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Question

A, B, C and D are four points such that AB=m(2^i6^j+2^k), BC=(^i2^j) and CD=n(6^i+15^j3^k). If ^CD intersects AB at some point E, then which of the following option(s) is/are correct?


A

m12

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B

n13

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C

m=n

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D

m14

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Solution

The correct options are
A

m12


B

n13


Let EB=p(AB) and (CE)=q(CD)

Then 0<p,q1

Since EB+BC+CE=0

pm(2^i6j+2^k)+(^i2j)+qn(6^i+15j3^k)=0

(2pm+16qn)^i+(6pm2+15qn)^j+(2pm3qn)^k=0

2pm6qn+1=0,6pm2+15qn=0,2pm3qn=0

p=12m,q=13n

0<12m1 and 0<13n1

m12,n13


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