Math, asked by learner911, 5 months ago

- Geometric Application of determinants: (Curves and surfaces through given points)
The idea is to get an equation from the vanishing of the determinant of a homogeneous linear
system as the condition for a nontrivial solution. We explain the trick for obtaining
such a system for the case of a line L through two given points P1 : (x1 yı)
and P2 : (x2.yz). The unknown line is ax +by+c=0, say. We write it as ax +by
+c.1=0. To get a nontrivial solution for a,b,c, the determinant of the
"coefficients” x,y, 1 must be zero. The system is
axtby +c. 1 =0 (Line L)
axı +by. +0.1=0 (P1 on L)
... (1)
ax2 +by2 +0.1=0 (P2 on L)
(a) Line through two given points: Derive from D=0 in (1) the familiar
formula
(2-1))
(1-x)
=
O'-1)
Ya-D
(b) Plane : Find the analog of (1) for a plane through three given points. Apply it when the
points are(1,1,1), (3,2,6), (5,0,5)
(c) Circle : Find a similar formula for a circle in the plane through three given points. Find
and sketch the circle through (2,6), (6,3), (7,1).
(d) Sphere: Find the analog of the formula in (C) for a sphere through four given points. Find
the sphere through (0,0,5), (4,0,1), (0,4,1), (0,0,-3) by the formula or by
inspection​

Answers

Answered by martincochindutyfree
0

Answer:

ko shnskslam skskakanab

bakakama a

jakakaksossm

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