Find a vector satisfying given relations with a given vector

Find a vector $B \in \mathbb{R}^2$ such that $B \cdot A = 0$ and $\lVert B \rVert = \lVert A \rVert$ for the following vectors $A$ ,

$A = (1,1)$ ;
$A = (1,-1)$ ;
$A = (2,-3)$ ;
$A = (a,b)$ .

If $A = (1,1)$ then letting $B = (b_1, b_2)$ we have
$B \cdot A = 0 \quad \implies \quad b_1 + b_2 = 0 \quad \implies \quad b_2 = -b_1.$

Since $\lVert A \rVert = \sqrt{2}$ we have $\lVert B \rVert = \sqrt{2}$ therefore,

$\sqrt{b_1^2 + b_2^2} =\sqrt{2} \quad \implies \quad 2b_1^2 = 2 \quad \implies \quad b_1 = \pm 1, \ b_2 = \mp 1.$

Therefore, $B = (1,-1)$ or $B = (-1,1)$ .
If $A = (1,-1)$ then letting $B = (b_1, b_2)$ we have
$B \cdot A = 0 \quad \implies \quad b_1 - b_2 = 0 \quad \implies \quad b_2 = b_1.$

Since $\lVert A \rVert = \sqrt{2}$ we have $\lVert B \rVert = \sqrt{2}$ therefore,

$\sqrt{b_1^2 + b_2^2} =\sqrt{2} \quad \implies \quad 2b_1^2 = 2 \quad \implies \quad b_1 = \pm 1, \ b_2 = \pm 1.$

Therefore, $B = (1,1)$ or $B = (-1,-1)$ .
If $A = (2,-3)$ then letting $B = (b_1, b_2)$ we have
$B \cdot A = 0 \quad \implies \quad 2b_1 - 3b_2 = 0 \quad \implies \quad b_1 = \frac{3}{2} b_2.$

Since $\lVert A \rVert = \sqrt{13}$ we have $\lVert B \rVert = \sqrt{13}$ therefore,

$\sqrt{b_1^2 + b_2^2} =\sqrt{13} \quad \implies \quad \sqrt{ \left( \frac{9}{4} \right) b_2^2 + b_2^2 } = \sqrt{13} \quad \implies \quad b_2 = \pm 2, \ b_2 = \pm 3.$

Therefore, $B = (3,2)$ or $B = (-3,-2)$ .
If $A = (a,b)$ then letting $B = (b_1, b_2)$ we have
$B \cdot A = 0 \quad \implies \quad ab_1 + bb_2 = 0 \quad \implies \quad b_1 = \frac{b}{a} b_2.$

Since $\lVert A \rVert = \sqrt{a^2 + b^2}$ we have $\lVert B \rVert = \sqrt{a^2 + b^2}$ therefore,

$\sqrt{b_1^2 + b_2^2} =\sqrt{2} \quad \implies \quad \sqrt{ \left( \frac{b}{a} \right) b_2^2 + b_2^2 } = \sqrt{a^2+b^2} \quad \implies \quad b_1= \pm b \quad b_2 = \mp a.$

(The final equalities follow from $b_2^2 = \pm (a^2+b^2) \left( \left(\frac{b}{a}\right)^2 + 1 \right)^{-1}$ .) Therefore we have $B = (b,-a)$ or $B = (-b,a)$ .