Given position vectors are
OA=a=2+3+4 OB=b=3+4+2 OC=c=4+4+2 OC=c=4+2+3 ∴|AB|=√12+12+22=√6 |BC|=√12+22+12=√6 and,
|CA|=√22+12+12=√6 Therefore,
△ABC is an equilateral triangle.
B. Given position vectors are
OA=a=+2+3 OB=b=3+4+7 OC=c=3−2−5 |AB|=√4+4+16=√24=2√6
|BC|=√36+36+144=216=6√6
and
|CA|=√16+16+64=√96=4√6
∵|AB|+|CA|=|BC|⇒A,B,C
are collinear points.
C. Given position vectors are
OA=a=2−+ OB=b=−3−5 and,
OC=c=−3−4−4 ∴|AB|=√1+4+36=√41 |BC|=√16+1+1=√18 and,
|CA|=√25+9+25=√59 ∵|CA|2=|AB|2+∣BC2=∆ABC
is right angled triangle.
D. Given position vectors,
OA=a=++ OB=b=+2+3 and,
OC=C=2−+ ∴|AB|=√0+1+4=√5,|BC|=√1+9+4=√14
and
|AC|=√1+4+0=√5 ∵|AB|=|BC| and
|AB|+|AC|>|BC| ∴△ABC is isosceles triangle.