Find-the-distance-from-the-point-S-1-1-5-to-the-line-L-x-1-t-y-3-t-z-2t-

Question Number 123876 by john_santu last updated on 29/Nov/20

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

p o i n t S (1, 1, 5) t o t h e l i n e

L : {\begin{cases} x = 1 + t \\ y = 3 - t \\ z = 2 t \end{cases} .

Answered by mr W last updated on 29/Nov/20

M e t h o d I

d^{2} = {(1 + t - 1)}^{2} + {(3 - t - 1)}^{2} + {(2 t - 5)}^{2}

= 6 t^{2} - 24 t + 29

\frac{d (d^{2})}{d t} = 12 t - 24 = 0

\Rightarrow t = 2

d_{m i n}^{2} = 6 \times 2^{2} - 24 \times 2 + 29 = 5

\Rightarrow d i s t a n c e = d_{m i n} = \sqrt{5}

Answered by mr W last updated on 29/Nov/20

M e t h o d I I

S (1, 1, 5)

P (1, 3, 0)

P Q = (1, - 1, 2)

P S = (0, 2, - 5)

d i s t a n c e = \sqrt{∣ P S ∣^{2} - {(\frac{P Q ∙ P S}{∣ P Q ∣})}^{2}}

= \sqrt{0^{2} + 2^{2} + {(- 5)}^{2} - \frac{{(1 \times 0 - 1 \times 2 - 2 \times 5)}^{2}}{1^{2} + {(- 1)}^{2} + 2^{2}}}

= \sqrt{29 - 24}

= \sqrt{5}

Answered by mr W last updated on 29/Nov/20

M e t h o d I I I

S (1, 1, 5)

P (1, 3, 0)

S P = (0, 2, - 5)

P Q = (1, - 1, 2)

d i s t a n c e = \frac{∣ S P \times P Q ∣}{∣ P Q ∣} = \frac{∣ (0, 2, - 5) \times (1, - 1, 2) ∣}{\sqrt{1^{2} + {(- 1)}^{2} + 2^{2}}}

= \frac{\sqrt{{(- 1)}^{2} + {(- 5)}^{2} + {(- 2)}^{2}}}{\sqrt{6}}

= \sqrt{\frac{30}{6}}

= \sqrt{5}

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