Vector Algebra
Orthocenter of Triangle
Grade 12
Question:
<p>Let position vectors of points A, B and C of triangle ABC respectively be <p>\(\vec{i} + \vec{j} + 2\vec{k}\), \(\vec{i} + 2\vec{j} + \vec{k}\) and \(2\vec{i} + \vec{j} + \vec{k}\)</p>. Let \(l_1\), \(l_2\) and \(l_3\) be the lengths of perpendiculars drawn from the orthocenter 'O' on the sides AB, BC and CA, then \((l_1 + l_2 + l_3)\) equals</p>
<p>(a) \(\frac{2}{\sqrt{6}}\)</p>
<p>(b) \(\frac{3}{\sqrt{6}}\)</p>
<p>(c) \(\frac{\sqrt{6}}{2}\)</p>
<p>(d) \(\frac{\sqrt{6}}{3}\)</p>
Step-by-Step Solution
Key Concept: For an equilateral triangle, the orthocenter coincides with the centroid, and the distance from orthocenter to any side equals the inradius.
Solution: Clearly, the triangle formed by the given points \(\hat{i} + \hat{j} + 2\hat{k}\), \(\hat{i} + 2\hat{j} + \hat{k}\) and \(2\hat{i} + \hat{j} + \hat{k}\) is equilateral as \(AB = BC = AC = \sqrt{2}\). ∴ Distance of orthocenter 'O' from the sides is equal to the inradius of the triangle. For an equilateral triangle with side \(a = \sqrt{2}\), the inradius \(r = \frac{a}{2\sqrt{3}} = \frac{\sqrt{2}}{2\sqrt{3}} = \frac{1}{\sqrt{6}}\) Therefore, \(l_1 + l_2 + l_3 = 3r = \frac{3}{\sqrt{6}} = \frac{\sqrt{6}}{2}\) ∴ Answer is (c).
Correct Answer: C