MHB *8s6.12.7 Find the lengths of the ides of the triangle PQR

[karush]

$\tiny{s6.12.7 Linkedin}$
a. Find the lengths of the ides of the triangle PQR.
b. Is it a right triangle? Is it an isosceles triangle?
$P(3,-2,-3), Q(7,0,1), R(1,2,1) PQ =\left|\sqrt{(3-7)^2 +(-2-0)^2+(-3-1)^2}\right|$
this is just an intro to vector calculus to get the basics of calc III

 

[Greg]

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?

 

[karush]

$\tiny{s6.12.7}\\$
$\textsf{7.Find the lengths of the ides of the triangle PQR.}\\$ $\textsf{Is it a right triangle? Is it an isosceles riangle?}
\\$ \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}
$\textit{isosceles}$