Find-the-shortest-distance-from-the-point-P-2-3-5-to-the-line-L-x-3-2-y-1-3-z-2-4-

Question Number 96240 by mr W last updated on 30/May/20

F i n d t h e s h o r t e s t d i s t a n c e f r o m t h e

p o i n t P (2, - 3, 5) t o t h e l i n e L

\frac{x + 3}{2} = \frac{y - 1}{- 3} = \frac{z - 2}{4}

Answered by 1549442205 last updated on 31/May/20

Denote Q be the point on L such that

PQ ⊥ L \Rightarrow Q (2 t - 3; - 3 t + 1; 4 t + 2) . Then

\vec{PQ} = (2 t - 5; - 3 t + 4; 4 t - 3) . The direction

vector of L is \vec{p} = (2; - 3; 4) perpemdicular

to \vec{PQ} . Hence, \vec{p} . \vec{PQ} = 0 \Leftrightarrow 2 (2 t - 5) + 3 (3 t - 4)

+ 4 (4 t - 3) \Leftrightarrow 29 t - 34 = 0 \Rightarrow t = \frac{34}{29} . From that

\vec{PQ} = (\frac{- 77}{29}; \frac{14}{29}; \frac{49}{29}) \Rightarrow PQ = \frac{7 \sqrt{1 \overset{2}{1} + 2^{2} + 7^{2}}}{29} =

\frac{7 \sqrt{174}}{29} is shortest distance from P to L

Commented by mr W last updated on 31/May/20

t h a n k y o u s i r!

Answered by john santu last updated on 31/May/20

let S (2 s - 3, - 3 s + 1, 5 s + 2) a point

on line L . take vector PS = (2 s - 5, - 3 s + 4, 5 s - 3)

PS ⊥ n = (2, - 3, 4)

4 s - 10 + 9 s - 12 + 20 s - 12 = 0

33 s = 34 \Rightarrow s = \frac{34}{33}

PS (- \frac{97}{33}, \frac{30}{33}, \frac{71}{33}) .

∣ PS ∣ = \frac{\sqrt{15, 350}}{33} \approx 3.754397483

Commented by mr W last updated on 31/May/20

t h a n k y o u s i r!

t y p o i n f i r s t l i n e 5 s + 2 ?

s = \frac{34}{29} ?

Commented by bobhans last updated on 31/May/20

mr santu typo in 5 s + 2

it should be 4 s + 2

Answered by mr W last updated on 31/May/20

M E T H O D I

a n y p o i n t o n l i n e Q :

x = - 3 + 2 k

y = 1 - 3 k

z = 2 + 4 k

d i s t a n c e P Q = d

D = d^{2} = {(2 + 3 - 2 k)}^{2} + {(- 3 - 1 + 3 k)}^{2} + {(5 - 2 - 4 k)}^{2}

D = {(2 k - 5)}^{2} + {(3 k - 4)}^{2} + {(4 k - 3)}^{2}

s u c h t h a t d i s m i n i m u m :

\frac{d D}{d k} = 2 (2 k - 5) 2 + 2 (3 k - 4) 3 + 2 (4 k - 3) 4 = 0

\Rightarrow 29 k = 34

\Rightarrow k = \frac{34}{29}

d_{m i n} = \sqrt{{(2 \times \frac{34}{29} - 5)}^{2} + {(3 \times \frac{34}{29} - 4)}^{2} + {(4 \times \frac{34}{29} - 3)}^{2}} = \frac{7 \sqrt{174}}{29}

M E T H O D I I

k n o w n p o i n t o n l i n e R (- 3, 1, 2)

\vec{P R} = (5, - 4, 3)

\vec{l} = (2, - 3, 4)

