CBSE Class 12-science Answered
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 | 01 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 | 02 Feb, 2009, 02:27: AM
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