Line where two planes meet; shortest distance from point to line; distance between two parallel lines.
1. DISTANCE FROM A LINE TO A PLANE
Let the point be s and the line be r = a + tb written so that b is a unit vector
Then the area of the parallelogram formed by the vectors s−a and b = shortest distance × 1
So shortest distance = | (s−a)×b |
If we have the equation of the line in form r = a + tb with b not necessarily a unit vector,then:
shortest distance = | (s−a)×b |/|b|
2. DISTANCE BETWEEN TWO PARALLEL LINES
Take any point on one line, and find the shortest distance from that point to the other line.
Distance = | (a1−a2)×b |
if b is the direction vector (same for both lines) adjusted to be a unit vector.
3. LINE WHERE TWO PLANES MEET (LINE OF INTERSECTION)
Write the planes in form r1.n1=p1 and r2.n2=p2
The direction v of the line where the planes meet is perpendicular to both n1 and n2, so can be described by v=n1×n2
So in example 29 on p.122, the direction of the line where the planes meet must be:
v=(2i−2j−k) × (i−3j+k), which is −5i−3j−4k
To make things easier, change that to 5i+3j+4k (same line, whether we count the direction positive or negative). This is a direction vector v for the line where the planes meet.
All we need now is a point a on the line where the planes meet, and we can write the equation of the line in form r=a+tv
Put z=0 in both plane equations: 2x − 2y = 2; x − 3y = 5. Solve: y=−2, x=−1
(-1,-2,0) will do for the point a we need. It is a point on both planes, and so on the line where the two planes meet.
(If putting z=0 doesn’t work, try y=0 or x=0. One of them must work).