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    #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} $\textit{this is just an intro to vector calculus to get the basics of calc III which starts in Jan }$ ☕
    Last edited by karush; December 10th, 2016 at 23:50.

  2. Perseverance
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    #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?

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    #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}
    $\textit{isosceles}$
    Last edited by karush; December 11th, 2016 at 00:50.

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