The scalar ` vec Adot ( ( vec B+ vec C)xx( vec A+ vec B+ vec C))` equals a.`0` b. `[ vec A vec B vec C]+[ vec B vec C vec A]` c. `[ vec A vec B vec C]` d. none of these

[vec a + vec b, vec b + vec c, vec c + vec a] = 2 [vec a, vec b, vec c]

If vec adot vec b= vec adot vec c\ a n d\ vec axx vec b= vec axx vec c ,\ vec a!=0, then vec b= vec c b. vec b=0 c. vec b+ vec c=0 d. none of these

vec a * {(vec b + vec c) xx (vec a + 2vec b + 3vec c)} = [vec with bvec c]

[vec a+vec b,vec b+vec c,vec c+vec a]=2[vec a,vec b,vec c]

If vec A xx vec B= vec C xx vec B , show that vec C need not be equal to vec A .

[[vec a + vec b-vec c, vec b + vec c-vec a, vec c + vec a-vec b is equal to

Given vec a=x hat i+y hat j+2 hat k , vec b= hat i- hat j+ hat k , vec c= hat i+2 hat j ; vec a_|_ vec b , vec adot vec c=4. Then [ vec a vec b vec c]^2=| vec a| b. [ vec a vec b vec c]^=| vec a| c. [ vec a vec b vec c]^=0 d. [ vec a vec b vec c]^=| vec a|^2

A B C D E is pentagon, prove that vec A B + vec B C + vec C D + vec D E+ vec E A = vec0 vec A B+ vec A E+ vec B C+ vec D C+ vec E D+ vec A C=3 vec A C

The length of the perpendicular form the origin to the plane passing through the point a and containing the line vec r= vec b+lambda vec c is a. ([ vec a vec b vec c])/(| vec axx vec b+ vec bxx vec c+ vec cxx vec a|) b. ([ vec a vec b vec c])/(| vec axx vec b+ vec bxx vec c|) c. ([ vec a vec b vec c])/(| vec bxx vec c+ vec cxx vec a|) d. ([ vec a vec b vec c])/(| vec cxx vec a+ vec axx vec b|)