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Centroid

The points A,...

Question

The points A, B and C with position vectors $\to a, \to b a n d \to c$ , and respectively lie on a circle centred at origin O. Let G and E be the centroid of $Δ$ ABC and $Δ$ ACD respectively where D is midpoint of AB.
If GE and OC are mutually perpendicular then orthocentre of $Δ A B C$ must lie on

A

median through A

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B

median through C

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C

angle bisector through A

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D

angle bisector through B

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Solution

The correct option is B median through C
$\to G E . \to O C = 0 \Rightarrow (\frac{3 \to a + \to b + 2 \to c}{6} - \frac{\to a + \to b + \to c}{3}) . \to c = 0$
On simplification; $(\to a - \to b) = 0 \Rightarrow \to A B ⊥ \to O C$
$\Rightarrow Δ A B C$ must be isosceles with base AB
$\Rightarrow$ Circumcentre and centroid lie on median through C
$\Rightarrow$ Orthocentre also lies on median through C

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Similar questions

Q. The points A, B and C with position vectors

\to a, \to b a n d \to c

, and respectively lie on a circle centred at origin O. Let G and E be the centroid of

Δ

Δ

ACD respectively where D is midpoint of AB.
If OE and CD are mutually perpendicular, then which of the following will be necessarily true?

Q. The position vectors of vertices

A, B, C

Δ A B C

\to a, \to b, \to c

respectively and

| \to c - \to a | = 3 | \to b - \to a |

C D

is drawn perpendicular to the bisector of

∠ A

, then which of the following is correct?

Q. Let the position vectors of the points

A, B

C

\to a, \to b

\to c

, respectively. Let

Q

be the point of intersection of the medians of the

Δ A B C

\to Q A + \to Q B + \to Q C

A B C

be a triangle whose circumcentre is at P. If the position vectors of

A, B, C

\to a, \to b, \to c

\frac{\to a + \to b + \to c}{4}

respectively, then the position vector of the orthocentre of this triangle, is:

Δ A B C

A \equiv (\to a), B \equiv (\to b)

C = (\to c),

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then the angle between the medians

¯ ¯¯¯¯¯¯¯¯ ¯ A M

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