1 6=?2011-04-24 21:51:13

1 6=?2011-04-24 21:51:13

seven
Lihot marketing 891

Given that the line L passes through point a (1,2,3) and is perpendicular to plane a, the equation of a is: 2x + 3Y + 51 = 0
Given that the line L passes through point a (1,2,3) and is perpendicular to plane a, the equation of a is: 2x + 3Y + 51 = 0
For specific solutions!

The normal vector of the plane is n = (2,3,0),
This is also the direction vector of the line, so the linear equation is (x-1) / 2 + (Y-3) / 3 = (Z-3) / 0
(the suspected plane equation is 2x + 3Y + 5Z + 1 = 0 or 2x + 3Y + 5z-1 = 0,
If so, the linear equation should be (x-1) / 2 = (Y-2) / 3 = (Z-3) / 5)

X × (1-2 / 5) = 51 equation

X×（1-2/5）=51
3X/5=51
X=51*5/3
X=85