Let be three vector of magnitude  respectively, satisfying  If   ,  then the value of k.

2081 Points
3 years ago
hello manohar, consider the following:
[a b c]= a.(b*c)= a.(|b||c|sin(alpha) n cap)= |b||c|sin(alpha)a.(n cap) (here n cap is unit vector perpendicular to both b and c)
= |b||c|sin(alpha)|a||n cap|.cos(beta)
but |n cap|= 1
so that |[a b c]|= |a||b||c| |sin(alpha)cos(beta)| whose max value is |a||b||c| (this is a well known property but i still provided the proof for clarity sake).
now, in our case |a|, |b|, |c| are 1, 1/root2, root2 so that max value of |[a b c]| is |a||b||c|= 1. however, we r also given that |[a b c]|= 1, so this max only occurs when |sin(alpha)|= |cos(beta)| = 1
so alpha = ± 90 deg so that b and c are perpendicular to each other. also, beta= 0 or 180 deg so that a is perpendicular to both b and c. this means that a b and c form a perpendicular system of vectors. so a.b= b.c= c.a= 0
now, as regards the monstrous term Q= (2a+b+c).(((a*c)*(a-c)) + b), let a*c= mb where m is real, and we have already established that a*c shall be linear with b.
also, write 2a+b+c= 2(a – c) + b + 3c and let a – c= v
so, Q= (2v+b+3c).(mb*v + b)= 2[v mb v] + 2v.b + [b mb v] + |b|^2 + 3[c mb v] + 3c.b
now we use the property of scalar triple products to simplify this expression
Q= 2*0 + 2(a.b – c.b) + 0 + ½ + 3[c mb v] + 3*0
= 0 + 0 + 0 + ½ + 3[c mb v] + 0
= ½ + 3[mb v c]
= ½ + 3mb.(v*c)
= ½ + 3(a*c).((a – c)*c)
= ½ + 3(a*c).(a*c – 0)
= ½ + 3(a*c).(a*c)
= ½ + 3|a*c|^2
= ½ + 3|a|^2|c|^2(sin±90)^2
= ½ + 3*1*2*1
= 13/2= 13/k
so k = 2
kindly approve :))