x-t-1-y-2t-3-z-t-2-x-3t-2-y-t-1-z-t-1-Are-these-two-lines-located-in-the-same-plane-or-not-where-t-R-

Question Number 182772 by Acem last updated on 14/Dec/22

χ {\begin{cases} x = t + 1 \\ y = 2 t - 3 \\ z = - t + 2 \end{cases} Δ {\begin{cases} x = 3 t + 2 \\ y = - t - 1 \\ z = t + 1 \end{cases} {\begin{cases} A r e t h e s e t w o l i n e s l o c a t e d \\ i n t h e s a m e p l a n e o r n o t ? \\ w h e r e t \in R \end{cases}

Answered by mr W last updated on 14/Dec/22

M e t h o d I

l i n e 1 : (1, - 3, 2) + s (1, 2, - 1)

l i n e 2 : (2, - 1, 1) + t (3, - 1, 1)

{i t}^{'} s t o s e e t h a t b o t h l i n e s a r e n o t

p a r a l l e l .

a s s u m e t h e y l i e i n t h e s a m e p l a n e,

t h e n t h e y s h o u l d i n t e r s e c t .

1 + s = 2 + 3 t \dots (i)

- 3 + 2 s = - 1 - t \dots (i i)

2 - s = 1 + t \dots (i i i)

f r o m (i) a n d (i i i) :

\Rightarrow t = 0, s = 1

p u t t h i s i n t o (i i) :

- 3 + 2 = - 1 - 0 ✓

t h a t m e a n s b o t h l i n e s i n t e r s e c t i n d e e d .

t h e y l i e i n t h e s a m e p l a n e .

Commented by mr W last updated on 14/Dec/22

M e t h o d I I

v e c t o r 1 : (1, 2, - 1)

v e c t o r 2 : (3, - 1, 1)

v e c t o r 3 : (1, 2, - 1) = \vec{P Q} *)

n_{1} = v_{1} \times v_{3} = (0, 0, 0)

n_{2} = v_{2} \times v_{3} = (- 1, 4, 7)

n_{1} / / n_{2} . t h e l i n e s a r e c o p l a n a r .

*) {i t}^{'} s t o s e e t h a t p o i n t Q o n l i n e 2

l i e s a l s o o n l i n e 1, b o t h l i n e s a r e i n

t h e s a m e p l a n e .

Commented by Acem last updated on 14/Dec/22

Y e s S i r! e x a c t l y t h a n k s s s s

a n d t h e i n t e r s e c t i o n p o i n t i s C (2, - 1, 1) \in P

H a v e a g o o d t i m e

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