# Thread: s6.12.7 Find the lengths of the ides of the triangle PQR

1. \tiny{s6.12.7}\\\textsf{7.Find the lengths of the ides of the triangle PQR.}\\ \textsf{Is it a right triangle? Is it an isosceles riangle?}\\ \begin{align}\displaystyle&P(3, -2, -3), \; Q(7,0,1), \; R(1,2,1)\\&\left| PQ \right|=\sqrt{(3-7)^2 +(-2-0)^2+(-3-1)^2}\end{align} \textit{this is just an intro to vector calculus to get the basics of calc III which starts in Jan } ☕ 2. I believe it is isosceles, with equal sides 6 units and third side 2\sqrt{10} units. Use (for example)\left|\vec{PQ}\right|=\sqrt{(P_1-Q_1)^2+(P_2-Q_2)^2+(P_3-Q_3)^2}$Can you finish? 3. Thread Author$\tiny{s6.12.7}\\\textsf{7.Find the lengths of the ides of the triangle PQR.}\\\textsf{Is it a right triangle? Is it an isosceles riangle?}
\\\begin{align} &P(3, -2, -3), \; Q(7,0,1), \; R(1,2,1)\\ \left|\vec{PQ}\right| &=\sqrt{(P_1-Q_1)^2+(P_2-Q_2)^2+(P_3-Q_3)^2}\\ &=\sqrt{(3-7)^2 +(-2-0)^2+(-3-1)^2}=6\\ \left|\vec{QR}\right| &=\sqrt{(7-1)^2 +(0-2)^2+(-1-1)^2}=\sqrt{40}\\ \left|\vec{RP}\right| &=\sqrt{(3-1)^2+(-2-2)^2+(-3-1)^2}=6 \end{align}\textit{isosceles}\$

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