2
x
x
−a
=
2
2
2
2
y
y
z
z
b
c
−
−
=
is
2
1
2
1
2
1
1
1
1
2
2
2
2
2
2
1 2
2 1
1
2
2 1
1 2
2 1
(
)
(
)
(
)
x
x
y
y
z
z
a
b
c
a
b
c
b c
b c
c a
c a
a b
a b
−
−
−
−
+
−
+
−
11 5 2 Distance between parallel lines
If two lines l1 and l2 are parallel, then they are coplanar |
1 | 5801-5804 | (
)
|
|
b
b
a
a
b
b
×
−
×
r
r
r
r
r
r
Cartesian form
The shortest distance between the lines
l1 :
1
1
x
x
−a
=
1
1
1
1
y
y
z
z
b
c
−
−
=
and
l2 :
2
2
x
x
−a
=
2
2
2
2
y
y
z
z
b
c
−
−
=
is
2
1
2
1
2
1
1
1
1
2
2
2
2
2
2
1 2
2 1
1
2
2 1
1 2
2 1
(
)
(
)
(
)
x
x
y
y
z
z
a
b
c
a
b
c
b c
b c
c a
c a
a b
a b
−
−
−
−
+
−
+
−
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
rr =
1a
b
+ λ
r
r |
1 | 5802-5805 | 5 2 Distance between parallel lines
If two lines l1 and l2 are parallel, then they are coplanar Let the lines be given by
rr =
1a
b
+ λ
r
r (1)
and
rr =
2a
b
+ µ
r
r
… (2)
where,
1ar is the position vector of a point S on l1 and
2
ar
is the position vector of a point T on l2 Fig 11 |
1 | 5803-5806 | 2 Distance between parallel lines
If two lines l1 and l2 are parallel, then they are coplanar Let the lines be given by
rr =
1a
b
+ λ
r
r (1)
and
rr =
2a
b
+ µ
r
r
… (2)
where,
1ar is the position vector of a point S on l1 and
2
ar
is the position vector of a point T on l2 Fig 11 9 |
1 | 5804-5807 | Let the lines be given by
rr =
1a
b
+ λ
r
r (1)
and
rr =
2a
b
+ µ
r
r
… (2)
where,
1ar is the position vector of a point S on l1 and
2
ar
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 = |TP | |
1 | 5805-5808 | (1)
and
rr =
2a
b
+ µ
r
r
… (2)
where,
1ar is the position vector of a point S on l1 and
2
ar
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 = |TP | Let θ be the angle between the vectors ST
uur
and b
r |
1 | 5806-5809 | 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 = |TP | Let θ be the angle between the vectors ST
uur
and b
r Then
ST
b ×
uur
r
=
ˆ
(|
b|| ST| sin )
n
θ
uur
r |
1 | 5807-5810 | 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 = |TP | Let θ be the angle between the vectors ST
uur
and b
r Then
ST
b ×
uur
r
=
ˆ
(|
b|| ST| sin )
n
θ
uur
r (3)
where nˆ is the unit vector perpendicular to the plane of the lines l1 and l2 |
1 | 5808-5811 | Let θ be the angle between the vectors ST
uur
and b
r Then
ST
b ×
uur
r
=
ˆ
(|
b|| ST| sin )
n
θ
uur
r (3)
where nˆ is the unit vector perpendicular to the plane of the lines l1 and l2 But
ST
uur
=
2
1
a
a−
r
r
Fig 11 |
1 | 5809-5812 | Then
ST
b ×
uur
r
=
ˆ
(|
b|| ST| sin )
n
θ
uur
r (3)
where nˆ is the unit vector perpendicular to the plane of the lines l1 and l2 But
ST
uur
=
2
1
a
a−
r
r
Fig 11 9
© NCERT
not to be republished
MATHEMATI CS
476
Therefore, from (3), we get
2
1
(
)
b
a
a
×
−
r
r
r =
ˆ
|
b| PT
n
r
(since PT = ST sin θ)
i |
1 | 5810-5813 | (3)
where nˆ is the unit vector perpendicular to the plane of the lines l1 and l2 But
ST
uur
=
2
1
a
a−
r
r
Fig 11 9
© NCERT
not to be republished
MATHEMATI CS
476
Therefore, from (3), we get
2
1
(
)
b
a
a
×
−
r
r
r =
ˆ
|
b| PT
n
r
(since PT = ST sin θ)
i e |
1 | 5811-5814 | But
ST
uur
=
2
1
a
a−
r
r
Fig 11 9
© NCERT
not to be republished
MATHEMATI CS
476
Therefore, from (3), we get
2
1
(
)
b
a
a
×
−
r
r
r =
ˆ
|
b| PT
n
r
(since PT = ST sin θ)
i e ,
2
1
|
(
)|
b
a
a
×
−
r
r
r
= |
b| PT 1
⋅
r
(as |
|
ˆn = 1)
Hence, the distance between the given parallel lines is
d =
2
1
(
)
| PT |
|
|
b
a
a
b
×
−
=
r
