Let A and B be points with position vectors ¯¯¯a and ¯¯b with respect to origin O. If the point C on OA is such that 2¯¯¯¯¯¯¯¯AC=¯¯¯¯¯¯¯¯CO, ¯¯¯¯¯¯¯¯¯CD is parallel to ¯¯¯¯¯¯¯¯OB and |¯¯¯¯¯¯¯¯¯CD|=3|¯¯¯¯¯¯¯¯OB| then AD is
A
¯¯b−a9
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B
3¯¯b−a3
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C
¯¯b−a3
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D
¯¯b+a3
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Solution
The correct option is B3¯¯b−a3 Given that 2−−→AC=−−→CO 2(→c−→a)=−→c 3→c=2→a →c=2→a3
Also given −−→CD||−−→OB ⇒→a−→c=k→b ⇒∣∣→d−→c∣∣=3∣∣→b∣∣ ⇒k∣∣→b∣∣=3∣∣→b∣∣ ⇒k=3 and →d=→c+3→b ⇒−−→AD=→d−→a=3→b+→c−→a ⇒−−→AD=3→b−→a3