Find the equations of the line passing through the point `(3,0,1)` parallel to the planes `x+2y=0` and `3y-z=0`.

Options

A. `(x-3)/(-2)=(y-0)/(1)=(z-1)/(3)`
B. `(x-3)/(1)=(y-0)/(-2)=(z-1)/(3)`
C. `(x-3)/(3)=(y-0)/(1)=(z-1)/(-2)`
D. None of these

Answer

A

Explanation

Let a,b, and c be the direction ratios of the required line .
Then, its equation is `(x-3)/(a)=(y-0)/(b)=(z-1)/(c)` . . .(i)
Sinec, Eq. (i) is parallel to the planes x+2y+0z=0 and 0x+3y-z=0.
Therefore, normal to the plane is perpendicular to the line
`:.a(1)+b(2)+c(0)=0anda(0)+b(3)+c(-1)=0`
`[becausea_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)=0]`
On solving these two equations by cross - multiplication
we get `(a)/((2)(-1)-(0)(3))=(b)/((0)(0)-(1)(-1))=(c)/((1)(3)-(0)(2))`
`rArr" "(a)/(-2)=(b)/(a)=(c)/(3)=lamda` [say]
`rArr" "a=-2lamda,b=lamdaandc=3lamda`
On substituting the values of a,b and c in Eq. (i), we get the equation of the required line as
`(x-3)/(-2)=(y-0)/(1)=(z-1)/(3)`