Question

The position vector of vertices of Triangle ABC are i , j, and k then find the position vector of its orthocenter. ​

Answer 1

Answer:

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

Step-by-step explanation:

❒ We have, the position vectors of the vertices of a triangle ABC are $\hat{i}$ $\hat{j}$ $\hat{k}$

❒ Let O be the fixed point

$\hat{i} = position\:\:vector\:\: of\: A = \vec{OA}$

$\hat{j} = position \: \:vector \: \: of \: \: B = \vec{OB}$

$\hat{k} = position\:\: vector\:\: of\:\: C = \vec{OC}$

❒ Let AD be the median of the triangle ABC.

❒ Since, D is the median point of BC.

❒ $\bold{Position\:of\:vector\:D = vector\:OD=} \frac{\vec{OB}\:+\:\vec{OC}}{2}$

❒ Now, position vector of G

$\hookrightarrow \frac{2(OD)+1(OA)}{3}$

$\hookrightarrow \frac{2(\frac{OB+OC}{2})+OA }{3}$

$\hookrightarrow \frac{\vec{OA} +\vec{OB}+\vec{OC}}{3}$

So,

$\hookrightarrow \frac{\hat{i}\:+\:\hat{j}\:+\:\hat{k}}{3}$

Hence, the position vector of the ortho centre is  $\frac{1}{3} (\hat{i}+\hat{j}+\hat{k})$