Question-188998

Question Number 188998 by SLVR last updated on 10/Mar/23

Commented by SLVR last updated on 10/Mar/23

k i n d l y h e l p m e

Commented by manxsol last updated on 12/Mar/23

Commented by manxsol last updated on 12/Mar/23

Answered by a.lgnaoui last updated on 10/Mar/23

Commented by SLVR last updated on 15/Mar/23

s i r \dots e x p l a n a t i o n i n e n g l i s h . .

w h e r e i c a n n o t r e m a i n i n g

Answered by manxsol last updated on 12/Mar/23

p t o g e n e r i c C^{'} (m e d i a n)

(x, y, z) = (2 t, t, 2 t)

p t o g e n e r i c B (b i s e c t)

(x, y, z) = (s, 2 s, 3 s)

\frac{A + B}{2} = C^{'} (A C^{'} B) c o l i n e a l

A + B = 2 C^{'}

(- 1, 1, 1) + (s, 2 s, 3 s) = (4 t, 2 t, 4 t)

- 1 + s = 4 t I s = 4 t + 1

1 + 2 s = 2 t I I

1 + 3 s = 4 t I I I

I I I - I I s = 2 t

2 t = 4 t + 1

t = \frac{- 1}{2} s = - 1

B = (- 1, - 2, - 3)

C^{'} = (- 2, - 1, - 2)

A = (- 1, 1, 1)

a n g u l o s

B A = (0, 3, 4)

b i s e c = (1, 2, 3)

B A . b i s e c =∣ B A ∣∣ b i s e c ∣∣ c o s θ

0.1 + 3.2 + 4.3 = 5 \sqrt{14} c o s θ

c o s θ = \frac{18}{5 \sqrt{14}}

B C = C - B

B C = (2 t, t, 2 t) + (1, 2, 3)

B C = (2 t + 1, t + 2, 2 t + 3)

B C ∙ b i s e c t =∣ B C ∣∣ b i s e c t ∣ \frac{18}{5 \sqrt{14}}

(2 t + 1, t + 2, 2 t + 3) . (1, 2, 3) =

\sqrt{9 t^{2} + 14 + 20 t} . \sqrt{14} \frac{18}{5 \sqrt{14}}

2 t + 1 + 2 t + 4 + 6 t + 9 = \frac{18 \sqrt{9 t^{2} + 14 + 20 t}}{5}

(10 t + 14) (5) = 18 \sqrt{9 t^{2} + 20 t + 14}

(5 t + 7) (5) = 9 \sqrt{9 t^{2} + 14 + 20 t} .

625 t^{2} + 1750 t + 1225 = 729 t^{2} + 1134 + 1620 t

104 t^{2} - 130 t - 91 = 0

t = - 0.5

t = 1.75

C = (2 t, t, 2 t)

C = (- 1, - \frac{1}{2}, - 1)

C = (\frac{7}{2}, \frac{7}{4}, \frac{7}{2})

B C = (2 t + 1, t + 2, 2 t + 3)

B C = (\frac{9}{2}, \frac{15}{4}, \frac{13}{2})

B C = (18, 15, 26)

L_{B C} :

(x, y, z) = (- 1, - 2, - 3) + t (18, 15, 26)

x + 1 = 18 t

y + 2 = 15 t

z + 3 = 26 t

\frac{x + 1}{18} = \frac{y + 2}{15} = \frac{z + 3}{26}

f o o t p e r p e r d i c u l a r P (s, 2 s, 3 s)

P A ∙ P B = 0

P A = (- 1, 1, 1) - (s, 2 s, 3 s)

P B = (- 1, - 2, - 3) - (s, 2 s, 3 s)

[- 1 - s, 1 - 2 s, 1 - 3 s) ∙ (- 1 - s, - 2 - 2 s, - 3 - 3 s

(1 + 2 s + s^{2}) + (4 s^{2} + 2 s - 2) + (9 s^{2} + 6 s - 3) = 0

14 s^{2} + 10 s - 4 = 0

7 s^{2} + 5 s - 2 = 0

7 s - 2

1 s + 1

s = - 1

s = \frac{2}{7}

P = (- 1, - 2, - 3) i s P t o B

P = (\frac{2}{7}, \frac{4}{7}, \frac{6}{7})

g r a p h i c y c o n c l u s i o n s

Commented by SLVR last updated on 15/Mar/23

s i r i c o u l d n o t f o l l o w w e l l .

w h i c h o p t i o n (s) c o r r e c t ?

i c a n f o l l o w e n g l i s h o n l y

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