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Mathematics 15 Online
OpenStudy (anonymous):

if A is any vector then prove that A=(A.i)i+(A.j)j+(A.k)k

OpenStudy (lalaly):

Let \[A = A_x i + A_y j + A_z k\] Ax, Ay and Az are components of the vector SO \[A.i = (A_x i + A_y j + A_z k) . i = A_x\] \[A .j = (A_x i + A_y j + A_z k) . j = A_y\] \[A.k = (A_x i + A_y j + A_z k) . k = A_z\] Therfore \[(A.i)i + (A.j) j + (A. k) k = A_x i + A_y j + A_z k=A\]

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