find k if the vector (1^� −2,k) in R^3 be a linear combination of the vectors (3,0,2) &(2,−1,−5)


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Question Number 93874 by i jagooll last updated on 15/May/20

$find k if the vector (\bar{1} - 2, k) in R^{3}$ $be a linear combination of the$ $vectors (3, 0, 2) & (2, - 1, - 5)$
Commented byi jagooll last updated on 16/May/20

Commented byi jagooll last updated on 16/May/20

$mr W and Pritwish . my way is$ $correct ?$
Commented byPRITHWISH SEN 2 last updated on 16/May/20

$yes sir .$
Commented byi jagooll last updated on 16/May/20

$thanks sir$
	Answered by mr W last updated on 15/May/20

	$\| \begin{matrix} 1 & - 2 & k \\ 3 & 0 & 2 \\ 2 & - 1 & - 5 \end{matrix} \| = 0$ $\| \begin{matrix} 0 & - 2 & k \\ 6 & 0 & 2 \\ 3 & - 1 & - 5 \end{matrix} \| = 0$ $\| \begin{matrix} 0 & - 2 & k \\ 0 & 1 & 6 \\ 3 & - 1 & - 5 \end{matrix} \| = 0$ $k + 2 \times 6 = 0$ $\Rightarrow k = - 12$
	Commented byi jagooll last updated on 16/May/20

	$it does mean \vec{u} \times (\vec{v} \times \vec{w}) sir ?$
	Commented bymr W last updated on 16/May/20

	$t h i s i s m y o w n u n d e r s t a n d i n g a n d$ $m e t h o d w h i c h i {d i d n}^{'} t l e a r n i n s c h o o l$ $o r r e a d i n a b o o k . b u t i k n o w i t i s$ $c o r r e c t .$ $s a y$ $v e c t o r V_{1} = (a_{1}, b_{1}, c_{1})$ $v e c t o r V_{2} = (a_{2}, b_{2}, c_{2})$ $v e c t o r V_{3} = (a_{3}, b_{3}, c_{3})$ ${l e t}^{'} s l o o k t h e e q n . s y s t e m$ $a_{1} x + b_{1} y + c_{1} z = 0 . . . (i)$ $a_{2} x + b_{2} y + c_{2} z = 0 . . . (i i)$ $a_{3} x + b_{3} y + c_{3} z = 0 . . . (i i i)$ $w e k n o w, i f (i i i) c a n b e o b t a i n e d$ $f r o m (i) a n d (i i) b y a d d i n g t h e m$ $a f t e r m u l t i p l i c a t e d b y s o m e v a l u e s,$ $t h e n t h e e q n . s y s t e m i s i n d e t e r m i n a t e,$ $i . e . h a s i n f i n i t e l y m a n y s o l u t i o n s,$ $i n t b i s c a s e w e h a v e :$ $\| \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} \| = 0 .$ $b u t t h i s m e a n s t h e s a m e a s t h a t t h e$ $v e c t o r V_{3} i s a l i n e a r c o m b i n a t i o n o f$ $t h e v e c t o r s V_{1} a n d V_{2} .$ $t h e r e f o r e, i f v e c t o r (a_{3}, b_{3}, c_{3}) i s a$ $l i n e a r c o m b i n a t i o n o f v e c t o r s$ $(a_{1}, b_{1}, c_{1}) a n d (a_{2}, b_{2}, c_{2}) t h e n$ $\| \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} \| = 0 .$
	Commented byPRITHWISH SEN 2 last updated on 16/May/20

	$Actually if three vectors form a linear combination$ $then the volume of the parallelepiped so formed$ $by the three vectors must be zero .$ $And since the determinant means the volume$ $of the parallelepiped formed by the three vectors$ $then it must be^{'} {zero}^{'} .$
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