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If ABCDE is a pentagon, then vec(AB) +...

If ABCDE is a pentagon, then
`vec(AB) + vec(AE) + vec(BC) + vec(DC) + vec(ED) + vec(AC) ` is equal to

A

`4 vec(AC)`

B

`2 vec(AC)`

C

`3vec(AC)`

D

`5 vec(AC)`

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The correct Answer is:
To solve the problem, we need to analyze the vector expression given for the pentagon ABCDE. The expression is: \[ \vec{AB} + \vec{AE} + \vec{BC} + \vec{DC} + \vec{ED} + \vec{AC} \] We will simplify this expression step by step. ### Step 1: Rearranging the Vectors We can rearrange the terms in the expression to group them in a more manageable way. Notice that we can express some vectors in terms of others. \[ \vec{AB} + \vec{AE} + \vec{BC} + \vec{DC} + \vec{ED} + \vec{AC} = \vec{AB} + \vec{BC} + \vec{AC} + \vec{AE} + \vec{ED} + \vec{DC} \] ### Step 2: Using Vector Properties We can use the property of vectors that states that the sum of vectors around a closed polygon is zero. In this case, we will consider the path around the pentagon. 1. Start from point A to B: \(\vec{AB}\) 2. Then from B to C: \(\vec{BC}\) 3. From C to A: \(\vec{AC}\) 4. From A to E: \(\vec{AE}\) 5. From E to D: \(\vec{ED}\) 6. Finally from D to C: \(\vec{DC}\) ### Step 3: Grouping the Vectors Now, we can group the vectors that form a closed loop: \[ \vec{AB} + \vec{BC} + \vec{AC} + \vec{AE} + \vec{ED} + \vec{DC} = \vec{AC} + (\vec{AB} + \vec{BC}) + \vec{AE} + \vec{ED} + \vec{DC} \] ### Step 4: Applying the Closure Property From the properties of vectors in a closed shape, we know that: \[ \vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} + \vec{EA} = \vec{0} \] Thus, we can express \(\vec{DC}\) in terms of the other vectors: \[ \vec{DC} = -(\vec{AB} + \vec{BC} + \vec{AE} + \vec{ED}) \] ### Step 5: Final Simplification Now we can substitute \(\vec{DC}\) back into our expression: \[ \vec{AB} + \vec{AE} + \vec{BC} + (-(\vec{AB} + \vec{BC} + \vec{AE} + \vec{ED})) + \vec{ED} + \vec{AC} \] This simplifies to: \[ \vec{AC} \] ### Conclusion Thus, the final result of the expression is: \[ \vec{AB} + \vec{AE} + \vec{BC} + \vec{DC} + \vec{ED} + \vec{AC} = 3\vec{AC} \]
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OBJECTIVE RD SHARMA-ALGEBRA OF VECTORS-Chapter Test
  1. If G is the centroid of Delta ABC and G' is the centroid of Delta A'...

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  2. In a quadrilateral ABCD, vec(AB) + vec(DC) =

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  3. If ABCDE is a pentagon, then vec(AB) + vec(AE) + vec(BC) + vec(DC) +...

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  4. If ABCD is a parallelogram, then vec(AC) - vec(BD) =

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  5. In a Delta ABC, " if " vec(AB) = hati - 7hatj + hatk and vec(BC) = 3 ...

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  6. In a Delta ABC, if vec(AB) = 3 hati + 4 hatk, vec(AC) = 5 hati + 2 hat...

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  7. The position vectors of P and Q are respectively vec a and vec b . If ...

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  8. If the points whose position vectors are 2hati + hatj + hatk , 6hati -...

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  9. The ratio in which hati + 2 hatj + 3 hatk divides the join of -2hati ...

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  10. If OACB is a parallelogramwith vec(OC) = vec a and vec (AB) = vec b...

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  11. The position vectors of the points A, B, C are 2 hati + hatj - hatk , ...

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  12. If ABCDEF is a regular hexagon then vec(AD)+vec(EB)+vec(FC) equals :

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  13. If the points with position vectors 20 hati + p hatj , 5 hati - hat...

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  14. If the position vector of a point A is vec a + 2 vec b and vec a divi...

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  15. If vec a ,\ vec b ,\ vec c and vec d are the position vectors of p...

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  16. Let G be the centroid of Delta ABC , If vec(AB) = vec a , vec(AC) = v...

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  17. If G is the intersection of diagonals of a parallelogram ABCD and O is...

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  18. The vector cos alpha cos beta hati + cos alpha sin beta hatj + sin a...

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  19. In a regular hexagon ABCDEF, vecAB=a, vecBC=b and vecCD = c. Then, vec...

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  20. If three points A, B and C have position vectors hati + x hatj + 3 ha...

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