Find distance of the pt (2,3,4) from the plane 3x+2y+2z+5=0, parallel to r=(-3i+2j)+m(3i+6j+2k)

Asked by  | 1st Feb, 2009, 02:14: PM

Expert Answer:

the line which is parallel to r=(-3i+2j)+m(3i+6j+2k) and passes through (2,3,4) is

r=(2i+3j+4k)+m(3i+6j+2k)

any point on this line is (2+3m,3+6m,4+2m)

now find the intersection point of this line with the given plane i.e. the general point will satisfy the equation of the plane

3(2+3m)+2(3+6m)+2(4+2m)+5=0

25m+25 = 0

m = -1

so the intersection point of the plane and the line is (-1,-3,2)

distance between (2,3,4) and (-1,-3,2) = 9+36+4 = 49 = 7

Answered by  | 2nd Feb, 2009, 02:27: AM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.