Don’t do as the book says, FP3 vectors

Line where two planes meet; shortest distance from point to line; distance between two parallel lines.

Click here to get “don’t do” notes as pdf

1. DISTANCE FROM A LINE TO A PLANE

point_to_line

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 sa and b = shortest distance × 1

So shortest distance = | (sab |

If we have the equation of the line in form r = a + tb with b not necessarily a unit vector,then:

shortest distance = | (sab |/|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 = | (a1a2b |

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)

line_where_2_planes_meet

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−2jk) × (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).