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let a=i+j and b=2i-k then point of intersection of the lines r×a=b×a and r×b=a×b is (1)-i+j+k (2)3i-j+k (3)3i+j-k (4)i-j-k

let a=i+j and b=2i-k then point of intersection of the lines r×a=b×a and r×b=a×b is (1)-i+j+k (2)3i-j+k (3)3i+j-k (4)i-j-k

Grade:12

1 Answers

Arun
25757 Points
5 years ago
Dear Akshat
 
r x a = b x a
(r – b) x a = 0
r – b is parallel to a
hence 
r – b = \lambda a
i.e.
r = b + \lambda a
similarly, r x b = a x b can be written as
r = a + \mu b
for point of intersection of these lines
we get
b + \lambda a = a + \mu b
\lambda = \mu = 1
 
hence the required point of intersection is
r = a+ b = i +j + 2 i – k = 3 i +j – k
 
hence option 3 is correct
 
regards
Arun (askIITians forum expert)

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