The perpendicular distance between the line `vecr = 2hati-2hatj+3hatk+lambda(hati-hatj+4hatk)` and the plane `vecr.(hati|5hatj|hatk) = 5` is :
The perpendicular distance between the line `vecr = 2hati-2hatj+3hatk+lambda(hati-hatj+4hatk)` and the plane `vecr.(hati|5hatj|hatk) = 5` is :
A. `(10)/(3sqrt3)`
B. `(10)/(9)`
C. `(10)/(3)`
D. `(3)/(10)`
4 views
1 Answers
Correct Answer - a
It is obvious that the given line and point are parallel. Given point on the line is A(2,-2,3). B(0,0,5) is a point on the plane. Therefore,
`vec(AB)=(2-0)hati+(-2-0)hatj+(3-5)hatk` ltbgt Then distance of B from the plane = projection of
`vec(AB)" on vector "hati+5hatj+hatk`
`p=|((2hati-2hatj-2hatk).(hati+5hatj+hatk))/(sqrt(1+25+1))|`
`=|(2-10-2)/(sqrt27)|=(10)/(3sqrt3)`
4 views
Answered