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If |vec (AO) +vec (OB)| =|vec(BO) + vec...

If `|vec (AO) +vec (OB)| =|vec(BO) + vec(OC)|` , then `A, B, C` form

A

equilateral triangle

B

collinear

C

non-collinear

D

none of these

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To solve the problem, we start with the given equation: \[ |\vec{AO} + \vec{OB}| = |\vec{BO} + \vec{OC}| \] ### Step 1: Rewrite the vectors We can express the vectors in terms of the position vectors of points A, B, C, and O. Let: - \(\vec{A}\) be the position vector of point A, - \(\vec{B}\) be the position vector of point B, - \(\vec{C}\) be the position vector of point C, - \(\vec{O}\) be the position vector of point O. Then we can rewrite the vectors as: - \(\vec{AO} = \vec{O} - \vec{A}\) - \(\vec{OB} = \vec{B} - \vec{O}\) - \(\vec{BO} = \vec{O} - \vec{B}\) - \(\vec{OC} = \vec{C} - \vec{O}\) ### Step 2: Substitute into the equation Now substituting these into the equation gives us: \[ |(\vec{O} - \vec{A}) + (\vec{B} - \vec{O})| = |(\vec{O} - \vec{B}) + (\vec{C} - \vec{O})| \] This simplifies to: \[ |-\vec{A} + \vec{B}| = |-\vec{B} + \vec{C}| \] Which can be rewritten as: \[ |\vec{B} - \vec{A}| = |\vec{C} - \vec{B}| \] ### Step 3: Interpret the equation The equation \( |\vec{B} - \vec{A}| = |\vec{C} - \vec{B}| \) implies that the distance between points A and B is equal to the distance between points B and C. ### Step 4: Conclusion about the points Since the distances are equal, points A, B, and C must be positioned such that they are collinear. This means they lie on a straight line. ### Final Answer Thus, the points A, B, and C form a collinear arrangement.

To solve the problem, we start with the given equation: \[ |\vec{AO} + \vec{OB}| = |\vec{BO} + \vec{OC}| \] ### Step 1: Rewrite the vectors We can express the vectors in terms of the position vectors of points A, B, C, and O. Let: ...
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OBJECTIVE RD SHARMA ENGLISH-ALGEBRA OF VECTORS-Chapter Test
  1. If |vec (AO) +vec (OB)| =|vec(BO) + vec(OC)| , then A, B, C form

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  2. If the vectors vec a =2hati + 3hatj +6hatk and vec b are collinear and...

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  3. If vec a , vec b , vec c are three non-zero vectors (no two of which ...

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  4. Vectors vec aa n d vec b are non-collinear. Find for what value of ...

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  5. If the diagonals of a parallelogram are 3 hati + hatj -2hatk and hati ...

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  6. If ABCD is a quadrilateral, then vec(BA) + vec(BC)+vec(CD) + vec(DA)=

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  7. The points with position vectors 60hati+3hatj,40hati-8hatj, ahati-52ha...

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  8. If ABCDEF is a regualr hexagon, then vec(AC) + vec(AD) + vec(EA) + ve...

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  9. In a regular hexagon ABCDEF, vec(AB)+vec(AC)+vec(AD)+vec(AE)+vec(AF)=k...

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  10. If P, Q , R are the mid-points of the sides AB, BC and CA of Delta AB...

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  11. If G is the centroid of the DeltaABC and if G' is the centroid of anot...

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

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

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

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

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  16. If vectors vec(AB) = -3hati+ 4hatk and vec(AC) = 5hati -2hatj+4hatk ar...

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

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

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

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  20. If OACB is a parallelogrma with vec( OC) = vec(a) and vec( AB) = vec(...

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

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