Find-distance-of-point-1-1-1-from-the-line-passing-through-2-3-4-amp-1-2-3-

Question Number 45259 by rahul 19 last updated on 11/Oct/18

F i n d d i s t a n c e o f p o i n t (1, 1, 1) f r o m

t h e l i n e p a s s i n g t h r o u g h (2, 3, 4) &

(- 1, 2, 3) ?

Commented by rahul 19 last updated on 11/Oct/18

A n o t h e r m e t h o d :

L e t P (1, 1, 1), Q (2, 3, 4), R (- 1, 2, 3)

a n d M b e f o o t o f p e r p e n d i c u l a r \dots

N o w i n Δ P Q R,

a r e a o f Δ = \frac{1}{2} ∣ P Q \times Q R ∣= \frac{1}{2} ∣ P M \times Q R ∣

\Rightarrow ∣ (i + 2 j + 3 k) \times (3 i + j + k) ∣= P M \times \sqrt{11}

\Rightarrow P M = \sqrt{\frac{90}{11}}

Answered by ajfour last updated on 11/Oct/18

l e t e q . o f l i n e b e : \bar{r} = \bar{a} + λ \bar{b}

\bar{b} = 3 \hat{i} + \hat{j} + \hat{k}; ∣ \bar{b} ∣= \sqrt{11}

\bar{d} = \frac{1}{\sqrt{11}} | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix} |

= \frac{1}{\sqrt{11}} (- \hat{i} + 8 \hat{j} - 5 \hat{k})

∣ \bar{d} ∣ = \sqrt{\frac{90}{11}} .

Commented by rahul 19 last updated on 11/Oct/18

Thank you sir !

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Answered by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

e q n o f s t l i n e p a s s i n g t h r o u g h (2, 3, 4) a n d (- 1, 2, 3)

i s \frac{x - 2}{- 1 - 2} = \frac{y - 3}{2 - 3} = \frac{z - 4}{3 - 4} = r

(- 3 r + 2, - r + 3, - r + 4) p o i n t l i e s o n s t l i n e a n d

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

d r s w n f r o m (1, 1, 1) o n t h e s t l i n e .

d i r e c t i o n r a t i o b e t w e e n (1, 1, 1) a n d (- 3 r + 2, - r + 3, - r + 4)

i s (- 3 r + 1, - r + 2, - r + 3)

n o w (- 3 r + 1) \times (- 3) + (- r + 2) \times (- 1) + (- r + 3) \times - 1 = 0

9 r - 3 + r - 2 + r - 3 = 0

11 r = 8 r = \frac{8}{11}

n o w d i s t a n c e b d t w e e n (1, 1, 1) a n d (- 3 r + 2, - r + 3, - r + 4)

\sqrt{{(- 3 r + 2 - 1)}^{2} + {(- r + 3 - 1)}^{2} + {(- r + 4 - 1)}^{2}}

= \sqrt{({(- 3 r + 1)}^{2} + {(- r + 2)}^{2} + {(- r + 3)}^{2}}

= \sqrt{{(\frac{- 24 + 11}{11})}^{2} + {(\frac{- 8 + 22}{11})}^{2} + {(\frac{- 8 + 33}{11})}^{2}}

= \sqrt{\frac{169}{121} + \frac{196}{121} + \frac{625}{121}}

= \sqrt{\frac{990}{121}} = \sqrt{\frac{90}{11}}

Commented by rahul 19 last updated on 11/Oct/18

thank you sir

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