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Question Number 63427 by Rio Michael last updated on 04/Jul/19

(1) A p l a n e c o n t a i n s t h e l i n e s \frac{x + 1}{2} = \frac{4 - y}{2} = \frac{z - 2}{3} a n d

r = (2 i + 2 j + 12 k) + t (- i + 2 j + 4 k) . f i n d

(a) t h e a n g l e b e t w e e n t h e s e l i n e s .

(b) A c a r t e s i a n e q u a t i o n o f t h e p l a n e .

(2) G i v e n t h e l i n e s l_{1} : \frac{x - 10}{3} = \frac{y - 1}{1} = \frac{z - 9}{4} l_{2} : r = (- 9 j + 13 k) + μ (i + 2 j - 3 k)

w h e r e μ i s a p a r a m e t e r; l_{3} : \frac{x + 10}{4} = \frac{y + 5}{3} = \frac{z + 4}{1} .

a) s h o w t h a t t h e p o i n t (4, - 1, 1) i s c o m m o n t o l_{1} a n d l_{2} . F i n d

b) t h e p o i n t o f i n t e r s e c t i o n o f l_{2} a n d l_{3} .

c) A v e c t o r p a r a m e t r i c e q u a t i o n o f t h e p l a n e c o n t a i n i n g t h e

l i n e s l_{2} a n d l_{3} .

s i r F o r k u m M i c h a e l

Commented by Tony Lin last updated on 04/Jul/19

(2) (b)

l e t {\begin{cases} l_{2} : \frac{x}{1} = \frac{y + 9}{2} = \frac{z - 13}{- 3} = μ \\ l_{3} : \frac{x + 10}{4} = \frac{y + 5}{3} = \frac{z + 4}{1} = s \end{cases}

\Rightarrow μ = - 10 + 4 s

- 9 + 2 μ = - 5 + 3 s

13 - 3 μ = - 4 + s

\Rightarrow n o s o l u t i o n t o t h e e q a t i o n s

\Rightarrow l_{2} & l_{3} a r e s k e w l i n e s

\Rightarrow t h e r e i s n o p o i n t o f i n t e r s e c t i o n o f l_{2} & l_{3}

(2) (c)

∵ l_{2} & l_{3} a r e s k e w l i n e s

∴ t h e r e i s n o p l a n e c o n t a i n i n g t h e l i n e s

Commented by Tony Lin last updated on 04/Jul/19

(1) (a)

(1) c o s θ = \frac{(2, - 2, 3) \cdot (- 1, 2, 4)}{\sqrt{2^{2} + {(- 2)}^{2} + 3^{2}} \sqrt{1^{2} + 2^{2} + 4^{2}}} = \frac{6}{\sqrt{357}}

\Rightarrow θ = {c o s}^{- 1} (\frac{6}{\sqrt{357}})

(1) (b)

{\overset{⇀}{n}}_{E} = (2, - 2, 3) \times (- 1, 2, 4) = (- 14, - 11, 2)

a n d (- 1, 4, 2), (2, 2, 12) i s o n t h e p l a n e

\Rightarrow E : 14 x + 11 y - 2 z = 26

Commented by Tony Lin last updated on 04/Jul/19

(2) (a)

l e t {\begin{cases} l_{1} : \frac{x - 10}{3} = \frac{y - 1}{1} = \frac{z - 9}{4} = t \\ l_{2} : \frac{x}{1} = \frac{y + 9}{2} = \frac{z - 13}{- 3} = μ \end{cases}

\Rightarrow 10 + 3 t = μ

1 + t = - 9 + 2 μ

9 + 4 t = 13 - 3 μ

\Rightarrow t = - 2, μ = 4

\Rightarrow (4, - 1, - 1)

i s t h e p o i n t o f i n t e r s e c t i o n o f l_{1} & l_{2}

Commented by Rio Michael last updated on 04/Jul/19

p e r f e c t .