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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

Arun
25763 Points
3 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