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Question 23633: I am trying to solve some challenge problems regarding perpendicular distances.. and I just cannot seem to make any sense of this one:
Calculate the perpendicular distance from A(2,3,-1) to the plane 2x + y - 2z + 9 = 0 using the following method.
By inspection, determine the coordinates of any point C on the plane. Calculate the projection AC projected onto n, where n is the normal vector to the plane. Calculate the magnitude of AC projected onto n.
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! the perpendicular distance from A(2,3,-1) to the plane 2x + y - 2z + 9 = 0.
The distance formula of a point A(xo,y,zo) to the plane ax + b y + cz+d = 0
is D = |axo + byo + czo + d|/ sqrt(a^2+b^2+c^2)...(**)
where N = ai + bj + ck (=(a,b,c))is normal to the plane and
|N| = sqrt(a^2+b^2+c^2).
The formula can be obtained by assuming the point P=(x,y,z) of A on the
plane and solve (x-xo,y-yo,z-zo) // (a,b,c) , ax+by+cz+d = 0 and
let k = (x-xo,y-yo,z-zo) /(a,b,c), (x,y,z)= (xo+ka,yo+kb,zo+kc).
a(xo+ka)+ b(yo+kb)+c(zo+kc)+d = 0, get k(a^2+b2+c^2) = -(axo+byo+czo+d).
D = |(x-xo,y-yo,z-zo)| = |k| |(a,b,c)| =
|axo + byo + czo + d|/ sqrt(a^2+b^2+c^2)
It is the same value as your projection of AC to N.
Kenny
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