Find the length and the foot of perpendicular from the point `(1,3/2,2)` to the plane `2x-2y+4z+5=0.`

Options

A. `sqrt(7) units`
B. `sqrt(2) units`
C. `sqrt(6) units`
D. `sqrt(3) units`

Answer

C

Explanation

Equation of the given plane is `2x-2y+4z+5=0`……….i
`implies vecn=2hati-2hatj+4hatk`
So,the equation of line through `(1, 3/2,2)` and parallel to `vecn` is given by
`(x-1)/2=(y-3//2)/(-2)=(z-2)/4=lamda`
`implies x=2lamda+1,y=-2lamda+3/2` and `z=4lamda+2`
If this point lies on the given plane then
`2(2lamda+1)-2(-2 lamda +3/2)+4(4lamda+2)5=0`
`implies 4lamda+2+4lamda-3+16lamda+8+5=0`
`implies 24lamda=-12implieslamda=(-1)/2`
`:.` Required foot of perpendicular
`=[2xx((-1)/2)+1,2xx((-1)/2)+3/2,4xx((-1)/2)+2] "i.e."(0,5/2,0)`
`:.` Required length of perpendicular`=sqrt((1-0)^(2)+(3/2-5/2)^(2)+(2-0)^(2))`
`=sqrt(1+1+4)=sqrt(6)` units.