Given $\vec{A_1}(1,2), \vec{A_2}(2,4), \vec{A_3}(3,b).$ find $b$ so that triangle $\triangle{A_1A_2A_3}$ will be a right-angled triangle

Question

Given $\vec{A_1}(1,2), \vec{A_2}(2,4), \vec{A_3}(3,b).$ find $b$ so that triangle $\triangle{A_1A_2A_3}$ will be a right-angled triangle.

I know that in order that $\triangle{A_1A_2A_3}$ will be right-angled ,

the angle $\theta$ between $A_i$ and $A_j$ ($i,j \in \{1,2,3\}, i \ne j)$ must be $90 \deg.$

this is my solution:

$|\vec{A_1}| = \sqrt{1^2 + 2^2} = \sqrt{5}.$

$|\vec{A_3}| = \sqrt{3^2 + b^2} = \sqrt{9 + b^2}.$

I Found the dot product between $\vec{A_1} \cdot \vec{A_3} = (3 + 2b)$.

$\cos\theta = \frac{3 + 2b}{\sqrt{5}\sqrt{9 + b^2}} = \frac{3 + 2b}{\sqrt{5(9+b^2)}}.$

$\theta = 90 \iff \cos \theta = 0$, so the only solution for $b = -1.5 = \frac{-3}{2}$

Back to out equation:

$cos\theta = \frac{3 + 2\cdot \frac{-3}{2}}{\sqrt{5(9 + b^2)}} = \frac{3 + (-3)}{\sqrt{5(9 + b^2)}} = \frac{0}{\sqrt{5(9 + b^2)}} = 0$.

$\theta = \arccos 0 = 90.$

So $b$ must equal to $-1.5$.

When I plotted it I didn't noticed any $90 \deg$ angel.

When is my problem? thanks in advance!

Answer 1

Hint: solve $\langle \overrightarrow{A_3A_1},\overrightarrow{A_3A_2}\rangle=0$, $\langle \overrightarrow{A_2A_1},\overrightarrow{A_2A_3}\rangle=0$, and $\langle \overrightarrow{A_1A_2},\overrightarrow{A_1A_3}\rangle=0$ for $b$. I've got $b=1$ and $b=5/2$.