A line is drawn through a point A (6, 2, 2) in the direction of the vector 
–2 
+ 2 
and another line through a point A ′
(–4, 0, – 1) in the direction of the vector 3 
– 2 
– 2 
.
(i) Then a unit vector in direction of the common
perpendicular to two lines is-
Text Solution
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Ans.
(i)
Sol.Let 
= 
– 2 
+ 2 
and 
= 3 
– 2 
+ 2 
, then the desired unit vector = ± 
= ± 
(2 
+ 2 
+ 
)
(ii)
Sol. 
= 6 
+ 2 
+ 2 
and 
′ = –4 
– 2 
⇒ 
– 
′ = 10 
+ 2 
+ 3 
∴ S.D. = ± 
( 
– 
′ ) = 9
(iii)
Sol. The equation of the plane containing first line common perpendicular is [ 
– 
, 
, 
× 
] = 0
[ 
– (6 
+ 2 
+ 2 
)] . ( 
– 2 
+ 2 
) × (2 
+ 2 
+ 
)... (1)
The equation of the first line is 
= (6 
+ 2 
+ 2 
)
t( 
– 2 
+ 2 
) ... (2)
Solving (1) and (2) we get = 5 
+ 4 
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