(1) A plane contains the lines ((x+1)/2)=((4−y)/2)=((z−2)/3) and r= (2i+2j + 12k)+t(−i+2j +4k). find (a) the angle between these lines. (b) A cartesian equation of the plane. (2) Given the lines l_1 :((x−10)/3)=((y−1)/1)=((z−9)/4) l_2 :r=(−9j+13k)+μ(i+2j−3k) where μ is a parameter; l_3 :((x+10)/4)=((y+5)/3)=((z+4)/1). a) show that the point (4,−1,1) is common to l_1 and l_2 . Find b) the point of intersection of l_2 and l_3 . c) A vector parametric equation of the plane containing the lines l_2 and l


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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
$(2)(b) let { ((l_2 :(x/1)=((y+9)/2)=((z−13)/(−3))=μ)),((l_3 :((x+10)/4)=((y+5)/3)=((z+4)/1)=s)) :} ⇒μ=−10+4s −9+2μ=−5+3s 13−3μ=−4+s ⇒no solution to the eqations ⇒l_2 &l_3 are skew lines ⇒there is no point of intersection of l_(2 ) & l_3 (2)(c) ∵l_2 & l_3 are skew lines ∴there is no plane containing the lines$
$(2) (b)$ $l e t {\begin{cases} l_{2} : \frac{x}{1} = \frac{y + 9}{2} = \frac{z - 13}{- 3} = μ \\ l_{3} : \frac{x + 10}{4} = \frac{y + 5}{3} = \frac{z + 4}{1} = s \end{cases}$ $\Rightarrow μ = - 10 + 4 s$ $- 9 + 2 μ = - 5 + 3 s$ $13 - 3 μ = - 4 + s$ $\Rightarrow n o s o l u t i o n t o t h e e q a t i o n s$ $\Rightarrow l_{2} & l_{3} a r e s k e w l i n e s$ $\Rightarrow t h e r e i s n o p o i n t o f i n t e r s e c t i o n o f l_{2} & l_{3}$ $(2) (c)$ $∵ l_{2} & l_{3} a r e s k e w l i n e s$ $∴ t h e r e i s n o p l a n e c o n t a i n i n g t h e l i n e s$
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
$(2)(a) let { ((l_1 :((x−10)/3)=((y−1)/1)=((z−9)/4)=t)),((l_2 :(x/1)=((y+9)/2)=((z−13)/(−3))=μ)) :} ⇒10+3t=μ 1+t=−9+2μ 9+4t=13−3μ ⇒t=−2, μ=4 ⇒(4, −1, −1) is the point of intersection of l_1 &l_2$
$(2) (a)$ $l e t {\begin{cases} l_{1} : \frac{x - 10}{3} = \frac{y - 1}{1} = \frac{z - 9}{4} = t \\ l_{2} : \frac{x}{1} = \frac{y + 9}{2} = \frac{z - 13}{- 3} = μ \end{cases}$ $\Rightarrow 10 + 3 t = μ$ $1 + t = - 9 + 2 μ$ $9 + 4 t = 13 - 3 μ$ $\Rightarrow t = - 2, μ = 4$ $\Rightarrow (4, - 1, - 1)$ $i s t h e p o i n t o f i n t e r s e c t i o n o f l_{1} & l_{2}$
Commented by Rio Michael last updated on 04/Jul/19

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