-3
$\begingroup$

can someone help me understand this?

enter image description here

https://gyazo.com/d828940ff01bd1daf1f4cd6b437d9529

$A_2$, $R_2$, $R_3$ are $3\times 1$ vectors and $d$ and $e$ are constants. $R_2$ and $R_3$ are orthogonal with length one.

  • 1
    Perhaps someone with the gift of divination, mind reading or something like that: the link you gave is lacking lines, so you better find them and write down the whole thing here. You can use http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – 2017-02-07
  • 0
    Vector of length one is a vector with length equal to one, e.g. in 2D $(1,0)$, $(0,1)$ or $(1/\sqrt{2},1/\sqrt{2})$. Orthogonal vectors are vectors that are perpendicular, e.g. in 2D $(1,0)$ and $(0,1)$. See for more: https://en.wikipedia.org/wiki/Orthogonality – 2017-02-07

1 Answers 1

2

Orthogonal means $R_2 \cdot R_3 = 0$, length one means $R_2 \cdot R_2 = 1$ etc., the dot means the scalar product. Note that we can write the standard scalar product also as matrix products: $$ a \cdot b = b \cdot a = a^T b = b^T a $$ (in case of column vectors).

So $$ A_2^T R_3 = A_2 \cdot R_3 = (d R_2 + e R_3) \cdot R_3 = d R_2 \cdot R_3 + e R_3 \cdot R_3 = d \cdot 0 + e \cdot 1 = e $$ similar $$ A_2^T R_2 = A_2 \cdot R_2 = (d R_2 + e R_3) \cdot R_2 = d R_2 \cdot R_2 + e R_3 \cdot R_2 = d \cdot 1 + e \cdot 0 = d $$

  • 0
    Thank you! I was confused by the transpose and didn't realize that $A_2^{T}R_3 = A_2*R_3$. Makes sense now – 2017-02-07