If |alpha vector + beta vector | = | alpha vector - beta vector|

The vertices A , B , C of triangle A B C have respectively position vectors vec a , vec b , vec c with respect to a given origin O . Show that the point D where the bisector of /_A meets B C has position vector vec d=(beta vec b+gamma vec c)/(beta+gamma), where beta=| vec c- vec a| and, gamma=| vec a- vec b|dot Hence, deduce that incentre I has position vector (alpha vec a+beta vec b+gamma vec c)/(alpha+beta+gamma) where alpha=| vec b- vec c|

Unit vectors vec a and vec b are perpendicular, and unit vector vec c is inclined at angle theta to both vec a and vec bdot If vec c = alpha vec a+beta vec b+gamma( vec axx vec b), then a=beta b. gamma^1=1-2alpha^2 c. gamma^2=-cos2theta d. beta^2=(1+cos2theta)/2

alpha + beta = 5 , alpha beta= 6 .find alpha - beta

Let a , b and c be three unit vectors out of which vectors b and c are non -parallel. If alpha " and " beta are the angles which vector a makes with vectors b and c respectively and axx (b xx c) = (1)/(2) b, Then |alpha -beta| is equal to

Unit vectors veca and vecb ar perpendicular , and unit vector vecc is inclined at an angle theta to both veca and vecb . If alpha veca + beta vecb + gamma (veca xx vecb) then.

If alpha ne beta but alpha^(2)= 5 alpha - 3 and beta ^(2)= 5 beta -3 then the equation having alpha // beta and beta // alpha as its roots is :

The vector equation of the plane passing through vec a ,\ vec b , vec c\ i s\ vec r=alpha vec a+beta vec b+gamma vec c provided that a. alpha+beta+gamma=0 b. alpha+beta+gamma=1 c. alpha+beta=gamma d. alpha^2+beta^2+gamma^2=1

If alpha != beta but, alpha^(2) = 4alpha - 2 and beta^(2) = 4beta - 2 then the quadratic equation with roots (alpha)/(beta) and (beta)/(alpha) is

Define unit vector, null vector and position vector .

If alpha and beta are acute such that alpha+beta and alpha-beta satisfy the equation tan^(2)theta-4tan theta+1=0 , then (alpha, beta ) =