Wrong answer using correct logic

[Nikitin]

Homework Statement

A plane has the equation aX + bY + cZ + d = 0. A line L goes from (0,0,0) and crosses the plane at some point. L and plane are orthogonal. express the coordinates of the crossing point P by: a, b, c and d.

Homework Equations

|d|/sqrt(a^2 + b^2 + c^2) is the distance between point P and origo.
[a,b,c] is the directional vector for the line and a normal vector on the plane.

The Attempt at a Solution

So, if I divide [a,b,c] by the length of the directional vector, sqrt(a^2 + b^2 + c^2) and then multiply it by the length of the OP vector, |d|/sqrt(a^2 + b^2 + c^2), I get:

a|d|/(a^2 + b^2 + c^2), b|d|/(a^2 + b^2 + c^2), c|d|/(a^2 + b^2 + c^2)=P.

This is the crossing point of the line and the plane (or a point completely opposite to the plane).

The answer is supposed to be (-ad/(a² + b² + c^²),-bd/(a² + b² + c²),-cd/(a² + b² + c²))=P

I know that my answer is completely wrong, because |d| will remain always positive, while [a,b,c] could be negative or positive, with no consequence to |d|.

Can somebody tell me how I can get to the correct answer?

 

[ehild]

Hi, Nikitin,

Your logic is not that correct. You made a vector of direction the same as the normal of the plane and length equal to the distance between the origin and the plane. This is OK, but is this vector a point of the plane? Try. Substitute the coordinates you got into the equation of the plane. You get |d|+d =0, is it correct?

Write the parametric equation of the line: It is (at, bt, ct). Substitute for x,y,z in the equation of the plane. You get an equation for t. Solve and get the coordinates of he crossing point using the value of t.

ehild

 

[Nikitin]

yes, I got help previously and was told that one could do it your way.

Tho I understand the fault in my reasoning.. the normal-vector must be pointing towards the plane.

thanks