Please answer 20th question Q 20 If ABCDEF is a regular hexagon, then AD + EB + FC equals - Maths - Vector Algebra

Question

Please answer 20th question
Q 20 If ABCDEF is a regular hexagon, then AD + EB + FC equals
(a) zero
(b) 2 AB
(c) 4 AB
(d) 3 AB

Neha Sethi · Accepted Answer

Dear student 4AB→ is the answer
<img alt="" src=
"https://img-nm%20mnimgs%20com/img/study_content/content_ck_images/images/94844%20png"
style="width: 262px; height: 208px;">

AD→=2BC→EB→=2FA→FC→=2AB→

AD→+EB→=2BC→+FA→ =2AO→+FA→           ∵  BC→=AO→

In triangle AOF,

FA→+AO→+FO→=0∴  FA→+AO→=−FO→∴  AD→+EB→=−2FO→

And AB→=−FO→

∴  AD→+EB→=2AB→∴  AD→+EB→+FC→=2AB→+2AB→=4AB→
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Please answer 20th question Q.20. If ABCDEF is a regular hexagon, then AD + EB + FC equals (a) zero (b) 2 AB (c) 4 AB (d) 3 AB

Please answer 20th question
Q.20. If ABCDEF is a regular hexagon, then AD + EB + FC equals
(a) zero
(b) 2 AB
(c) 4 AB
(d) 3 AB