Linear algebra study realm

Find the closest point to a line.
2 methods: Use the projection or the dot product. Suppose you have line: X = tu + A, a point P, and closest point on the line to P is Q
The projection method would give you: Proj_u AP + A = Q.
The dot product method gives: PQ*u=0, since Q is on the line, you can write Q in form of the line's equation.

Find the closest point to a plane.
Suppose you have a plane: ax+by+cz = d and a point P. The closest point form the point to the plane is Q.
Therefore, Q is the intersection of the line PQ and the plane. Line PQ can be written as (x,y,z) = t(a,b,c) + P. Substitute PQ equation into the plane's equation and solve for t. Then, you get the point.

Find the closest distance between two parallel planes.
Provides that the two plane are parallel, thus they have the same normal. For two planes ax+by+cz = d1 and ax+by+cz = d2, the formula is |d2-d1|/(a^2 + b^2 + c^2)^(1/2) = dist.

Find the closest distance between two lines.
Suppose that you have a line X1 = tu + A and X2 = sv + B. Define two plane. Use u x v for their normal and find d1 and d2 with A and B. Then use |d2-d1|/(a^2 + b^2 + c^2)^(1/2) = dist.

Find the closest point between two skew-lines.
Suppose you have a line X1 = tu + A = t(u1,u2,u3)T + (a1,a2,a3) which contains the point P that is the closest to Q on the line X2 = sv + B = s(v1,v2,v3)T + (b1,b2,b3)T.
We know that PQ*v = PQ*u = 0, solve for s or t.
P(tu1 + a1, tu2 + a2, tu3 + a3)
Q(sv1 + b1, sv2 + b2, sv3 + b3)
PQ = (sv1 + b1 - tu1 - a1, sv2 + b2 - tu2 - a2, sv3 + b3 - tu3 - a3)T (Let it be = (i,j,k)T
PQ*v = iv1 + jv2 + kv3 = 0
PQ*u = iu1 + iu2 + iu3 = 0
2 equations, 2 unknowns, solve for s if you want to find Q, solve for t if you want to find P.

Find the closest distance between a point and a line.
Suppose you have a line X= tu + A and a point P, use the formula: ||AP x u||/||u||

Find the intersection of two planes.
You have 2 equations, solve the system.

Find the intersection of two lines.
Suppose, in the first line, you have
x = at + d,
y = bt + e,
z = ct + f,
and in the second line, you have
x = 2as + d,
y = 2bs + e,
z = 2cs + f.
Solve t or s. Let's say you want to solve for s, you get
s = (x-d)/2a,
s = (y-e)/2b and
s = (z-f)/2c.
Substitute the first line into it, you get
s = ((at + d)-d)/2a, s = ((bt + e)-e)/2b and s = ((ct + f)-f)/2c
See if the equation is consistent according to t.
If if it is for t=k, such as k is a specific value, then the lines intersect at t=k.
If if is consistent for all t, then they are the same line.
If it's never consistent for any t, then they do not intersect.

Find the intersection of a line and a plane.
Substitute the equation of the equation of the line into the plane. Similar to "Find the closest point to a plane."

Find two orthogonal planes which intersect at a given line.
Use dot product. Suppose that the line is X = t(u1,u2,u3)T + (p1,p2,p3)T. The normal n = (a,b,c)T of the plane is orthogonal to the vector u = (u1,u2,u3)T. Thus, n*u = 0 = au1 + bu2 + cu3 = 0,
Thus, n = s(-u2/u1, 1, 0)T + t(-u3/u1, 0, 1)T
Take any n and use n x u to find another normal for the second plane. Use the point (p1,p2,p3)T to find d.

Find a plane which contains a given line.
Just do the first part of "Find two orthogonal planes which intersect at a given line."

Find a plane which contains a given line and which is parallel to a specific axis.
The normal of the plane is perpendicular from the axis and the vector of the line. Use the cross product and use the point on the line to find d.
i.e.: Suppose that the line is X1 = tu + A and the axis is X2= sv
The normal n = u x v
Then substitute A to find d.

Find two parallel plane each containing one of the given skew-lines.
The normal of two planes is the same, use cross product of the two vectors in the lines. Use the points on each line to find each d.
i.e.: Suppose that the lines are X1 = tu + A and X2 = sv + B.
The normal n = u x v for both plane.
Find d1 by using A and d2 by using B.

Find a plane with contain a given point and a line.
Suppose that the line is X = tu + A and the point is P. The normal n = u x AP. Use A or P to find d.

Find a plane containing two intersecting lines.
Suppose that the lines are X1 = tu + A and X2 = sv + B. Normal n = u x v and use A or B to find d.

Find a plane containing two parallel lines.
Suppose that the lines are X1 = tu + A and X2 = sv + B. Since they are parallel, u = kv, thus we cannot use the cross product. Use normal n = u x AB or n = v x AB and use A or B to find d.

Find the line of intersection of two planes.
You've already asked that. =.=

Find a plane that passes through the origin and orthogonal to two given planes.
The normal of the questioning plane is orthogonal to the normals of the given planes, i.e. use cross product. Use the origin (0,0,0)T to find d.

Find a plane passing through a given point and parallel to a specific axis.
Find the closest point A on the axis from the given point P. AP is the normal, use P to find d.

Find a plane which contains a given line and orthogonal to a given plane.
Suppose that the line is X = tu + A and the plane is ax + by + cz = d. The normal n = u x (a,b,c)T and use A to find d.

Find two planes which intersect at a given point.
Do they really ask this question?
Just take any normals and substitute the point to find the two d.

Find a line that is perpendicular to a given line.
Suppose that the given line is X1 = tu + A and the line you're looking for is X2 = sv + B. u x v = 0 solve for v. Use A for B.