$3 \hat{i}-2 \hat{j}-\hat{k},-2 \hat{i}-\hat{j}+3 \hat{k}$ and $-\hat{i}+3 \hat{j}-2 \hat{k}$ are the…

1. $2 \overrightarrow{\mathrm{SA}}$
2. $\overrightarrow{0}$
3. $2 \overrightarrow{\mathrm{AB}}$
4. $\hat{i}+\hat{j}+\hat{k}$

Solution

Let $H(\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k})$ be the orthocentre of the $\triangle \mathrm{ABC}$. Since $\overrightarrow{A H} \cdot \overrightarrow{B C}=0$ $\{\because \overrightarrow{A H} \perp \overrightarrow{B C}\}$ $\Rightarrow \alpha+4 \beta-5 \gamma=0$...(1)
$ \begin{aligned} & \text { Since } B H \perp \overrightarrow{A C} \Rightarrow \overrightarrow{B H} \cdot \overrightarrow{A C}=0 \\ & \Rightarrow 5 \alpha-\beta-4 \gamma=0 ...(2)\end{aligned} $ Since $\overrightarrow{C H} \perp \overrightarrow{A B}$ Hence $\overrightarrow{C H} \cdot \overrightarrow{A B}=0$ $ \Rightarrow 5 \alpha-\beta-4 \gamma=0...(3) $ Solving equation (1), (2) \& (3) we get $ \alpha=\beta=\gamma=0 $ Hence orthocentre is $\mathrm{H}(0,0,0)$ $ \begin{aligned} & \text { Hence } \overrightarrow{H A}+\overrightarrow{H B}+\overrightarrow{H C}=(3 \hat{i}-2 \hat{j}-\hat{k})+(-2 \hat{i}-\hat{j}+3 \hat{k}) \\ & +(-\hat{i}+3 \hat{j}-2 \hat{k}) \\ & =\overrightarrow{0} \end{aligned} $

Asked in: AP EAMCET 2023 (15 May Shift 1)

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