If A(0,3,4), B(1,5,6), C(-2,0,-2) are the ver | vedclass

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(B) First,calculate the lengths of sides $AB$ and $AC$ using the distance formula: $AB = \sqrt{(1-0)^2 + (5-3)^2 + (6-4)^2} = \sqrt{1^2 + 2^2 + 2^2} =

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$\frac{\sqrt{42}}{10}$

Vedclass · Answer

(B) First,calculate the lengths of sides $AB$ and $AC$ using the distance formula: $AB = \sqrt{(1-0)^2 + (5-3)^2 + (6-4)^2} = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3$. $AC = \sqrt{(-2-0)^2 + (0-3)^2 + (-2-4)^2} = \sqrt{(-2)^2 + (-3)^2 + (-6)^2} = \sqrt{4+9+36} = \sqrt{49} = 7$. By the Angle Bisector Theorem,the bisector of $\angle A$ divides the side $BC$ in the ratio $AB:AC = 3:7$. Thus,the coordinates of $D$ are given by the section formula: $D = \left( \frac{3(-2) + 7(1)}{3+7}, \frac{3(0) + 7(5)}{3+7}, \frac{3(-2) + 7(6)}{3+7} \right) = \left( \frac{-6+7}{10}, \frac{0+35}{10}, \frac{-6+42}{10} \right) = \left( \frac{1}{10}, \frac{35}{10}, \frac{36}{10} \right) = (0.1, 3.5, 3.6)$. Now,calculate the length $AD$: $AD = \sqrt{(0.1-0)^2 + (3.5-3)^2 + (3.6-4)^2} = \sqrt{(0.1)^2 + (0.5)^2 + (-0.4)^2} = \sqrt{0.01 + 0.25 + 0.16} = \sqrt{0.42} = \sqrt{\frac{42}{100}} = \frac{\sqrt{42}}{10}$.

If $A(0,3,4), B(1,5,6), C(-2,0,-2)$ are the vertices of a triangle $ABC$ and the bisector of angle $A$ meets the side $BC$ at $D$,then $AD=$

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