Since the positions of the points #A & B,# w.r.t. the Origin #O# are
#i-7j# and #4i+kj,# we find that, the position vectors of the pts.
#A & B# are given by,
#vec(OA)=i-7j, and, vec(OB)=4i+kj, k# is a scalar.
Then, #vec(AB)=4i+kj-(i-7j)=3i+(k+7)j#
Hence, the unit vector along #vec(AB)=hat(AB)=(vec(AB))/|AB|,#
where, #|AB|=sqrt{3^2+(k+7)^2}=sqrt{9+(k+7)^2}#
#:. hat(AB)=(3i)/sqrt{9+(k+7)^2}+((k+7)j)/sqrt{9+(k+7)^2}...(1)#.
But, #hat(AB)=(6i)/10+(8j)/10.................(2)....(given)#.
Comparing #(1) and (2),# we get,
#3/sqrt{9+(k+7)^2}=6/10, &, (k+7)/sqrt{9+(k+7)^2}=8/10#
Dividing the last #2^(nd)# eqn. by the #1^(st)#, we have,
#(k+7)/3=8/6=4/3 rArr k=-3#.
Alternatively,
From the last #1^(st) eqn., 9+(k+7)^2=(((3)(10))/6)^2=25#
#rArr (k+7)^2=16 rArr k+7=+-4 rArr k=+-4-7#,
i.e., #k=-3, or, -11#
But, #k=-11, and, (1)," gives "hat(AB)=3/5i-4/5j=6/10i-8/10j,#
which contradicts the data, &, hence, #k=-11# is not admissible.
Therefore, #k=-3#.
Enjoy Maths!
