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One vector of the line should be a direction vector, and this is v. The second point Po can be anywhere on the line that together with the direction vector, will satisfy the line equation that will together with plane equation will create an intersection point (see above).

ex. p = (0,0,0) + t(4,3,5) in this case Po = origin (0,0,0) and direction vector (4,3,5) create a ray from origin outward.

Parametric eq from the ray: X = 4t +0 Y = 3t+0 Z = 5t+0

Plane Equation: 4x + 3y - 1z = 4 solve for t 4(4t) + 3(3t) -1(5t) = 4 16t + 9t -5t = 4 t = 4/20 = 1/5

Find x,y,z X = (4)(1/5) = 4/5 y = (3)(1/5) = 3/5 Z = (5)(1/5) = 5/5 = 1.

If you changed Po then the direction vector would change.

Alternative Plane Equation n.p + d = 0 sub for p = p0+tv n.(po+tv)+d = 0 n.po+n.tv+d = 0 t(n.v) = -d-n.po t = -d-n.po/n.v t = -(d+n.po)/n.v solve for t

One vector of the line should be a direction vector, and this is v. The second point Po can be anywhere on the line that together with the direction vector, will satisfy the line equation that will together with plane equation will create an intersection point (see above).

ex. p = (0,0,0) + t(4,3,5) in this case Po = origin (0,0,0) and direction vector (4,3,5) create a ray from origin outward.

Parametric eq from the ray: X = 4t +0 Y = 3t+0 Z = 5t+0

Plane Equation: 4x + 3y - 1z = 4 solve for t 4(4t) + 3(3t) -1(5t) = 4 16t + 9t -5t = 4 t = 4/20 = 1/5

Find x,y,z X = (4)(1/5) = 4/5 y = (3)(1/5) = 3/5 Z = (5)(1/5) = 5/5 = 1.

If you changed Po then the direction vector would change.

Alternative Plane Equation n.p + d = 0 sub for p = p0+tv n.(po+tv)+d = 0 n.po+n.tv+d = 0 t(n.v) = -d-n.po t = -d-n.po/n.v t = -(d+n.po)/n.v solve for t