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

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

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Answer 1 · 2018-04-06T09:02:13

See below.

Explanation:

Calling

${\to v}_{1}, {\to v}_{2}, {\to v}_{3}$ $vec v_1, vec v_2, vec v_3$ the unit vectors with the direction of $A - P, B - P$ $A-P, B-P$ and $C - P$ $C-P$ we have

$A = P + a {\to v}_{1}$ $A = P + a vec v_1$
$B = P + b {\to v}_{2}$ $B = P + b vec v_2$
$C = P + c {\to v}_{3}$ $C = P + c vec v_3$

Here

$a = 10$ $a = 10$
$b = 6$ $b = 6$

then

$B - A = b {\to v}_{2} - a {\to v}_{1}$ $B-A = b vec v_2-a vec v_1$
$C - B = c {\to v}_{3} - b {\to v}_{2}$ $C-B = c vec v_3 - b vec v_2$

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

$⟨ B - A, C - B ⟩ = ⟨ b {\to v}_{2} - a {\to v}_{1}, c {\to v}_{3} - b {\to v}_{2} ⟩ = 0$ $<< B-A, C-B >> = << b vec v_2-a vec v_1, c vec v_3 - b vec v_2 >> = 0$

$b c ⟨ {\to v}_{2}, {\to v}_{3} ⟩ - b^{2} {∥ ∥ {\to v}_{2} ∥ ∥}^{2} - a c ⟨ {\to v}_{1}, {\to v}_{3} ⟩ + a b ⟨ {\to v}_{1}, {\to v}_{2} ⟩ = 0$ $b c << vec v_2, vec v_3 >> - b^2 norm(vec v_2)^2-ac << vec v_1, vec v_3 >>+a b << vec v_1, vec v_2 >> =0$

and then

$c = \frac{b^{2} - a b ⟨ {\to v}_{1}, {\to v}_{2} ⟩}{b ⟨ {\to v}_{2}, {\to v}_{3} ⟩ - a ⟨ {\to v}_{1}, {\to v}_{3} ⟩} = 33$ $c = (b^2-ab << vec v_1, vec v_2>>)/(b << vec v_2, vec v_3 >> - a << vec v_1, vec v_3>>) = 33$

with

$⟨ {\to v}_{1}, {\to v}_{2} ⟩ = - \frac{1}{2}$ $<< vec v_1, vec v_2 >> =-1/2$
$⟨ {\to v}_{2}, {\to v}_{3} ⟩ = - \frac{1}{2}$ $<< vec v_2, vec v_3 >> =-1/2$
$⟨ {\to v}_{1}, {\to v}_{3} ⟩ = - \frac{1}{2}$ $<< vec v_1, vec v_3 >> = -1/2$

NOTE

$⟨ \cdot, \cdot ⟩$ $<< cdot, cdot >>$ represents the scalar product of two vectors.
$∥ \cdot ∥$ $norm(cdot)$ represents the norm operation.

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

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