Calculating spread Rational trigonometry
1 calculating spread
1.1 trigonometric
1.2 vector/slope (two-variable)
1.3 cartesian (three-variable)
calculating spread
trigonometric
suppose 2 lines, l1 , l2, intersect @ point shown @ right. choose point b ≠ on l1 , let c foot of perpendicular b l2. spread s is
s
(
ℓ
1
,
ℓ
2
)
=
q
(
b
,
c
)
q
(
a
,
b
)
=
q
r
.
{\displaystyle s(\ell _{1},\ell _{2})={\frac {q(b,c)}{q(a,b)}}={\frac {q}{r}}.}
vector/slope (two-variable)
like angle, spread depends on relative slopes of 2 lines (constant terms being eliminated) , invariant under translation (i.e. preserved when lines moved keeping parallel themselves). given 2 lines equations are
a
1
x
+
b
1
y
=
constant
and
a
2
x
+
b
2
y
=
constant
{\displaystyle a_{1}x+b_{1}y={\text{constant}}\qquad {\text{and}}\qquad a_{2}x+b_{2}y={\text{constant}}}
we may rewrite them 2 lines meet @ origin (0, 0) equations
a
1
x
+
b
1
y
=
0
and
a
2
x
+
b
2
y
=
0
{\displaystyle a_{1}x+b_{1}y=0\qquad {\text{and}}\qquad a_{2}x+b_{2}y=0}
in position point (−b1, a1) satisfies first equation , (−b2, a2) satisfies second , 3 points (0, 0), (−b1, a1) , (−b2, a2) forming spread give 3 quadrances:
q
1
=
(
b
1
2
+
a
1
2
)
,
q
2
=
(
b
2
2
+
a
2
2
)
,
q
3
=
(
b
1
−
b
2
)
2
+
(
a
1
−
a
2
)
2
{\displaystyle {\begin{aligned}q_{1}&=\left(b_{1}^{2}+a_{1}^{2}\right),\\q_{2}&=\left(b_{2}^{2}+a_{2}^{2}\right),\\q_{3}&=\left(b_{1}-b_{2}\right)^{2}+\left(a_{1}-a_{2}\right)^{2}\end{aligned}}}
the cross law – see below – in terms of spread is
1
−
s
=
(
q
1
+
q
2
−
q
3
)
2
4
q
1
q
2
.
{\displaystyle 1-s={\frac {(q_{1}+q_{2}-q_{3})^{2}}{4q_{1}q_{2}}}.}
which becomes:
1
−
s
=
(
a
1
2
+
a
2
2
+
b
1
2
+
b
2
2
−
(
b
1
−
b
2
)
2
−
(
a
1
−
a
2
)
2
)
2
4
(
a
1
2
+
b
1
2
)
(
a
2
2
+
b
2
2
)
.
{\displaystyle 1-s={\frac {\left(a_{1}^{2}+a_{2}^{2}+b_{1}^{2}+b_{2}^{2}-(b_{1}-b_{2})^{2}-(a_{1}-a_{2})^{2}\right)^{2}}{4\left(a_{1}^{2}+b_{1}^{2}\right)\left(a_{2}^{2}+b_{2}^{2}\right)}}.}
this simplifies, in numerator, (2a1a2 + 2b1b2), giving:
1
−
s
=
(
a
1
a
2
+
b
1
b
2
)
2
(
a
1
2
+
b
1
2
)
(
a
2
2
+
b
2
2
)
.
{\displaystyle 1-s={\frac {\left(a_{1}a_{2}+b_{1}b_{2}\right)^{2}}{\left(a_{1}^{2}+b_{1}^{2}\right)\left(a_{2}^{2}+b_{2}^{2}\right)}}.}
(note: 1 − s expression cross, square of cosine of either angle between pair of lines or vectors, gives name cross law.)
then, using brahmagupta–fibonacci identity
(
a
2
b
1
−
a
1
b
2
)
2
+
(
a
1
a
2
+
b
1
b
2
)
2
=
(
a
1
2
+
b
1
2
)
(
a
2
2
+
b
2
2
)
,
{\displaystyle \left(a_{2}b_{1}-a_{1}b_{2}\right)^{2}+\left(a_{1}a_{2}+b_{1}b_{2}\right)^{2}=\left(a_{1}^{2}+b_{1}^{2}\right)\left(a_{2}^{2}+b_{2}^{2}\right),}
the standard expression spread in terms of slopes (or directions) of 2 lines becomes
s
=
(
a
1
b
2
−
a
2
b
1
)
2
(
a
1
2
+
b
1
2
)
(
a
2
2
+
b
2
2
)
.
{\displaystyle s={\frac {\left(a_{1}b_{2}-a_{2}b_{1}\right)^{2}}{\left(a_{1}^{2}+b_{1}^{2}\right)\left(a_{2}^{2}+b_{2}^{2}\right)}}.}
in form (and in cartesian equivalent follows) spread ratio of square of determinant of 2 vectors (numerator) product of quadrances (denominator)
cartesian (three-variable)
this replaces (−b1, a1) (x1, y1), (−b2, a2) (x2, y2) , origin (0, 0) (as point of intersection of 2 lines) (x3, y3) in previous result:
s
=
(
(
y
1
−
y
3
)
(
x
2
−
x
3
)
−
(
y
2
−
y
3
)
(
x
1
−
x
3
)
)
2
(
(
y
1
−
y
3
)
2
+
(
x
1
−
x
3
)
2
)
(
(
y
2
−
y
3
)
2
+
(
x
2
−
x
3
)
2
)
.
{\displaystyle s={\frac {{\bigl (}(y_{1}-y_{3})(x_{2}-x_{3})-(y_{2}-y_{3})(x_{1}-x_{3}){\bigr )}^{2}}{{\bigl (}(y_{1}-y_{3})^{2}+(x_{1}-x_{3})^{2}{\bigr )}{\bigl (}(y_{2}-y_{3})^{2}+(x_{2}-x_{3})^{2}{\bigr )}}}.}
^ cite error: named reference horizons invoked never defined (see page).
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