if A(5,4,6)B(1,-1,3)and C(4,3,2) are the vertices of a triangle and the internal bisector of <BAC meets BC at D find the coordinates of D. Also find the coordinates of D. also find the length of AD
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Please find below the solution to the asked query:

Let ABC be a triangle and ADis the angle bisector.Applying angle bisector we get:ABAC=BDCDA5,4,6, B1,-1,3 and C4,3,2Applying section formula we getAB=1-52+-1-42+3-62=16+25+9=50=52 unitsAC=4-52+3-42+2-62=1+1+16=32 units ABAC=BDCD=5232BDCD=53Thus, point Ddivides the line segment joining the points B1,-1,3 and C4,3,2in the ratio 5:3Let the coordinates of D be Dx,y,zHence by section formula we get;x=mx2+nx1m+n=5×4+3×15+3=238y=my2+ny1m+n=5×3-3×15+3=128=32z=mz2+nz1m+n=5×2+3×35+3=198Hence Coordinates of Dare D238,32,198Length of ADwill be given byAD=5-238+4-322+6-1982=40-238+8-322+48-1982=1782+522+2982=18289+400+841=181530Hence AD=181530 units

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