r
r
uuur
r
Example 11 Find the shortest distance between the lines l1 and l2 whose vector
equations are
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
(2
)
i
j
i
j
k
+
+ λ
−
+ |
1 | 5812-5815 | 9
© NCERT
not to be republished
MATHEMATI CS
476
Therefore, from (3), we get
2
1
(
)
b
a
a
×
−
r
r
r =
ˆ
|
b| PT
n
r
(since PT = ST sin θ)
i e ,
2
1
|
(
)|
b
a
a
×
−
r
r
r
= |
b| PT 1
⋅
r
(as |
|
ˆn = 1)
Hence, the distance between the given parallel lines is
d =
2
1
(
)
| PT |
|
|
b
a
a
b
×
−
=
r
r
r
uuur
r
Example 11 Find the shortest distance between the lines l1 and l2 whose vector
equations are
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
(2
)
i
j
i
j
k
+
+ λ
−
+ (1)
and
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
(3
5
2
)
i
j
k
i
j
k
+
−
+ µ
−
+ |
1 | 5813-5816 | e ,
2
1
|
(
)|
b
a
a
×
−
r
r
r
= |
b| PT 1
⋅
r
(as |
|
ˆn = 1)
Hence, the distance between the given parallel lines is
d =
2
1
(
)
| PT |
|
|
b
a
a
b
×
−
=
r
r
r
uuur
r
Example 11 Find the shortest distance between the lines l1 and l2 whose vector
equations are
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
(2
)
i
j
i
j
k
+
+ λ
−
+ (1)
and
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
(3
5
2
)
i
j
k
i
j
k
+
−
+ µ
−
+ (2)
Solution Comparing (1) and (2) with rr =
1
1
a
b
+ λ
r
r
and
2
2
b
a
r
r
r
r
+µ
=
respectively,,
we get
1ar =
1
ˆ
ˆ
ˆ
ˆ
ˆ
,
2
i
j
b
i
j
k
+
=
−
+
r
2
ar
= 2 ˆi + ˆj – ˆk and
2b
r
= 3 ˆi – 5 ˆj + 2 ˆk
Therefore
2
1
a
r−a
r =
ˆ
ˆi
k
−
and
1
2
b
b
×
r
r
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
( 2
)
( 3
5
2
)
i
j
k
i
j
k
−
+
×
−
+
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
1
1
3
7
3
5
2
i
j
k
i
j
k
−
=
−
−
−
So
1
2
|
|
b
b
×
r
r
=
9
1
49
59
+ +
=
Hence, the shortest distance between the given lines is given by
d =
|
|
)
) |
1 | 5814-5817 | ,
2
1
|
(
)|
b
a
a
×
−
r
r
r
= |
b| PT 1
⋅
r
(as |
|
ˆn = 1)
Hence, the distance between the given parallel lines is
d =
2
1
(
)
| PT |
|
|
b
a
a
b
×
−
=
r
r
r
uuur
r
Example 11 Find the shortest distance between the lines l1 and l2 whose vector
equations are
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
(2
)
i
j
i
j
k
+
+ λ
−
+ (1)
and
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
(3
5
2
)
i
j
k
i
j
k
+
−
+ µ
−
+ (2)
Solution Comparing (1) and (2) with rr =
1
1
a
b
+ λ
r
r
and
2
2
b
a
r
r
r
r
+µ
=
respectively,,
we get
1ar =
1
ˆ
ˆ
ˆ
ˆ
ˆ
,
2
i
j
b
i
j
k
+
=
−
+
r
2
ar
= 2 ˆi + ˆj – ˆk and
2b
r
= 3 ˆi – 5 ˆj + 2 ˆk
Therefore
2
1
a
r−a
r =
ˆ
ˆi
k
−
and
1
2
b
b
×
r
r
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
( 2
)
( 3
5
2
)
i
j
k
i
j
k
−
+
×
−
+
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
1
1
3
7
3
5
2
i
j
k
i
j
k
−
=
−
−
−
So
1
2
|
|
b
b
×
r
r
=
9
1
49
59
+ +
=
Hence, the shortest distance between the given lines is given by
d =
|
|
)
) (
(
2
1
1
2
2
1
b
b
a
a
b
b
r
r
r
r
r
r
×
−
×
59
10
59
7|
0
|3
=
+
−
=
Example 12 Find the distance between the lines l1 and l2 given by
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
4
( 2
3
6
)
i
j
k
i
j
k
+
−
+ λ
+
+
and
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
3
5
( 2
3
6
)
i
j
k
i
j
k
+
−
+ µ
+
+
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
477
Solution The two lines are parallel (Why |
1 | 5815-5818 | (1)
and
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
(3
5
2
)
i
j
k
i
j
k
+
−
+ µ
−
+ (2)
Solution Comparing (1) and (2) with rr =
