The direction cosines of a vector `hati+hatj+sqrt(2) hatk` are:-

Answer

(i) `B_(x)=6m, B_(y)=3m`, (ii) `sqrt(45)m` (iii) `tan^(-1)(1/2)`

Explanation

`:' A_(x)=4, A_(y)=6` so `A_(x)+B_(x)=10` and `A_(y)+B_(y)=9`
(i) `B_(x)=10-4=6m` and `B_(y)=9-6=3 m`
(ii) length `=sqrt(B_(x)^(2)+B_(y)^(2))=sqrt(36+9)=sqrt(45) m`
(iii) `theta=tan^(-1) (B_(y)/B_(x))=tan^(-1)(3/6)=tan^(-1)(1/2)`