# What is the length PC? How did you solve this?

Apr 6, 2018

See below.

#### Explanation:

Calling

${\vec{v}}_{1} , {\vec{v}}_{2} , {\vec{v}}_{3}$ the unit vectors with the direction of $A - P , B - P$ and $C - P$ we have

$A = P + a {\vec{v}}_{1}$
$B = P + b {\vec{v}}_{2}$
$C = P + c {\vec{v}}_{3}$

Here

$a = 10$
$b = 6$

then

$B - A = b {\vec{v}}_{2} - a {\vec{v}}_{1}$
$C - B = c {\vec{v}}_{3} - b {\vec{v}}_{2}$

but $B - A$ and $C - B$ are perpendicular then

$\left\langleB - A , C - B\right\rangle = \left\langleb {\vec{v}}_{2} - a {\vec{v}}_{1} , c {\vec{v}}_{3} - b {\vec{v}}_{2}\right\rangle = 0$

$b c \left\langle{\vec{v}}_{2} , {\vec{v}}_{3}\right\rangle - {b}^{2} {\left\lVert {\vec{v}}_{2} \right\rVert}^{2} - a c \left\langle{\vec{v}}_{1} , {\vec{v}}_{3}\right\rangle + a b \left\langle{\vec{v}}_{1} , {\vec{v}}_{2}\right\rangle = 0$

and then

$c = \frac{{b}^{2} - a b \left\langle{\vec{v}}_{1} , {\vec{v}}_{2}\right\rangle}{b \left\langle{\vec{v}}_{2} , {\vec{v}}_{3}\right\rangle - a \left\langle{\vec{v}}_{1} , {\vec{v}}_{3}\right\rangle} = 33$

with

$\left\langle{\vec{v}}_{1} , {\vec{v}}_{2}\right\rangle = - \frac{1}{2}$
$\left\langle{\vec{v}}_{2} , {\vec{v}}_{3}\right\rangle = - \frac{1}{2}$
$\left\langle{\vec{v}}_{1} , {\vec{v}}_{3}\right\rangle = - \frac{1}{2}$

NOTE

$\left\langle\cdot , \cdot\right\rangle$ represents the scalar product of two vectors.
$\left\lVert \cdot \right\rVert$ represents the norm operation.