\cos θ = \frac{P R \cdot l}{∣ P R ∣∣ l ∣}

d =∣ P R ∣ \sin θ =∣ P R ∣ \sqrt{1 - {(\frac{P R \cdot l}{∣ P R ∣∣ l ∣})}^{2}}

d = \sqrt{∣ P R ∣^{2} - \frac{{(P R \cdot l)}^{2}}{∣ l ∣^{2}}}

d = \sqrt{5^{2} + 4^{2} + 3^{2} - \frac{{(5 \times 2 + 4 \times 3 + 3 \times 4)}^{2}}{2^{2} + 3^{2} + 4^{2}}}

d = \sqrt{50 - \frac{34^{2}}{29}} = \frac{7 \sqrt{174}}{29}

M E T H O D I I I

p l a n e p e r p e n d i c u l a r t o l i n e L i s

2 x - 3 y + 4 z + s = 0

p o i n t P i s o n t h e p l a n e :

2 \times 2 - 3 \times (- 3) + 4 \times 5 + s = 0

s = - 33

\Rightarrow 2 x - 3 y + 4 z - 33 = 0

i n t e r s e c t i o n o f l i n e w i t h p l a n e a t Q :

x = - 3 + 2 k

y = 1 - 3 k

z = 2 + 4 k

2 (- 3 + 2 k) - 3 (1 - 3 k) + 4 (2 + 4 k) - 33 = 0

29 k - 34 = 0

\Rightarrow k = \frac{34}{29}

P Q = \sqrt{{(2 + 3 - 2 k)}^{2} + {(- 3 - 1 + 3 k)}^{2} + {(5 - 2 - 4 k)}^{2}}

P Q = \sqrt{{(5 - 2 k)}^{2} + {(- 4 + 3 k)}^{2} + {(3 - 4 k)}^{2}}

P Q = \sqrt{{(5 - 2 \times \frac{34}{29})}^{2} + {(- 4 + 3 \times \frac{34}{29})}^{2} + {(3 - 4 \times \frac{34}{29})}^{2}}

d_{m i n} = P Q = \frac{7 \sqrt{174}}{29}

M E T H O D I V

p o i n t o n l i n e Q (- 3 + 2 k, 1 - 3 k, 2 + 4 k)

\vec{l} = (2, - 3, 4)

\vec{P Q} = (2 + 3 - 2 k, - 3 - 1 + 3 k, 5 - 2 - 4 k)

\vec{P Q} = (5 - 2 k, - 4 + 3 k, 3 - 4 k)

s u c h t h a t \vec{P Q} ⊥ \vec{l} :

(5 - 2 k) \times 2 \times (- 4 + 3 k) \times (- 3) + (3 - 4 k) \times 4 = 0

\Rightarrow k = \frac{34}{29}

d_{m i n} =∣ P Q ∣= \sqrt{{(5 - 2 \times \frac{34}{29})}^{2} + {(- 4 + 3 \times \frac{34}{29})}^{2} + {(3 - 4 \times \frac{34}{29})}^{2}}

= \frac{7 \sqrt{174}}{29}

M E T H O D V

k n o w n p o i n t o n l i n e R (- 3, 1, 2)

\vec{P R} = (5, - 4, 3)

\vec{l} = (2, - 3, 4)

∣ \vec{l} ∣= \sqrt{2^{2} + 3^{2} + 4^{2}} = \sqrt{29}

\vec{P R} \times \vec{l} = (5, - 4, 3) \times (2, - 3, 4) = (- 7, - 14, - 7)

∣ \vec{P R} \times \vec{l} ∣= \sqrt{7^{2} + 14^{2} + 7^{2}} = 7 \sqrt{6}

d = \frac{∣ P R \times \vec{l} ∣}{∣ \vec{l} ∣} = \frac{7 \sqrt{6}}{\sqrt{29}} = \frac{7 \sqrt{174}}{29}

Commented by bobhans last updated on 31/May/20

waw ==== great

Commented by mr W last updated on 31/May/20

M e t h o d I I i s m y f a v o u r i t e m e t h o d

d_{m i n} = \sqrt{∣ \vec{R P} ∣^{2} - \frac{{(\vec{R P} ∙ \vec{l})}^{2}}{∣ \vec{l} ∣^{2}}}

t h i s f o r m u l a i s n o t t e a c h e d i n s c h o o l

a n d i n b o o k .

Commented by mr W last updated on 31/May/20

Commented by bobhans last updated on 31/May/20

whether the point R is arbitrarily provided it lies on the L line

Commented by bobhans last updated on 31/May/20

this method from Pytagorean theorem ?

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