In an octagon A B C D E F G H of equal side, what is the sum of AB+AC+AD+AE+AF+AG+AH \;{ if, }\;{A O}=2{i}+3{j}-4{k} ? [25 Feb 2021 Shift 1]

Correct Answer is : 16i^+24j^−32k¹6{i}+24{j}-32{k}16i^+24j^−32k^

Correct Answer is : 16i^+24j^−32k¹6{i}+24{j}-32{k}16i^+24j^−32k^

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Vector Algebra

Section: Physics

Question : 41 of 47

Marks: +1, -0

In an octagon

A B C D E F G H

AB+AC+AD+AE+AF+AG+AH

\;\text{ if, }\;\overline{A O}=2\hat{i}+3\hat{j}-4\hat{k} ?

Solution:

Given,

A O=2\hat{i}+3\hat{j}-4\hat{k}

By using triangle law,

\;\text{ Similarly, }\;\;A C=A O+O B

A D\;\;=A O+O C

A E\;\;=A O+O E

A F\;\;=A O+O F

A G\;\;=A O+O G

A H\;\;=A O+O H

Now, adding all vectors

\;A B+A C+A D+A E+A F+A G+A H

\;\;=7 A O+(O B+O C+O D+O E+O F+O G+O H)

. . . (i)

By using cyclic vector,

OA+OB+OC+OD+OE+OF+OG+OH=0

\Rightarrow OB+OC+OD+OE+OF+OG+OH

=0-OA=0+AO

Substituting in Eq. (i), we get

A B+A C+A D\;+A E+A F+A G+A H=7 A O+A O=8 A O

\;\;=8(2\hat{i}+3\hat{j}-4\hat{k})

\;\;=16\hat{i}+24\hat{j}-32\hat{k}

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