The position vectors of the vertices A, B and | vedclass

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(D) The position vectors of the vertices are $A(2, -3, 3)$,$B(2, 2, 3)$,and $C(-1, 1, 3)$.First,calculate the lengths of sides $AB$ and $AC$:$AB = \sq

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(D) The position vectors of the vertices are $A(2, -3, 3)$,$B(2, 2, 3)$,and $C(-1, 1, 3)$.First,calculate the lengths of sides $AB$ and $AC$:$AB = \sqrt{(2-2)^2 + (2 - (-3))^2 + (3-3)^2} = \sqrt{0^2 + 5^2 + 0^2} = 5$.$AC = \sqrt{(-1-2)^2 + (1 - (-3))^2 + (3-3)^2} = \sqrt{(-3)^2 + 4^2 + 0^2} = \sqrt{9 + 16} = 5$.Since $AB = AC = 5$,the triangle $ABC$ is an isosceles triangle.In an isosceles triangle,the angle bisector $AD$ of the vertex angle $\angle BAC$ is also the median to the base $BC$.Therefore,$D$ is the midpoint of $BC$.$D = \left( \frac{2 + (-1)}{2}, \frac{2 + 1}{2}, \frac{3 + 3}{2} ight) = \left( \frac{1}{2}, \frac{3}{2}, 3 ight)$.The length $l$ of the angle bisector $AD$ is the distance between $A(2, -3, 3)$ and $D\left(\frac{1}{2}, \frac{3}{2}, 3ight)$:$l = \sqrt{\left(2 - \frac{1}{2}ight)^2 + \left(-3 - \frac{3}{2}ight)^2 + (3 - 3)^2}$$l = \sqrt{\left(\frac{3}{2}ight)^2 + \left(-\frac{9}{2}ight)^2 + 0^2} = \sqrt{\frac{9}{4} + \frac{81}{4}} = \sqrt{\frac{90}{4}} = \sqrt{\frac{45}{2}}$.Thus,$l^2 = \frac{45}{2}$.Therefore,$2 l^2 = 2 	imes \frac{45}{2} = 45$.

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