The position vector of the vertices of a triangle A B C are \hat{i} , \hat{j} , \hat{k} then the position vector of its orthocentre is

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Question

The position vector of the vertices of a triangle

A B C

are

\hat{i}, \hat{j}, \hat{k}

then the position vector of its orthocentre is

\hat{i} + \hat{j} + \hat{k}

2 (\hat{i} + \hat{j} + \hat{k})

\frac{1}{3} (\hat{i} + \hat{j} + \hat{k})

\frac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k})

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We have, the position vectors of the vertices of a triangle

A B C

are

\hat{i}, \hat{j}, \hat{k}

Let

O

be the fixed point.

\hat{i}

= position vector of

A = \vec{O A}

\hat{j}

= position vector of

B = \vec{O B}

\hat{k}

= position vector of

C = \vec{O C}

Let

A D

be the median ofthe triangle

A B C

Since,

D

is the mis point of

B C

Position vector of

D =

vector

O D = \frac{(\vec{O B} + \vec{O C})}{2}

Now, position vector of

G

= \frac{2 (\vec{O D}) + 1 (\vec{O A})}{3}

= \frac{2 (\frac{\vec{O B} + \vec{O C}}{2}) + \vec{O A}}{3}

= \frac{\vec{O A} + \vec{O B} + \vec{O C}}{3}

So,

= \frac{\hat{i} + \hat{j} + \hat{k}}{3}

Hence, the position vector of the orthocentre is

\frac{1}{3} (\hat{i} + \hat{j} + \hat{k})

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