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$3 \hat{i}-2 \hat{j}-\hat{k},-2 \hat{i}-\hat{j}+3 \hat{k}$ and $-\hat{i}+3 \hat{j}-2 \hat{k}$ are the position vectors of the vertices $A, B$ and $C$ of a $\triangle A B C$ respectively.
If $\mathrm{H}$ is its orthocenter, then $\overrightarrow{\mathrm{HA}}+\overrightarrow{\mathrm{HB}}+\overrightarrow{\mathrm{HC}}=$
Options:
If $\mathrm{H}$ is its orthocenter, then $\overrightarrow{\mathrm{HA}}+\overrightarrow{\mathrm{HB}}+\overrightarrow{\mathrm{HC}}=$
Solution:
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Verified Answer
The correct answer is:
$\overrightarrow{0}$
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 $\mathrm{H}(0,0,0)$
$$
\begin{aligned}
& \text { Hence } \overrightarrow{H A}+\overrightarrow{H B}+\overrightarrow{H C}=(3 \hat{i}-2 \hat{j}-\hat{k})+(-2 \hat{i}-\hat{j}+3 \hat{k}) \\
& +(-\hat{i}+3 \hat{j}-2 \hat{k}) \\
& =\overrightarrow{0}
\end{aligned}
$$
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 $\mathrm{H}(0,0,0)$
$$
\begin{aligned}
& \text { Hence } \overrightarrow{H A}+\overrightarrow{H B}+\overrightarrow{H C}=(3 \hat{i}-2 \hat{j}-\hat{k})+(-2 \hat{i}-\hat{j}+3 \hat{k}) \\
& +(-\hat{i}+3 \hat{j}-2 \hat{k}) \\
& =\overrightarrow{0}
\end{aligned}
$$
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