what-is-the-line-passing-through-2-2-1-and-parallel-to-2i-j-k-

Question Number 78522 by jagoll last updated on 18/Jan/20

w h a t i s t h e

l i n e p a s s i n g t h r o u g h (2, 2, 1)

a n d p a r a l l e l t o 2 \hat{i} - \hat{j} - \hat{k} ?

Commented by mr W last updated on 18/Jan/20

\frac{x - 2}{2} = \frac{y - 2}{- 1} = \frac{z - 1}{- 1}

Commented by jagoll last updated on 18/Jan/20

h o w a b o u t i f l i n e o r t h o g o n a l s i r ?

Commented by mr W last updated on 18/Jan/20

\frac{x - 2}{10} = \frac{y - 2}{13} = \frac{z - 1}{7}

Commented by jagoll last updated on 19/Jan/20

how get (10, 13, 7) sir ?

Commented by mr W last updated on 19/Jan/20

l i n e 1 (g i v e n) :

v e c t o r 2 i - j - k

i . e . \frac{x}{2} = \frac{y}{- 1} = \frac{z}{- 1}

l i n e 2 t h r o u g h p o i n t (2, 2, 1) a n d ⊥ l i n e 1 :

s a y \frac{x - 2}{a} = \frac{y - 2}{b} = \frac{z - 1}{c}

l i n e 2 ⊥ l i n e 1 :

2 \times a - 1 \times b - 1 \times c = 0

\Rightarrow a = \frac{b + c}{2}

a p o i n t s h o u l d b o t h o n l i n e 1 a n d o n

l i n e 2 :

p o i n t o n l i n e 1 : \frac{x}{2} = \frac{y}{- 1} = \frac{z}{- 1} = t

\Rightarrow x = 2 t

\Rightarrow y = - t

\Rightarrow z = - t

p o i n t o n l i n e 2 : \frac{x - 2}{a} = \frac{y - 2}{b} = \frac{z - 1}{c} = s

\Rightarrow x = 2 + s a = 2 t

\Rightarrow y = 2 + s b = - t

\Rightarrow z = 1 + s c = - t

\Rightarrow s c = - t - 1

\Rightarrow s b = - t - 2

\Rightarrow s a = 2 t - 2

a = \frac{b + c}{2} \Rightarrow s a = \frac{s b + s c}{2} \Rightarrow 2 t - 2 = \frac{- t - 2 - t - 1}{2}

\Rightarrow 4 t - 4 = - 2 t - 3

\Rightarrow t = \frac{1}{6}

\Rightarrow s a = 2 \times \frac{1}{6} - 2 = - \frac{10}{6}

\Rightarrow s b = - \frac{1}{6} - 2 = - \frac{13}{6}

\Rightarrow s c = - \frac{1}{6} - 1 = - \frac{7}{6}

\Rightarrow a : b : c = 10 : 13 : 7

\Rightarrow l i n e 2 : \frac{x - 2}{10} = \frac{y - 2}{13} = \frac{z - 1}{7}

Commented by mr W last updated on 19/Jan/20

c l e a r ?

m a y b e t h e r e a r e o t h e r e a s i e r w a y s .

Commented by jagoll last updated on 19/Jan/20

line passing trought point

(2, 2, 1) \Rightarrow \frac{x - 2}{a} = \frac{y - 2}{b} = \frac{z - 1}{c}

orthogonal to vector 2 \hat{i} - \hat{j} - \hat{k}

\bar{u} \times \bar{v} = | \begin{matrix} a b c \\ 2 - 1 - 1 \end{matrix} | = (- b + c) \hat{i} + (a + 2 c) \hat{j} - (a + 2 b) \hat{k}

\Rightarrow - 2 b + 2 c + 2 a + 4 c - a - 2 b = 0

a - 4 b + 6 c = 0

Commented by jagoll last updated on 19/Jan/20

what wrong my work ?

Commented by mr W last updated on 19/Jan/20

t o t a l l y w r o n g .

y o u a r e l o o k i n g f o r a v e c t o r u w h i c h i s

p e r p e n d i c u l a r t o v, n o t l o o k i n g f o r

a v e c t o r w h i c h i s p e r p e n d i c u l a r t o u

a n d v! . u \times v i s a v e c t o r w h i c h i s

p e r p e n d i c u l a r t o u a n d v!

u = (a, b, c)

v = (2, - 1, - 1)

s i n c e u ⊥ v, \Rightarrow a \times 2 + b \times (- 1) + c \times (- 1) = 0

Commented by jagoll last updated on 19/Jan/20

oo yes sir . i understand . thanks you sir

oo yes sir . i understand . thanks you sir

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