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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!

Gerry Myerson
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Billie
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  • You write about points $A_1,A_2,A_3$, but about triangle $ABC$. Can you edit for consistency, please? – Gerry Myerson Nov 19 '13 at 11:51
  • @GerryMyerson Im so sorry. I started to write this question with vectors A,B,C, but after that I changed to $A_1A_2A_3$. – Billie Nov 19 '13 at 11:53
  • Your dot product calculation will make the line from the origin to $A_1$ perpendicular to the line from the origin to $A_3$, but that's not what you want. – Gerry Myerson Nov 19 '13 at 11:54
  • Consider the line segment $L$ joining $(1,2)$ and $(2,4)$. Find the equation of a line $M$ perpendicular to $L$ and passing through one of the endpoints. Then find $b$ such that $M$ intersects $x=3$ at $(3,b)$. – Gerry Myerson Nov 19 '13 at 11:56
  • Tip: apply a translation: $B_{i}=A_{i}-A_{1}$ for $i=1,2,3$. Then $B_{1}=\left(0,0\right)$, $B_{2}=\left(1,2\right)$ and $B_{3}=\left(2,b'\right)$ for $b'=b-2$. Then triangle $A_{1}A_{2}A_{3}$ will be right-angled if and only if triangle $B_{1}B_{2}B_{3}$is right-angled and the second triangle is easyer to handle. – drhab Nov 19 '13 at 12:29

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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 $b=5/2$.

Michael Hoppe
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