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

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Arun · Answer

Dear Akshat r x a = b x a (r – b) x a = 0 r – b is parallel to a hence r – b = a i.e. r = b + a similarly, r x b = a x b can be written as r = a + b for point of intersection of these lines we get b + a = a + b = = 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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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

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