What are the direction cosines of the line represented by `3x+y+2z=7,x+2y+3z=5` ?

Options

A. `(-1,-7,5)`
B. `(-1,7,5)`
C. `(-1/sqrt75,-7/sqrt75,5/sqrt75)`
D. `(-1/sqrt75,7/sqrt75,5/sqrt75)`

Answer

C

Explanation

let direction ratio of the line be, a, b, c, ltbRgt This line is contained by both plane,
`3x+y+2z=7and x+2y+3z=5`.
`rArr3a+b+2c=0 …(1)`
and `a+2b+3c=0 …(2)`
Solving these two equations:
`{:a/abs((1,2),(2,3))=(-b)/abs((3,2),(1,3))=c/abs((3,1),(1,2))=k:}`
Let
`rArr a/(-1)=(-b)/7=c/5=k`
`a=-k,b=-7k,c=5k`.
Direction cosines are:
`a/sqrt(a^2+b^2+c^2),b/sqrt(a^2+b^2+c^2),c/sqrt(a^2+b^2+c^2)`
`sqrt(a^2+b^2+c^2)=sqrt((-k)^2+(-7k)^2+(5k)^2)`
`=sqrt(k^2+49k^2+25^2=ksqrt75`
So, Direction cosines are `((-k)/(ksqrt75),((-7k)/(5sqrt75),(5k)/(ksqrt75)).`