Question:

\(3\mathbf{\overline{i}} - 2\mathbf{\overline{j}} - \mathbf{\overline{k}}, -2\mathbf{\overline{i}} - \mathbf{\overline{j}} + 3\mathbf{\overline{k}}, -\mathbf{\overline{i}} + 3\mathbf{\overline{j}} - 2\mathbf{\overline{k}}\) are the position vectors of the vertices \( A \), \( B \), and \( C \) of a triangle \( ABC \)respectively. If \( H \) is its orthocenter, then find \( \overline{HA} + \overline{HB} + \overline{HC} \).

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For any triangle, the sum of the vectors from the orthocenter to the vertices is always zero.
Updated On: May 15, 2025
  • \( 2 \overline{SA} \)
  • \( \overline{0} \)
  • \( 2 \overline{AB} \)
  • \( \mathbf{\overline{i}} + \mathbf{\overline{j}} + \mathbf{\overline{k}} \)
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The Correct Option is B

Solution and Explanation

In geometry, the orthocenter of a triangle is the point of intersection of the altitudes of the triangle. One important property of the orthocenter is that the sum of the vectors from the orthocenter \( H \) to the vertices of the triangle is always zero. Specifically, for a triangle \( ABC \), we have the following relation: \[ \overline{HA} + \overline{HB} + \overline{HC} = \overline{0}. \] This property holds true for any triangle, as long as the point \( H \) is the orthocenter.
Thus, the correct answer is \( \boxed{\overline{0}} \).
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