For any two vectors ` -> a`and ` -> b`we always have `| -> adot -> b|lt=| -> a|| -> b|`(Cauchy-Schwartz inequality).

For any two sets A and B, A' – B'=

For any two matrices A and B , we have

For any two vectors vec a\ a n d\ vec b write when | vec a+ vec b|=| vec a|+| vec b| holds.

For any vector vec a\ a n d\ vec b prove that | vec a+ vec b|lt=| vec a|+| vec b|dot

For any two vectors vec aa n d vec b , prove that | vec a+ vec b|lt=| vec a|+| vec b| (ii) | vec a- vec b|lt=| vec a|+| vec b| (iii) | vec a- vec b|geq| vec a|-| vec b|

For any two vectors vec a\ a n d\ vec b , fin d\ ( vec axx vec b). vecbdot

For any two vectors vec a and vec b , show that : ( vec a+ vec b)dot( vec a- vec b)=0, when | vec a|=| vec b|dot

For any two vectors vec a and vec b , prove that: | vec a+ vec b|^2=| vec a|^2+| vec b|^2+2 vec adot vec b , | vec a- vec b|^2=| vec a|^2+| vec b|^2-2 vec adot vec b , | vec a+ vec b|^2+| vec a- vec b|^2=2(| vec a|^2+| vec b|^2) and | vec a+ vec b|^2=| vec a- vec b|^2 iff vec a_|_ vec bdot Interpret the result geometrically.

For two vectors vec A and vec B | vec A + vec B| = | vec A- vec B| is always true when.

If vec aa n d vec b are two vectors and angle between them is theta, then | vec axx vec b|^2+( vec adot vec b)^2=| vec a|^2| vec b|^2 | vec axx vec b|=( vec adot vec b),iftheta=pi//4 vec axx vec b=( vec adot vec b) hat n ,(w h e r e hat n is unit vector, ) if theta=pi//4 ( vec axx vec b)dot( vec a+ vec b)=0