If A-O- - O-B-B-O- - O-C- then

It is given that A⃗O⃗ + O⃗B⃗=B⃗O⃗ + O⃗C⃗. Let’s take O as the origin. A⃗O⃗= O⃗A⃗ = position vector of point A B⃗O⃗= O⃗B⃗ = position vector of point B So A⃗O⃗ ...

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Byju's Answer

Standard XII

Mathematics

Test for Collinearity of Points

If A⃗O⃗ + O⃗B...

Question

If $\to A O + \to O B = \to B O + \to O C$ , then

A, B and C are collinear

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A, B and C are not collinear

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Nothing can be said about Collinearity from this information

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\to A, \to B, \to C

form a triangle

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Solution

The correct option is A A, B and C are collinear
It is given that $\to A O + \to O B = \to B O + \to O C$ .
Let’s take O as the origin.
$\to A O = - \to O A$ = -position vector of point A
$\to B O = - \to O B$ = -position vector of point B
So $\to A O + \to O B = \to B O + \to O C$
$\Rightarrow - \to O A + \to O B = - \to O B + \to O C$
$\to O B - \to O A = \to O C - \to O B$
$\to O B - \to O A = \to A B$
$\to O C - \to O B = \to B C$ [Using triangle law of addition]
So $\to A B = \to B C$
$\Rightarrow \to A B i s p a r a l l e l t o \to B C$ .
But $\to A B$ and $\to B C$ have a point $\to B$ in common.
$\Rightarrow \to A B a n d \to A C a r e s a m e l i n e s$
$\Rightarrow \to A B a n d \to A C a r e c o l l i n e a r v e c t o r s$
$\Rightarrow A, B a n d C a r e c o l l i n e a r$ .

Suggest Corrections

Similar questions

Q. If

\to A O + \to O B = \to B O + \to O C

, then

A, B

and

C

are (where

O

is the origin)

Q. Assertion :The resultant of the three vector

\to O A, \to O B

and

\to O C

as shown in the figure is

R (1 + \sqrt{2})

.Here

R

is the radius of the circle.(Here

∠ A O B = ∠ B O C = 45^{\circ}

) Reason:

\to O A + \to O C

is along

\to O B

and

(\to O A + \to O C) + \to O B

is along

\to O B

Q. If

A B C D

is a rhombus whose diagonals cut at the origin

O

, then

\to O A + \to O B + \to O C + \to O D

equals to

\to O A = 3 i + 4 j + 12 k, \to O B = 12 i + 3 j + 4 k,

\to O C = 4 i + 12 j + 3 k .

Find the angle between

\to O A

and

\to O C .

Four vertices O, A, B, C of a tetrahedron satisfy

⎡ ⎢ ⎢ ⎣ \begin{matrix} \to O A \times \to O B =^i -^j +^k \to O B \times \to O C =^i \to O C \times \to O A = -^i +^j \end{matrix} ⎤ ⎥ ⎥ ⎦

where 'O' is the the origin.

Then

∣ ∣ \to C A \times \to C B ∣ ∣

has the value equal to

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If →AO+→OB=→BO+→OC, then

If $\to A O + \to O B = \to B O + \to O C$ , then