Search results
Results from the WOW.Com Content Network
The theorem is: [14] In a projective plane, every non-collinear set of n points determines at least n distinct lines. As the authors pointed out, since their proof was combinatorial, the result holds in a larger setting, in fact in any incidence geometry in which there is a unique line through every pair of distinct points.
The construction of orthogonality of vectors is motivated by a desire to extend the intuitive notion of perpendicular vectors to higher-dimensional spaces. In the Cartesian plane, two vectors are said to be perpendicular if the angle between them is 90° (i.e. if they form a right angle).
The proof is completed by showing this construction is always possible: If both chords are diameters, they intersect.(at the center of the boundary circle) If only one of the chords is a diameter, the other chord projects orthogonally down to a section of the first chord contained in its interior, and a line from the pole orthogonal to the ...
In the mathematical fields of linear algebra and ... it is called the perp, short for perpendicular complement. It is a ... [3] [proof 1] that ...
A twist is a screw used to represent the velocity of a rigid body as an angular velocity around an axis and a linear velocity along this axis. All points in the body have the same component of the velocity along the axis, however the greater the distance from the axis the greater the velocity in the plane perpendicular to this axis.
If this property holds in the affine plane defined by a ternary ring, then there is an equivalence relation between "vectors" defined by pairs of points from the plane. [14] Furthermore, the vectors form an abelian group under addition; the ternary ring is linear and satisfies right distributivity: (+) = +.
Given a line and a point P not on that line, construct a line, t, perpendicular to the given one through the point P, and then a perpendicular to this perpendicular at the point P. This line is parallel because it cannot meet ℓ {\displaystyle \ell } and form a triangle, which is stated in Book 1 Proposition 27 in Euclid's Elements . [ 15 ]
More specifically, let A, B, C and D be four points on a circle such that the lines AC and BD are perpendicular. Denote the intersection of AC and BD by M. Drop the perpendicular from M to the line BC, calling the intersection E. Let F be the intersection of the line EM and the edge AD. Then, the theorem states that F is the midpoint AD.