If `OP'` be the external bisector of `angleOAB`, then `OP'` may be regarded as the internal bisector of the angle
between the lines which are parallel to `hat(a)` and `- hat(b)`, . Hence its equaiton is

`vec(r) =t (hat(a) -hat(b) )` or ` vec(r) =t ( (hat(a)/a) -(hat(b) /a) )`

Corollary : If the the lines intersect atE having position vector ` vec(alpha)`
, then the above equations becomes

`vec(r)=vec(alpha)+ t (hat(a) +hat(b)) and `vec(r)= vec(alpha) + t( hat(a) - hat(b))` respectively.

1. If vec(a) , vec(b),vec(c) be position vectors of three points `A,B` and `C` respectively and `x, y, z` be three scalars so that
all are not zero, then the necessary and sufficient conditions for three points to be collinear is that

`x vec(a) +y vec(b) +zvec(c) =0` where `x+y+z =0`

2. Three points `A, B` and `C` are collinear, if `vec(AB) = lambda vec(BC)`

1. Vector area of plane figures :

With every closed bound surface which has been described in a certain specific manner and whose
boundaries do not cross, it is possible to associate a directed line segment `vec(c)`such that



(i) `|vec(c) | `= no. of units of area enclosed by the plane figure.
(ii) The support of `vec(c)` is perpendicular to the area and
(iii) The sense of description of the boundaries and the direction of `vec(c)` is in accordance with the R.H.
screw rule.

Vector area of `Delta OAB ` is `vec(Delta) =1/2 (vec(a) xx vec(b))`

If `vec(a),vec(b),vec(c)` are the position vectors then the vector area of `Delta ABC` is

`vec(Delta) =1/2 [ (vec(c) - vec(b)) xx (vec(a) - vec(b)) ]`

`vec(Delta) =1/2 ( (vec(a) xx vec(b)) + (vec(b) xx vec(c)) xx (vec(c) xx vec(a)) )`


Note:

(i) lf 3 points with position vectors `vec(a) ,vec(b)` and `vec(c)` are collinear then `vec(a) xx vec(b) + vec(b) xx vec(c)+ vec(c) xx vec(a)=0`
(ii) Unit vector perpendicular to the plane of the `Delta ABC` when `vec(a),vec(b),vec(c)` are the p.v. of its angular point is

` hat(n) = pm ( vec(a) xx vec(b) + vec(b) xx vec(c) + vec(c) xx vec(a) )/(2Delta)` ,where `vec(a),vec(b),vec(c)` are the position vectors of the angular points of the triangle `ABC` .

(iii) Vector Area of a quadrilateral `ABCD` = Vector area of `Delta ABC+ vector area of Delta ACD`

`= 1/2 (vec(AB) xx vec(AC)) +1/2 (vec(AC) xx vec(AD))`

`=1/2 ( vec(AB) xx vec(AC) - vec(AD) xx vec(AC)) =1/2 (vec(AB) - vec(AD)) xx vec(AC)`

`= 1/2 (vec(AB) + vec(DA))xx vec(AC) =1/2 vec(DB) xx vec(AC)`

`:.` Area of square ABCD `=1/2 |vec(DB) xx vec(AC) | =1/2 |vec(AC) xx vec(BD) |`

Note:
(i) 2 lines in a plane if not II must intersect and 2 lines in a plane if not intersecting must be parallel.
Convertely 2 intersecting or parallel lines must be coplanar.

(ii) In space, however we come across situation when two lines neither intersect nor II , Two such lines
(like the flight paths of two planes) in space are known as skew lines or non coplanar lines.

(iii) S.D. between two such skew lines is the segment intercepted betweeen the two lines and perpendicular
to both.

Method 1: Two ways to determine the S D

`L_1 : vec(r) = vec(a) +lambda vec(p)`
`L_2 : vec(r) = vec(b) + mu vec(q)`

`vec(n) =vec(p) xxvec(q)`

`vec(AB) = (vec(b) - vec(a))`

SD = | Projection of vec(AB) on vec(n)| = |(vec(AB)* vec(n))/(|vec(n)|) | = | ( ( vec(b) - vec(a)) *(vec(p) xx vec(q)))/( |vec(p)xxvec(q) |) |`


If S.D. `= 0 =>` lines are intersecting and hence coplanar.

p.v. of `N_1 = vec(a) + lambda vec(p)` ; p.v. of `N_2 = vec(b) + mu vec(q)`

`vec(N_1N_2) = (vec(b) -vec(a)) + (mu vec(q) -lambda vec(p))`

Now `vec(N_N_2) *vec(p) =0` and `vec(N_N_2) *vec(q) =0` (two linear equations to get the unique values of `lambda` and `mu` )

One p.v's of `N _1` and `N_2` are known we can also determine the equation to the line of shortest di stance

and the S.D.

Shortest Distance between two parallel lines :

`d = |vec(a) -vec(b) | sin theta => |( (vec(a) -vec(b)) xx vec(c))/(|vec(c)|) |`