|
|
let the line x/1=y-1/2=z-2/3=k
any point on this line is (k,2k+1,3k+2)
direction ratios of the line joining this point and (1,6,3) is
(k-1,2k-5,3k-1)
this line and the given line are perpendicular
therefore l1l2+m1m2+n1n2=0
1.(k-1)+2(2k-5)+3(3k-3)=0
solving we get k=1
the point on the line is 1,3,5
thi point is the midpoint of the reqd point and given point,
so put the section formula and the answer will be(1,0,7)
|
Moderation Team
![[Post New]](/templates/default/images/icon_minipost_new.gif){kind=icon}
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
54
