Find-the-shortest-distance-between-the-lines-L-1-4-2-N-1-3-2-and-r-1-1-1-1-2-1-

Question Number 56697 by Tawa1 last updated on 21/Mar/19

Find the shortest distance between the lines

L = (1, 4, 2) + N (1, 3, 2) and

r = (- 1, 1, - 1) + λ (1, 2, - 1)

Answered by mr W last updated on 21/Mar/19

m a n y m e t h o d s t o s o l v e .

d^{2} = {[1 + N - (- 1 + λ)]}^{2} + {[4 + 3 N - (1 + 2 λ)]}^{2} + {[2 + 2 N - (- 1 - λ)]}^{2}

d^{2} = {(2 + N - λ)}^{2} + {(3 + 3 N - 2 λ)}^{2} + {(3 + 2 N + λ)}^{2} \dots (i i i)

\frac{\partial (d^{2})}{\partial N} = 2 (2 + N - λ) + 2 \times 3 (3 + 3 N - 2 λ) + 2 \times 2 (3 + 2 N + λ) = 0

\Rightarrow 14 N - 5 λ + 17 = 0 \dots (i)

\frac{\partial (d^{2})}{\partial λ} = - 2 (2 + N - λ) - 2 \times 2 (3 + 3 N - 2 λ) + 2 (3 + 2 N + λ) = 0

\Rightarrow 5 N - 6 λ + 5 = 0 \dots (i i)

s o l v e (i) a n d (i i) :

\Rightarrow N = - \frac{77}{59}

\Rightarrow λ = - \frac{15}{59}

p u t t h i s i n t o (i i i) :

d_{m i n}^{2} = {(\frac{56}{59})}^{2} + {(\frac{- 24}{59})}^{2} + {(\frac{8}{59})}^{2}

\Rightarrow d_{m i n} = \frac{\sqrt{56^{2} + 24^{2} + 8^{2}}}{59} = \frac{8}{\sqrt{59}} = 1.0415

Commented by Tawa1 last updated on 22/Mar/19

Wow, God bless you sir .

Answered by mr W last updated on 23/Mar/19

v e c t o r m e t h o d

n o r m a l t o L a n d r :

(1, 3, 2) \times (1, 2, - 1) = (- 7, 3, - 1)

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

(2, 3, 3) ∙ (- 7, 3, - 1) = - 14 + 9 - 3 = - 8

d = \frac{∣ - 8 ∣}{\sqrt{{(- 7)}^{2} + 3^{2} + {(- 1)}^{2}}} = \frac{8}{\sqrt{59}} \approx 1.0415

Commented by Tawa1 last updated on 24/Mar/19

God bless you sir . I appreciate .

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