The position vectors of the vertices A,B,C of a triangle are 2i^−3j^+3k^, 2i^+2j^+3k^, and −i^+j^+3k^ respectively. Let l denote the length of the angle bisector AD of ∠BAC, where D is on segment BC. Then 2l2 equals:
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Official solution
\[\vec{AB}=(0,5,0),\ |AB|=5;\qquad \vec{AC}=(-3,4,0),\ |AC|=5.\]
Since AB=AC, the bisector foot D is the midpoint of BC:
\[D=\left(\frac{2-1}{2},\frac{2+1}{2},3\right)=\left(\frac12,\frac32,3\right).\]
\[\vec{AD}=\left(-\frac32,\frac92,0\right)\ \left(\text{i.e. } D-A=\left(-\tfrac32,\tfrac92,0\right)\right).\]
\[l^2=\frac94+\frac{81}{4}=\frac{90}{4}=22.5,\qquad 2l^2=45.\]