AP EAMCET · Maths · Three Dimensional Geometry
If \(\mathrm{A}(0,1,2), \mathrm{B}(2,-1,3)\) and \(\mathrm{C}(1,-3,1)\) are the vertices of a triangle, then the distance between its circumcentre and orthocentre is
- A \(\frac{3}{\sqrt{2}}\)
- B \(\frac{3}{2}\)
- C \(3\)
- D \(\frac{9}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{3}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\( \vec{AB} = (2, -2, 1), \vec{BC} = (-1, -2, -2) \) \( |\vec{AB}|^2 = 2^2 + (-2)^2 + 1^2 = 9 \) \( |\vec{BC}|^2 = (-1)^2 + (-2)^2 + (-2)^2 = 9 \) \( \vec{AB} \cdot \vec{BC} = (2)(-1) + (-2)(-2) + (1)(-2) = -2 + 4 - 2 = 0 \) Triangle ABC is right-angled at B. Orthocentre…
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