Consider a triangle ABC where A(1,3,2),B(-2,8 | vedclass

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(A) Given vertices are $A(1,3,2)$,$B(-2,8,0)$,and $C(3,6,7)$.First,calculate the lengths of sides $AB$ and $AC$:$AB = \sqrt{(-2-1)^2 + (8-3)^2 + (0-2)

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$\frac{37}{2 \sqrt{38}}$

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(A) Given vertices are $A(1,3,2)$,$B(-2,8,0)$,and $C(3,6,7)$.First,calculate the lengths of sides $AB$ and $AC$:$AB = \sqrt{(-2-1)^2 + (8-3)^2 + (0-2)^2} = \sqrt{(-3)^2 + 5^2 + (-2)^2} = \sqrt{9 + 25 + 4} = \sqrt{38}$.$AC = \sqrt{(3-1)^2 + (6-3)^2 + (7-2)^2} = \sqrt{2^2 + 3^2 + 5^2} = \sqrt{4 + 9 + 25} = \sqrt{38}$.Since $AB = AC$,the triangle $ABC$ is isosceles,and the angle bisector $AD$ of $\angle BAC$ is also the median to $BC$. Thus,$D$ is the midpoint of $BC$.$D = \left( \frac{-2+3}{2}, \frac{8+6}{2}, \frac{0+7}{2} \right) = \left( \frac{1}{2}, 7, \frac{7}{2} \right)$.Now,find the vector $\overrightarrow{AD}$:$\overrightarrow{AD} = \left( \frac{1}{2}-1 \right) \hat{i} + (7-3) \hat{j} + \left( \frac{7}{2}-2 \right) \hat{k} = -\frac{1}{2} \hat{i} + 4 \hat{j} + \frac{3}{2} \hat{k}$.Find the vector $\overrightarrow{AC}$:$\overrightarrow{AC} = (3-1) \hat{i} + (6-3) \hat{j} + (7-2) \hat{k} = 2 \hat{i} + 3 \hat{j} + 5 \hat{k}$.The length of the projection of $\overrightarrow{AD}$ on $\overrightarrow{AC}$ is given by $\left| \frac{\overrightarrow{AD} \cdot \overrightarrow{AC}}{|\overrightarrow{AC}|} \right|$.$\overrightarrow{AD} \cdot \overrightarrow{AC} = \left( -\frac{1}{2} \right)(2) + (4)(3) + \left( \frac{3}{2} \right)(5) = -1 + 12 + 7.5 = 18.5 = \frac{37}{2}$.$|\overrightarrow{AC}| = \sqrt{2^2 + 3^2 + 5^2} = \sqrt{38}$.Projection length $= \left| \frac{37/2}{\sqrt{38}} \right| = \frac{37}{2 \sqrt{38}}$.

Consider a $\triangle ABC$ where $A(1,3,2)$,$B(-2,8,0)$,and $C(3,6,7)$. If the angle bisector of $\angle BAC$ meets the line $BC$ at $D$,then the length of the projection of the vector $\overrightarrow{AD}$ on the vector $\overrightarrow{AC}$ is:

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