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CBSE Class 12 Math 2008 Solved Paper

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Question : 21
Total: 29
Find the shortest distance between the following lines:
x−3
1
 =
y−5
−2
 =
z−7
1
 and
x+1
7
 =
y+1
−6
 =
z+1
1
OR
Find the point on the line
x+2
3
 =
y+1
2
 =
z−3
2
 at a distance 3√2 from the point (1 , 2 , 3)
Solution:  
x−3
1
 =
y−5
−2
 =
z−7
1
The vector form of this equation is:
→
r
 = 3
^
i
+5
^
j
+7
^
k
 + λ (
^
i
−2
^
j
+
^
k
)
→
r
 =
→
a
1+λ
→
b1
 ... (1)
x+1
7
 =
y+1
−6
 =
z+1
1
The vector form of this equation is:
→
r
 = -
^
i
−
^
j
−
^
k
 + λ (7
^
i
−6
^
j
+
^
k
)
→
r
 =
→
a2
+λ
→
b2
Therefore,
→
a1
 = 3
^
i
+5
^
j
+7
^
k
 ,
→
b1
 =
^
i
−2
^
j
+
^
k
 ,
→
a2
 = -
^
i
−
^
j
−
^
k
 and
→
b2
 = 7
^
i
−6
^
j
+
^
k
Now, the shortest distance between these two lines is given by:
d = |
→
b1
×
→
b2
.
→
a2
−
→
a1
|
→
b1
×
→
b1
|
|
→
b1
×
→
b2
 = |
^
i
^
j
^
k
1−21
7−61
|
=
^
i
(2+6) -
^
j
(1−7) +
^
k
(−6+14)
= 4
^
i
+6
^
j
+8
^
k
|
→
b1
×
→
b2
| = √42+62+82 = √116
→
a2
−
→
a1
 = −
^
i
−
^
j
−
^
k
 - 3
^
i
+5
^
j
+7
^
k
= −4
^
i
−6
^
j
−8
^
k
∴ d =
|
4
^
i
+6
^
j
+8
^
k
.−4
^
i
−6
^
j
−8
^
k
√116
|
= |
−16−36−64
√116
| = |
−116
√116
| = √116
OR
Let
x+2
3
 =
y+1
2
 =
z−3
2
 = λ
x = 2 + 3 λ ,y = - 1 + 2 λ ,z = 3 + 2 λ
Therefore, a point on this line is: {(-2+3λ), (-1 + 2λ), (3 + 2λ)}
The distance of the point{(-2+3λ), (-1 + 2λ), (3 + 2λ)} from point (1, 2, 3) = 3√2
∴
√−2+3λ−12+(−1)+2λ−22+3+2λ−32
= 3√2
⇒ - 3 + 3λ2 + (-3) + 2λ+2λ2 = 18
⇒ 9 + 9λ2 - 18λ + 9 + 4λ2 - 12λ + 4λ2 = 18
17λ2 - 30λ = 0
λ = 0 , λ =
30
17
When λ =
30
17
x = - 2 + 3λ = - 2 + 3 (
30
17
) = - 2 +
90
17
 =
56
17
y = - 1 + 2λ = - 1 + 2 (
30
17
) = - 1 +
60
17
 =
43
17
z = 3 + 2λ = 3 + 2 (
30
17
) =
51+60
17
 =
111
17
Thus, when λ =
30
17
 , the point is (
56
17
,
43
17
,
111
17
) and when λ = 0 , the point is (- 2 , - 1 , 3)
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