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`

The vector equation of the plane passing through vec a,vec b,vec c is vec r=alphavec a+betavec b+gammavec c provided that alpha+beta+gamma=0 b.alpha+beta+gamma=1 c.alpha+beta=gamma dalpha^(2)+beta^(2)+gamma^(2)=1

If vec alpha+vec beta+vec gamma=0, prove the given equation

If |vec alpha + vec beta |=| vec alpha - vec beta|, then :

vec vec alpha + vec beta + vec gamma = 0, provet vec vec alphavec beta = vec betavec x gamma = vec gamma xvec alpha

If A=[alpha beta gamma-alpha] is such that A^(2)=I, then 1+alpha^(2)+beta gamma=0( b) 1-alpha^(2)+beta gamma=0 (c) 1-alpha^(2)-beta gamma=0 (d) 1+alpha^(2)-beta gamma=0

If A=[[alpha ,beta],[ gamma , -alpha]] is such that A^2=I , then A) 1+alpha^2+beta gamma=0 B) 1-alpha^2+beta gamma=0 C) 1-alpha^2-beta gamma=0^ D) 1+alpha^2-beta gamma=0

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

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

If quad vec r = alpha (vec b xxvec c) + beta (vec c xxvec a) + gamma (vec a xxvec b) and [vec with bvec c] = 2 then alpha + beta + gamma

Evaluate: |[alpha, beta,gamma],[alpha^2,beta^2,gamma^2],[beta+gamma,gamma+alpha,alpha+beta]|