1 Answer
Solution :-
Given:
The vertices of a triangle are A (1, 2, 3), B (0, 4, 1), C (-1, -1, -3)
By using the distance formula,
The distance between A(1, 2, 3) and B (0, 4, 1) is AB
\( \sqrt{(1-0)^2+(2-4)^2+(3-1)^2}\)
\( \sqrt{9}=3\)
The distance between A(1, 2, 3) and C(-1, -1, -3) is AC
\( \sqrt{(1-(-1))^2+(2-(-1))^2+(3-(-3))^2}\)
\( \sqrt{49}=7\)
So, \( \dfrac{AB}{AC}\) = \( \dfrac{3}{7}\)
AB: AC = 3:7
BD: DC = 3:7
Then, m = 3 and n = 7
B(0, 4, 1) and C(-1, -1, -3)
Coordinates of D by using section formula is given as
The coordinates of D are (\( \dfrac{-3}{10}\), \( \dfrac{5}{2}\), \( \dfrac{-1}{5}\)).
Answered by Aaryan | 1 year agoRelated Questions
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