i) ABC are in straight line, then are linearly dependents vec(AB)= kvec(BC)−−→AB=k−−→BC
Calculate vec(AB)=(2-0,5-2,-2+3)=(2,3,1)−−→AB=(2−0,5−2,−2+3)=(2,3,1) and
vec(BC)=(1,p-5,q+2)−−→BC=(1,p−5,q+2). Then
(2,3,1)=k(1,p-5,q+2)=(k,kp-5k,kq+2k)(2,3,1)=k(1,p−5,q+2)=(k,kp−5k,kq+2k) from here
2=k2=k
3=kp-5k=2p-10=3=kp−5k=2p−10= then p=13/2p=132
1=q+21=q+2 then q=-1q=−1
ii) If BAC is 90º, then vec(AB)·vec(AC)=0−−→AB⋅−−→AC=0
vec(AB)=(2,3,1)−−→AB=(2,3,1)
vec(AC)=(3,p-2,q+3)−−→AC=(3,p−2,q+3)
vec(AB)·vec(AC)=6+3p-6+q+3=3p+q+3=0−−→AB⋅−−→AC=6+3p−6+q+3=3p+q+3=0 then
q=-3-3pq=−3−3p
iii)If p=3p=3 the modvec(AB)=sqrt(4+9+1)=sqrt14−−→AB=√4+9+1=√14
vec(AC)=(3,1,q+3)−−→AC=(3,1,q+3) then mod vec(AC)=sqrt(9+1+(q+3)^2)−−→AC=√9+1+(q+3)2
Then squaring both sides 14=10+(q+3)^2=10+q^2+6q+414=10+(q+3)2=10+q2+6q+4 or
q^2+6q=q(q+6)=0q2+6q=q(q+6)=0 which give q=0q=0 and q=-6q=−6