Question

if d is the distance from the origin and l,m,n are.The direction cosines of the normal to the plane through the origin,then find the foot of the perpendicular of the plane ?​

Answer 1

The foot of the perpendicular of the plane is $(ad/(a^2 + b^2 + c^2),bd/(a^2 + b^2 + c^2),cd/(a^2 + b^2 + c^2))$

The direction cosines of a vector are the cosines of the angles between the vector and the positive x, y and z-axes.

A plane through the origin can be represented by the equation:

ax + by + cz = d

Where (a,b,c) are the direction cosines of the normal to the plane and d is the distance from the origin to the plane.

The foot of the perpendicular from the origin to the plane is the point of intersection of the line passing through the origin and perpendicular to the plane.

The direction vector of this line passing through the origin is normal to the plane (a,b,c)

So the direction vector of the line is (a,b,c) and passing through the origin (0,0,0)

The point of intersection of line and plane is given by

x = 0 + ta

y = 0 + tb

z = 0 + t*c

where t is a scalar variable.

Substituting the equation of plane we get

$t(ax + by + c*z) = d$

$t(a^2 + b^2 + c^2) = d$

$t = d / (a^2 + b^2 + c^2)$

So the coordinates of the foot of the perpendicular of the plane are (at,bt,ct)

So, the answer is $(ad/(a^2 + b^2 + c^2),bd/(a^2 + b^2 + c^2),cd/(a^2 + b^2 + c^2))$

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