Let \vec{P} is the p.v of the orthocentre and \vec{g} is the p.v of the centroid of the triangle ABC, where circumcentre is the origin. If \vec{P}= K\vec{g}\: then \: K

  • Option 1)

    3

  • Option 2)

    2

  • Option 3)

    \frac{1}{3}

  • Option 4)

    \frac{2}{3}

 

Answers (1)
V Vakul

Use the concept of

Position vector -

Let O be a fixed origin, then position vector of P is \overrightarrow{OP}

- wherein

 

 

Centroid is intersection is \frac{x_{1}+x_{2}+x_{3}}{3} , \frac{y_{1}+y_{2}+y_{3}}{3}

orthocentre: intersection of altitudes.

so that in a triangle ABC the orthocentre H, centroid G and circumcentre M are collinear ang G divides HM internally in the ration 2:1

\therefore g=\frac{2\times M+1\times P}{3}

\Rightarrow p=3g

\therefore k=3                 \left [ \because M=\left ( 0,0,0 \right ) \right ]


Option 1)

3

Correct option

Option 2)

2

Incorrect option

Option 3)

\frac{1}{3}

Incorrect option

Option 4)

\frac{2}{3}

Incorrect option

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