How Do You Determine the Intersection Point of a Line Orthogonal to a Plane?

[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

I know that [a,b,c] is the directional vector for the line and that

|d|/sqrt(a^2 + b^2 + c^2) is the distance between point P and origo. I know and understand this at least.

The Attempt at a Solution

I'm lost lol

 

[Nikitin]

ah damn, I haven't learned about directional vectors this year. but i was reminded about that stuff by a friend who helped me with this assignment.

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.

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

*sigh* does this have something to do with all the fools sometimes saying a plane's equation is ax+by+cz -d = 0 instead of ax+by+cz +d = 0 ? or that d is negative if the directional vector is posetive?

has been a really long day at school + after 4 me and I'm sleepy as hell.. hope somebody can fill me in tomorrow

 

[HallsofIvy]

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.

Let the line be given by x= ut, y= vt, z= wt (since (0,0,0) satifies that). If the plane and line are orthogonal, then <u, v, w> must be orthogonal to the plane and so must be parallel to <a, b, c>. That is u= ka, v=kb, w= kc. x= kat, y= kbv, z= kct. Putting that into the equation of the plane, k(a^2+ b^2+ c^2)t+ d= 0. t= -d/(a^2+ b^2+ c^2). Put that into the equation of the line and solve for x, y, and z.

Homework Equations

I know that [a,b,c] is the directional vector for the line and that

|d|/sqrt(a^2 + b^2 + c^2) is the distance between point P and origo. I know and understand this at least.

The Attempt at a Solution

I'm lost lol

 

[Nikitin]

hmm thanks but could you, or somebody else, answer the question in post 2?

 

[Nikitin]

ah damn, I haven't learned about directional vectors this year. but i was reminded about that stuff by a friend who helped me with this assignment.

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.

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

*sigh* does this have something to do with all the fools sometimes saying a plane's equation is ax+by+cz -d = 0 instead of ax+by+cz +d = 0 ? or that d is negative if the directional vector is positive?

has been a really long day at school + after 4 me and I'm sleepy as hell.. hope somebody can fill me in tomorrow

Can somebody explain to me the logic of why the answer should be negative [like this: (-ad/a^2 + b^2 + c^2,-bd/a^2 + b^2 + c^2,-cd/a^2 + b^2 + c^2) = P], and not in absolutes [like this (-ad/a^2 + b^2 + c^2,-bd/a^2 + b^2 + c^2,-cd/a^2 + b^2 + c^2) = P]?

I believe the absolutes answer is plain wrong.