Find the distance from the point S(1,1,5) to the line L : { ((x=1+t)),((y=3−t )),((z=2t)) :}.


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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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