find-k-if-the-vector-1-2-k-in-R-3-be-a-linear-combination-of-the-vectors-3-0-2-amp-2-1-5-

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 by i jagooll last updated on 16/May/20

Commented by i jagooll last updated on 16/May/20

mr W and Pritwish . my way is

correct ?

Commented by PRITHWISH SEN 2 last updated on 16/May/20

yes sir .

Commented by i 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 by i jagooll last updated on 16/May/20

it does mean \vec{u} \times (\vec{v} \times \vec{w}) sir ?

Commented by mr 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 \dots (i)

a_{2} x + b_{2} y + c_{2} z = 0 \dots (i i)

a_{3} x + b_{3} y + c_{3} z = 0 \dots (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 by PRITHWISH 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}^{'} .