The values of a, for which points A,B,C with position vectors 2i^−j^+k^, i^−3j^−5k^ and ai^−3j^+k^ respectively are the vertices of a right angled triangle with C=2π, are:
Solutions · 1
Official solution
CA=(2−a)i^+2j^, CB=(1−a)i^−6k^. Since CA⊥CB:
\[\overrightarrow{CA}\cdot\overrightarrow{CB}=0 \implies (2-a)(1-a)=0 \implies a=2,1\]