1
1
a
b
+ λ
r
r
and
2
2
b
a
r
r
r
r
+µ
=
respectively,,
we get
1ar =
1
ˆ
ˆ
ˆ
ˆ
ˆ
,
2
i
j
b
i
j
k
+
=
−
+
r
2
ar
= 2 ˆi + ˆj – ˆk and
2b
r
= 3 ˆi – 5 ˆj + 2 ˆk
Therefore
2
1
a
r−a
r =
ˆ
ˆi
k
−
and
1
2
b
b
×
r
r
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
( 2
)
( 3
5
2
)
i
j
k
i
j
k
−
+
×
−
+
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
1
1
3
7
3
5
2
i
j
k
i
j
k
−
=
−
−
−
So
1
2
|
|
b
b
×
r
r
=
9
1
49
59
+ +
=
Hence, the shortest distance between the given lines is given by
d =
|
|
)
) (
(
2
1
1
2
2
1
b
b
a
a
b
b
r
r
r
r
r
r
×
−
×
59
10
59
7|
0
|3
=
+
−
=
Example 12 Find the distance between the lines l1 and l2 given by
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
4
( 2
3
6
)
i
j
k
i
j
k
+
−
+ λ
+
+
and
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
3
5
( 2
3
6
)
i
j
k
i
j
k
+
−
+ µ
+
+
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
477
Solution The two lines are parallel (Why ) We have
1ar =
ˆ
ˆ
2ˆ
4
i
j
k
+
−
,
2ar
=
ˆ
ˆ
ˆ
3
3
5
i
j
k
+
−
and b
r
=
ˆ
ˆ
ˆ
2
3
6
i
j
k
+
+
Therefore, the distance between the lines is given by
d =
2
1
(
)
| |
b
a
a
b
×
−
r
r
r
r
=
ˆ
ˆ
ˆ
2
3
6
2
1
1
4
9
36
i
j
k
−
+
+
or
=
ˆ
ˆ
ˆ
|
9
14
4
|
293
7293
49
49
i
j
k
−
+
−
=
=
EXERCISE 11 |
1 | 5816-5819 | (2)
Solution Comparing (1) and (2) with rr =
1
1
a
b
+ λ
r
r
and
2
2
b
a
r
r
r
r
+µ
=
respectively,,
we get
1ar =
1
ˆ
ˆ
ˆ
ˆ
ˆ
,
2
i
j
b
i
j
k
+
=
−
+
r
2
ar
= 2 ˆi + ˆj – ˆk and
2b
r
= 3 ˆi – 5 ˆj + 2 ˆk
Therefore
2
1
a
r−a
r =
ˆ
ˆi
k
−
and
1
2
b
b
×
r
r
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
( 2
)
( 3
5
2
)
i
j
k
i
j
k
−
+
×
−
+
=
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
1
1
3
7
3
5
2
i
j
k
i
j
k
−
=
−
−
−
So
1
2
|
|
b
b
×
r
r
=
9
1
49
59
+ +
=
Hence, the shortest distance between the given lines is given by
d =
|
|
)
) (
(
2
1
1
2
2
1
b
b
a
a
b
b
r
r
r
r
r
r
×
−
×
59
10
59
7|
0
|3
=
+
−
=
Example 12 Find the distance between the lines l1 and l2 given by
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
4
( 2
3
6
)
i
j
k
i
j
k
+
−
+ λ
+
+
and
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
3
5
( 2
3
6
)
i
j
k
i
j
k
+
−
+ µ
+
+
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
477
Solution The two lines are parallel (Why ) We have
1ar =
ˆ
ˆ
2ˆ
4
i
j
k
+
−
,
2ar
=
ˆ
ˆ
ˆ
3
3
5
i
j
k
+
−
and b
r
=
ˆ
ˆ
ˆ
2
3
6
i
j
k
+
+
Therefore, the distance between the lines is given by
d =
2
1
(
)
| |
b
a
a
b
×
−
r
r
r
r
=
ˆ
ˆ
ˆ
2
3
6
2
1
1
4
9
36
i
j
k
−
+
+
or
=
ˆ
ˆ
ˆ
|
9
14
4
|
293
7293
49
49
i
j
k
−
+
−
=
=
EXERCISE 11 2
1 |
1 | 5817-5820 | (
(
2
1
1
2
2
1
b
b
a
a
b
b
r
r
r
r
r
r
×
−
×
59
10
59
7|
0
|3
=
+
−
=
Example 12 Find the distance between the lines l1 and l2 given by
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
4
( 2
3
6
)
i
j
k
i
j
k
+
−
+ λ
+
+
and
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
3
5
( 2
3
6
)
i
j
k
i
j
k
+
−
+ µ
+
+
© NCERT
not to be republished
THREE D IMENSIONAL G EOMETRY
477
Solution The two lines are parallel (Why ) We have
1ar =
ˆ
ˆ
2ˆ
4
i
j
k
+
−
,
2ar
=
ˆ
ˆ
ˆ
3
3
5
i
j
k
+
−
and b
r
=
ˆ
ˆ
ˆ
2
3
6
i
j
k
+
+
Therefore, the distance between the lines is given by
d =
2
1
(
)
| |
b
a
a
b
×
−
r
r
r
r
=
ˆ
ˆ
ˆ
2
3
6
2
1
1
4
9
36
i
j
k
−
+
+
or
=
ˆ
ˆ
ˆ
|
9
14
4
|
293
7293
49
49
i
j
k
−
+
−
=
=
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 |
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