A line l passing through the origin is perpendicular to the lines
l 1 : (3 + t) 
+ (– 1 + 2t) 
+ (4 + 2t) 
, – ∞ < t < ∞
l 2 : (3 + 2s) 
+ (3 + 2s) 
+ (2 + s) 
, – ∞ < s < ∞
Then, the coordinate(s) of the point(s) on l 2 at a distance of 
from the point of intersection of l and l 1 is(are)
Text Solution
Verified by ExpertsA
(b,d)
Let equation of line is
: 
= 
= 
= k
This line is perpendicular to given line 1 and 2 .
Hence a + 2b + 2c = 0
2a + 2b + c = 0
= 
= 
Hence equation of is 
= 
= 
= k 1 , k 2
↓ ↓
for 1 for 2
Now A(–2k 1 , 3k 1 , –2k 1 ) B(–2k 2 , 3k 2 , –2k 2 )
Point A satisfied 1
–2k 1 
+ 3k 1 
– 2k 1 
= (3 + t) 
+ (–1 + 2t) 
+ (4 + 2t) 
3 + t = –2k 1 .......
–1 + 2t = 3k 1 .......
4 + 2t = –2k 1 .......
& –5 = 5k 1 ⇒ k 1 = –1 ⇒ A (2, –3, 2)
Let any point on 2 (3 + 2S, 3 + 2S, 2 + S)
Given 
= 
9S 2 + 28S + 37 = 17
9S 2 + 28S + 20 = 0
9S 2 + 18S + 10S + 20 = 0
9S(S + 2) + 10 (S + 2) = 0
S = –2, –10/9
Hence (–1, –1, 0) , (7/9, 7/9, 8/9)
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