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

Let the point be __s__ and the line be __r__ = __a__ + t__b__ *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__ + t__b__ 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 = | (__a___{1}−__a___{2})×__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 r_{1}.n_{1}=p_{1} and r_{2}.n_{2}=p_{2}

The direction v of the line where the planes meet is perpendicular to both n_{1} and n_{2}, so can be described by v=n_{1}×n_{2}

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).