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सरकारी जागिर
यो एउटा निश्चित radius भएको circle हो जुन ५८ वर्षसम्म त्यसको center लाई परिवर्तन गर्न सकिदैन । radius लाई क्षमता अनुसार बढाउन त सकिन्छ तर जति नै बढ्यो उति नै भार थप्छ ।
तर निजी जागिरको centre जता पनि लान सकिन्छ यसकारण जुन क्षेत्र चाहिएको हो सानो radius को circle बनाएर त्यो क्षेत्र आफ्नो बनाउन सक्छौ ।
तर निजी जागिरको centre जता पनि लान सकिन्छ यसकारण जुन क्षेत्र चाहिएको हो सानो radius को circle बनाएर त्यो क्षेत्र आफ्नो बनाउन सक्छौ ।
Saturday, April 4, 2020
Equation of a plane through the intersection of two planes
In
the above figure, thick black and thin
black line represents the 2D form of two planes through which a black dotted
plane passes . and the equation of the thick black line and thin line plane are
a1x+b1y+c1z+d1=0 and a2x+b2y+c2z+d2=0
respectively. Here we have to find the
equation of black dotted plane. Let , OP , PQ and OR are constructed so that
they are normals of black dottted, thin black and thick black plane respectively.
Let the coordinate of O, R, P are (x3,y3,z3),
(x2,y2,z2)
and (x1,y1,z1) repectively. Then since
by definition of plane , the a1, b1 and c1 represents the direction ratio of the normal
of thick black plane. As we assume that OR is also the normal then direction
ratio of OR is
(x3-x2, y3-y2, z3-z2) so OR is also normal so it is parallel to any
normal of that plane so we have following relation
Then we have , x3-x2=ka1,
y3-y2=kb1, z3-z2=kc1
Here PQ is the normal of the thin black
plane which is also parallel to PR so PR is parallel to any normal of that
plane so by similar as above we get
Now the equation of black dotted plane
since it passes through (x3,y3,z3)
We have a(x-x3)+b(y-y3)+c(z-z3)=0 ………..(1)where a, b, c are the direction ratio
of the normal i.e OP of the required plane.
But By figure the direction ratio of OP is (x3-x1,
y3-y1, z3-z1 )
\a= x3-x1=x3-x2+x2-x1=ka1+ma2, similarly b= kb1+mb2 and c= kc1+mc2
Now from (1)
(ka1+ma2 )(x-x3)+
(kb1+mb2 )(y-y3)+ (kc1+mc2
)(z-z3)=0
Or,
(a1+λa2 )(x-x3)+ (b1+λb2
)(y-y3)+ (c1+λc2 )(z-z3)=0 where λ=m/k
Or,
a1x-a1x3+λa2(x-x3)+b1y-b1y3+λb2(y-y3)+c1z-c1z3+λc2(z-z3)=0
Or,
a1x+b1y+c1z+λ(a2x+b2y+c2z)-{(a1x3+b1y3+c1z3)+λ(a2x3+b2y3+c2z3)}=0……..(2)
Now
since (x3,y3,z3) also passes through two
planes a1x+b1y+c1z+d1=0 and a2x+b2y+c2z+d2=0
so
a1x3+b1y3+c1z3+d1=0 ………..´1
add above we get(a1x3+b1y3+c1z3)+λ(a2x3+b2y3+c2z3)=-(d1+λd2)……..(3)
now from (2) and (3) we get
a1x+b1y+c1z+λ(a2x+b2y+c2z)+
(d1+λd2)=0
\ a1x+b1y+c1z+d1+λ(a2x+b2y+c2z+d2)=0 is the required plane .
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