...mean value of the product of their projections on any 'fixed line in the plane containing them is half the product of their lengths multiplied by the cosine of the angle between them. The application of this proposition to problems in alternating currents is extremely simple. The power...

...angle between the directions of OQ and OP respectively; hence the scalar product of two vectors is the product of their lengths multiplied by the cosine of the angle between their respective directions. Let the vector OP be represented algebraically by — a + kb and the vector...

...by the condition of perpendicularity (26'4). In the elementary vector., theory, the scalar-product of two vectors is equal to the product of their lengths...multiplied by the cosine of the angle between them. Accordingly in the general theory the angle Q between two vectors A^ and B^ is defined by cos* —...

...by definition, we, find t'2 = t2 — Z(ta) . (ta) -\- (ta)"a* = t2 — (to)2 ; remembering that the product of two vectors is equal to the product of their lengths into the cosine of their angle, we finally obtain for our curvature | t' | = | i | sin >j/, that is:...

...in vector notation. Definition. — The scalar (or dot) product of two vectors is a scalar quantity equal to the product of their lengths multiplied by the cosine of the angle between them ; that is, AB = ABCos(A, B) (22) Bold-face letters denote vectors in magnitude and phase position;...

..." \si cosf sin9 sin<#>/ = |X| |U|cos(0 ' " ' v (17) Thus, the dot product of two nonzero vectors is the product of their lengths multiplied by the cosine of the angle between them. We may use the dot product to calculate the angle between two nonzero vectors just by writing form...

...the scalar product <O\, OB>: <OA, OB> = II OA II . II OB II cos (OA, OB) which is stated as follows: the scalar product of two vectors is equal to the product of the lengths of these vectors multiplied by the cosine of the angle formed by the two half-lines having...

...(1.22) and (1.23), we obtain B |A |cos 6 > 0 iBlcosO <0 |A |cos 6 < 0 AB = | A | |BI cos 6 (1.24) Thus the scalar product of two vectors is equal to the product of their magnitudes and the cosine of the angle between then. If we examine the plane formed by the vectors...

...scalar product is useful for determining the angle between two vectors. The Scalar Product Theorem The scalar product of two vectors is equal to the product of the magnitudes of the two vectors and the cosine of their included angle: ab = |a| |b| cos 6 ab cos...

...total; the profiles are the points visualized in CA. • scalar product — for two point vectors, the product of their lengths multiplied by the cosine of the angle between them; directly proportional to the projection of one point on the vector defined by the other. J-dimensional...