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Question

The lines ¯¯¯r=(66s)¯¯¯a+(4s4)¯¯b+(48s)¯¯c and ¯¯¯r=(2t1)¯¯¯a+(4t2)¯¯b2(t+3)¯¯c intersect at

A
4¯¯c
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B
4¯¯c
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C
3¯¯c
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D
2¯¯c
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Solution

The correct option is B 4¯¯c
Since vectors have intersection point so equate
(66s)a+(4s4)b+(48s)c=(2t1)a+(4t2)b2(t+3)c
66s=2t1 __________(1)
4s4=4t2 __________(2)
2×(1)(2)
s=1 , t=12
so r=4c

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