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If the vertices a (1. - 1,2), B (5, - 6,2) C (1,3, - 1) of △ ABC are known, then the length of high BD on the edge of AC is equal to
The direction vector of the line AC is AC = (0,4, - 3),
The direction vectors of AB and BC are ab = (4, - 5,0), BC = (- 4,9, - 3), respectively,
So the vector perpendicular to AB and BC is ab × BC = (15,12,16),
So the vector collinear with BD is n = AC × (AB × BC) = (100, - 45, - 60),
The height of edge AC is the absolute value of the projection of AB on n,
That is, H = | AB * n / | n | = | 400 + 225 | / √ (10000 + 2025 + 3600) = 5
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