If origin is the ortho-center of an equilateral triangle whose vertices are $\bar{a}, \bar{b}, \bar{c}$ then

If origin is the ortho-center of an equilateral triangle whose vertices are $\bar{a}, \bar{b}, \bar{c}$ then
  1. $\bar{a}+\bar{b}=\bar{c}$
  2. $\bar{a}+\bar{b}=-\bar{c}$
  3. $|\bar{a}|^2=|\bar{b}|^2=|\bar{c}|^2$
  4. $\bar{a}=\bar{b}=\bar{c}$

Solution

Given that origin is orthocentre with slides of triangle are $\vec{a}, \vec{b}, \vec{c}$ $ \begin{array}{ll} \text { so, } \frac{\vec{a}+\vec{b}+\vec{c}}{3}=0 & \\ \vec{a}+\vec{b}+\vec{c}=0 & \begin{array}{l} \text { \{By orthocentre = circumcentre } \\ \text { for equilateral triangle \} } \end{array} \\ \vec{a}+\vec{b}=-\vec{c} & \end{array} $

Asked in: AP EAMCET 2022 (06 Jul Shift 1)

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