Find the image of the line `(x-1)/2=(y+1)/(-1)=(z-3)/4` in the plane `3x-3y+10 z-26=0.`

Options

A. `(x-5//2)/9=(y-1//2)/(-1)=(z-2)/(-3)`
B. `(x+5//2)/9=(y+1//2)/(-1)=(z+2)/(-3)`
C. `(x-5//2)/9=(y+1//2)/(-1)=(z+2)/(-3)`
D. None of these

Answer

1

Explanation

`(x-1)/(9)=(y-2)/(-1)=(z+3)/(-3)=1" ".......(1)`
` 3x-3y+10z-26=0" "......(2)`

point M is the image of point A in plane (2)
(i.e . Foot of the perpendicular )
`therefore ` (DR's of the normal of the plane (2))--(DR's of the line AM )=3,-310
then equation of line
`AM implies (x-1)/(3)=(y-2)/(-3)=(z+3)/(10)=lamda`
` M (x_(1),y_(1),z_(1))=(3lamda +1,-3lamda +2,10lamda -3)`
`:' `point M lies in the plane (2)
`then 3 x_(1)-3y_(1)+10z_(1) -26 =0`
`3(3lamda +1)-3(-3lamda+2)+10(10 lamda -3)-26=0`
`118lamda =59 implieslamda=1//2`
`therefore M{(5)/(2),(1)/(2),2}`
`therefore ` equation of line B M
` implies((x-(5)/(2)))/(9)=((y-(1)/(2)))/(-1)=((z-2))/(-3)`