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CBSE Class 12 Maths 2010 Solved Paper

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Question : 29
Total: 29
Write the vector equations of the following lines and hence determine the distance between them:
x−1
2
 =
y−2
3
 =
z−4
6
 ;
x−3
4
 =
y−3
6
 =
z+5
12
Solution:  
Given equation of line is
x−1
2
 =
y−2
3
 =
z−4
6
This can also be written in the standard form as
x−1
2
 =
y−2
3
 =
z−(−4)
6
The vector form of the above equation is,
→
r
 = (
^
i
+2
^
j
−4k) + λ(2
^
i
+3
^
j
+6
^
k
)
⇒
→
r
 =
→
a
1+λ
→
b
 ... (i)
where,
→
a
 =
^
i
+2
^
j
−4
^
k
 and
→
b
 = 2
^
i
+3
^
j
+6
^
k
The second equation of line is
x−3
4
 =
y−3
6
 =
z+5
12
The above equation can also be written as
x−3
4
 =
y−3
6
 =
z−(−5)
12
The vector form of this equation is
→
r
 = (3
^
i
+3
^
j
−5k) + µ(4
^
i
+6
^
j
+12
^
k
)
⇒
→
r
 = (3
^
i
+3
^
j
−5k) + 2µ(2
^
i
+3
^
j
+6
^
k
)
⇒
→
r
 =
→
a
2+2µ
→
b
 ... (ii)
where
→
a
2 = (3
^
i
+3
^
j
−5k) and
→
b
 = 2
^
i
+3
^
j
+6
^
k
Since
→
b
 is same in equations (1) and (2), the two lines are parallel. Distance d, between the two parallel lines is given by the formula,
d = |
→
b
×(
→
a2
−
→
a1
|b|
|
Here,
→
b
 = 2
^
i
+3
^
j
+6
^
k
 ,
→
a
2 = (3
^
i
+3
^
j
−5k) and
→
a
1 =
^
i
+2
^
j
−4
^
k
On substitution, we get
d =
|
(2
^
i
+3
^
j
+6
^
k
)×(3
^
i
+3
^
j
−5
^
k
−(
^
i
+2
^
j
−4
^
k
)
√4+9+36
|

=
1
√49
|(2
^
i
+3
^
j
+6
^
k
)×(2
^
i
+
^
j
−
^
k
)|

=
1
7
|
^
i
^
j
^
k
236
21−1
|
=
1
7
 |
^
i
 (- 3 - 6) -
^
j
 (- 2 - 12) +
^
k
 (2 - 6)|
=
1
7
|−9
^
i
+14
^
j
−4
^
k
|
=
1
7
|√81+196+16|
=
√293
7
Thus, the distance between the two given lines is
√293
7
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