The perpendicular distance between the line `vecr = 2hati-2hatj+3hatk+lambda(hati-hatj+4hatk)` and the plane `vecr.(hati|5hatj|hatk) = 5` is :

Options

A. `(10)/(3)`
B. `(3)/(10)`
C. `(10)/(3sqrt(3))`
D. `(10)/(9)`

Answer

C

Explanation

Line is parallel to plane as
`( hati -hat j+4hatk)*(hati+5hatj+hatk)=0`
General point on the line is
`(lambda +2, -lambda-2, 4lambda+3) "for "lambda=0`
A point on this line is (2, -2, 3) and distance from that point to the plane
`r*(hati+5hatj+hatk)=5`
or `x+5y+z=5` is
`d=(|2+5(-2)+3-5|)/(sqrt(1+25+1))`
`=(|-10|)/(3sqrt(3))=(10)/(3sqrt(3))`