if $\vec v_1$ and $\vec v_2 $ are both $3$ by $1$ matrix ,now i new vector $f_A$

$\vec f_A=\alpha \vec v_1+\beta \vec v_2$,and if i want to normalize the $\vec f_A$ to let $\vec f_A ^T\vec f_A=1$,how do i rewrite the formula

$\vec n_{f_A}=\frac{\vec f_A}{||\vec f_A||}=\frac{\alpha \vec v_1+\beta \vec v_2}{\sqrt{\alpha^2 \vec v_1^T\vec v_1+\beta^2 \vec v_2 ^T\vec v_2}}?$

Because i found that $\vec n_{f_A} \vec n_{f_A}^T \neq 1 ,$so it must be wrong somewhere in my normalization of $\vec f_A$

  • $\begingroup$ Why is $\| f_A \| = \sqrt{\alpha^2v_1^tv_1+\beta^2v_2^tv_2}$? $\endgroup$
    – azif00
    Oct 8 '19 at 2:08
  • $\begingroup$ @Azif00 because i want to find the length of $f_A$,that is ,the square of the inner product of $f_A$ $\endgroup$
    – XM551
    Oct 8 '19 at 2:25

It should be $$\hat{\vec {f}}_A=\frac{\vec{f}_A}{\sqrt{(\vec{f}_A)^T \vec{f}_A}},$$ where $T$ denotes transpose. In denominator $v_1^T v_2$ and $v_2^T v_1$ will also be there. In case, these two are orthogonal these terms will vanish.


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