A 'simple' vector problem - where a line meets a plane

[Snoopey]

Hi all,
I am a little stuck on a problem I'm trying to solve for something I'm programming.
I'm trying to find the point at which a line meets a plane.

The line is defined as \vec{x} = \vec{a}+d\vec{l}
where \vec{a} is a point on the line, \vec{l} is a unit vector defining the direction of the line and d is the distance along the line.

The plane is defined using a point \vec{x_{0}} and normal \vec{n} as \vec{n}.\left(\vec{x}-\vec{x_{0}}\right)=0

I want to sub in my line equation into my plane equation and solve for d to get \vec{x} but my vector algebra is very rusty and I can't for the life of me figure out how to get my d out.

The funny thing is I used a similar method to find where a line intersects a sphere with equation \left|\vec{x}-\vec{c}\right|^{2}=R^{2} (\vec{c} = centre, R = radius) and subbed in no problems. But that dot product in the plane equation is just confusing me.

Has anyone got any suggestions for me to follow?
Many thanks!

 

[HallsofIvy]

Two vectors are perpendicular if and only if their dot product is 0. If x and x₀ are both points in the plane, then the vector x- x₀ lies in the plane so the dot product is just saying that the normal, n and the vector x- x₀ are perpendicular- which is the definition of "normal vector"!) Suppose the normal vector, n, is given by <A, B, C>, the generic point x by (x, y, z), and the point x[sub0[/sub by <x₀, y₀, z₀>. Then x- x<sub>0 is <x- x0, y- y0, z- z0> so the dot product is A(x- x0)+ B(y- y0), C(z- z0)= 0.

Now suppose \vec{a}= &lt;a_1, a_2, a_3&gt; and \vec{l}= &lt;l_1, l_2, l_3&gt; so that the line \vec{x}= &lt;a_1, a_2, a_3&gt;+ d&lt;l_1, l_2, l_3&gt;= &lt;a_1+ dl_1, a_2+ dl_2, a_3+ dl_3&gt;. Okay, put those components in for x, y, and z in the equation of the plane: A(a_1+ dl_1x_0)+ B(a_2+ dl_2y_0)+ C(a_3+ dl_3z_0)= 0

Multiplying that out, Aa_1+ Adl_1x_0+ Ba_2+ Bdl_2y_0+ Ca_3+ Cdl_3z_0= 0

Moving everything that does not involve "d" to the right, and factoring d out,
(Al_1x_0+ Bl_2y_0+ Cl_3z_0)d= -(Aa_1+ Ba_2+ Ca_3)

Now, of course, divide both sides by (Al_1x_0+ Bl_2y_0+ Cl_3z_0) to solve for d.

Finally, put that value of d int0 the formula for the line to determine the point of intersection.</sub>

 

[Snoopey]

Brilliant! Thank you for this, not sure why I didn't think of splitting into the different compenents :)

edit: For anyone interested I simplified the end result into

d= \frac{\vec{n}.(\vec{x_{0}}-\vec{a})}{\vec{n}.\vec{l}}