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Vector Algebra question

2021 · 25 Feb · Shift 1 · Q61
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Vector Algebra question

2021 · 25 Feb · Shift 1 · Q61

JEE MainPhysicsVector AlgebraMCQ+4 / −1
In an octagon ABCDEFGH of equal side, what is the sum of AB→+AC→+AD→+AE→+AF→+AG→+AH→\overrightarrow {AB} + \overrightarrow {AC} + \overrightarrow {AD} + \overrightarrow {AE} + \overrightarrow {AF} + \overrightarrow {AG} + \overrightarrow {AH}AB+AC+AD+AE+AF+AG+AH, if, AO→=2i^+3j^−4k^\overrightarrow {AO} = 2\widehat i + 3\widehat j - 4\widehat kAO=2i+3j​−4k JEE Main 2021 (Online) 25th February Morning Shift Physics - Vector Algebra Question 32 English
  1. A
    −16i^−24j^+32k^- 16\widehat i - 24\widehat j + 32\widehat k−16i−24j​+32k
  2. B
    16i^+24j^−32k^16\widehat i + 24\widehat j - 32\widehat k16i+24j​−32k
  3. C
    16i^+24j^+32k^16\widehat i + 24\widehat j + 32\widehat k16i+24j​+32k
  4. D
    16i^−24j^+32k^16\widehat i - 24\widehat j + 32\widehat k16i−24j​+32k
View written solutionFree

Correct answer: B

  1. Use the position-vector idea with the center of the regular octagon

Since the octagon ABCDEFGHABCDEFGHABCDEFGH has equal sides, it is a regular octagon. Let OOO be its center.

We are given

AO→=2i^+3j^−4k^\overrightarrow{AO}=2\hat i+3\hat j-4\hat kAO=2i^+3j^​−4k^

We need to find

AB→+AC→+AD→+AE→+AF→+AG→+AH→.\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}+\overrightarrow{AG}+\overrightarrow{AH}.AB+AC+AD+AE+AF+AG+AH.
  1. Express each vector through the center OOO

For any vertex XXX,

AX→=AO→+OX→.\overrightarrow{AX}=\overrightarrow{AO}+\overrightarrow{OX}.AX=AO+OX.

So,

∑X=BHAX→=∑X=BH(AO→+OX→).\sum_{X=B}^{H}\overrightarrow{AX} =\sum_{X=B}^{H}(\overrightarrow{AO}+\overrightarrow{OX}).X=B∑H​AX=X=B∑H​(AO+OX).

There are 777 terms, hence

∑X=BHAX→=7AO→+∑X=BHOX→.\sum_{X=B}^{H}\overrightarrow{AX}=7\overrightarrow{AO}+\sum_{X=B}^{H}\overrightarrow{OX}.X=B∑H​AX=7AO+X=B∑H​OX.
  1. Use symmetry of the regular octagon

For a regular octagon, the sum of position vectors of all vertices from the center is zero:

OA→+OB→+OC→+OD→+OE→+OF→+OG→+OH→=0⃗.\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}+\overrightarrow{OE}+\overrightarrow{OF}+\overrightarrow{OG}+\overrightarrow{OH}=\vec 0.OA+OB+OC+OD+OE+OF+OG+OH=0.

Since OX→=OX→\overrightarrow{OX}=\overrightarrow{OX}OX=OX, we get

OA→+∑X=BHOX→=0⃗.\overrightarrow{OA}+\sum_{X=B}^{H}\overrightarrow{OX}=\vec 0.OA+X=B∑H​OX=0.

Thus,

∑X=BHOX→=−OA→.\sum_{X=B}^{H}\overrightarrow{OX}=-\overrightarrow{OA}.X=B∑H​OX=−OA.

But

OA→=−AO→.\overrightarrow{OA}=-\overrightarrow{AO}.OA=−AO.

Therefore,

∑X=BHOX→=AO→.\sum_{X=B}^{H}\overrightarrow{OX}=\overrightarrow{AO}.X=B∑H​OX=AO.
  1. Substitute back

Hence,

∑X=BHAX→=7AO→+AO→=8AO→.\sum_{X=B}^{H}\overrightarrow{AX}=7\overrightarrow{AO}+\overrightarrow{AO}=8\overrightarrow{AO}.X=B∑H​AX=7AO+AO=8AO.

Now use the given value of AO→\overrightarrow{AO}AO:

8AO→=8(2i^+3j^−4k^)8\overrightarrow{AO}=8(2\hat i+3\hat j-4\hat k)8AO=8(2i^+3j^​−4k^) =16i^+24j^−32k^.=16\hat i+24\hat j-32\hat k.=16i^+24j^​−32k^.
  1. Match with the options

This is exactly

16i^+24j^−32k^,16\hat i+24\hat j-32\hat k,16i^+24j^​−32k^,

which is Option B.

Final Answer: 16i^+24j^−32k^\boxed{16\hat i+24\hat j-32\hat k}16i^+24j^​−32k^​ (Option B)

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