Saved Bookmarks
| 1. |
Let a=2hati+hat j and b=hat i+2hatj+hatk be two vectors. Consider a vector c=alphaa +betab,alpha,beta in R. If the projection of c on the vector (a+b) is 3sqrt(2), then the minimum value of (c-(axxb))).c equals........... |
|
Answer» and `b=hati+2hatj+hatk` so, `a+b=3hati+3hatjimplies|a+b|=3sqrt(2)` Since, it is given that projection of `C=alpha a+betab` on the vector `(a+b)` is `3sqrt(2)`, then `((a+b).c)/(|a+b|)=3sqrt(2)` `implies(a+b).(alpha a+betab)=18` `impliesalpha(a.a)+BETA(a.b)+alpha(b.a)+beta(a.b)=18` `implies6alpha+3beta+3alpha+6beta=18` `implies9alpha+9beta=18implies(alpha+beta)=2` . . (i) Now, for minimum value of `(c-(axxb)).c` `=(alpha a+betab-(axxb)).(alpha+betab)` `=alpha^(2)(a.a)+alphabeta(a.b)+alphabeta(a.b)+beta^(2)(b.b)""[because(axxb).a=0=(axxb).b]` `=6alpha^(2)+6alphabeta+6beta^(2)=6(alpha^(2)+beta^(2)+alphabeta)` `=6[(alpha+beta)^(2)-alphabeta]=6[4-alphabeta]=6[4-alpha(2-alpha)]` `=6[4-alpha+alpha^(2)]` the minimum value of `6(4-2alpha+alpha^(2))=6(3)=18` [As minimum value of `ax^(2)+bx+c=-(D)/(4a')`, if `a GT 0`] |
|