A vector ` vec r` has length 21 and its direction ratios are proportional to `2,-3,6.` Find the direction cosines and components of ` vec r` , is given that ` vec r` Makes an acute angle with `x-a xi sdot`

Explanation

Here, `|vecr|=14,veca=2k,vecb=3kandvecc=-6k`
:. Direction cosines l, m and n are
`l=(veca)/(|vecr|)=(2k)/(14)=(k)/(7)`
`m=(vecb)/(|vecr|)=(3k)/(14)`
and `n=(vecc)/(|vecr|)=(-6k)/(14)=(-3k)/(7)`
Also, we know that
`l^(2)+m^(2)+n^(2)=1`
`implies(k^(2))/(49)+(9k^(2))/(196)+(9k^(2))/(49)=1`
`implies(4k^(2)+9k^(2)+36^(2))/(196)=1`
`implies(4k^(2)+9k^(2)+36^(2))/(196)=1`
`impliesk=+-2`
So, the direction cosines (l, m, n) are `(2)/(7),(3)/(7)and(-6)/(7)`
[since, `vecr` makes an acute angle with X-axis]
`:'vecr=vecr.|vecr|`
`:.vecr=(lhati+mhatj+nhatk)|vecr|`
`=((+2)/(7)hati+(3)/(7)hatj-(6)/(7)hatk).14`
`=+4hati+6hatj-12hatk`