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

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

$$\mathrm{find}\:\mathrm{k}\:\mathrm{if}\:\mathrm{the}\:\mathrm{vector}\:\left(\bar {\mathrm{1}}−\mathrm{2},\mathrm{k}\right)\:\mathrm{in}\:\mathbb{R}^{\mathrm{3}} \\ $$ $$\mathrm{be}\:\mathrm{a}\:\mathrm{linear}\:\mathrm{combination}\:\mathrm{of}\:\mathrm{the} \\ $$ $$\mathrm{vectors}\:\left(\mathrm{3},\mathrm{0},\mathrm{2}\right)\:\&\left(\mathrm{2},−\mathrm{1},−\mathrm{5}\right)\: \\ $$

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?

$$\mathrm{mr}\:\mathrm{W}\:\mathrm{and}\:\mathrm{Pritwish}.\:\mathrm{my}\:\mathrm{way}\:\mathrm{is} \\ $$ $$\mathrm{correct}? \\ $$

Commented byPRITHWISH SEN 2 last updated on 16/May/20

yes sir.

$$\mathrm{yes}\:\mathrm{sir}. \\ $$

Commented byi jagooll last updated on 16/May/20

thanks sir

$$\mathrm{thanks}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 15/May/20

 determinant ((1,(−2),k),(3,0,2),(2,(−1),(−5)))=0   determinant ((0,(−2),k),(6,0,2),(3,(−1),(−5)))=0   determinant ((0,(−2),k),(0,1,6),(3,(−1),(−5)))=0  k+2×6=0  ⇒k=−12

$$\begin{vmatrix}{\mathrm{1}}&{−\mathrm{2}}&{{k}}\\{\mathrm{3}}&{\mathrm{0}}&{\mathrm{2}}\\{\mathrm{2}}&{−\mathrm{1}}&{−\mathrm{5}}\end{vmatrix}=\mathrm{0} \\ $$ $$\begin{vmatrix}{\mathrm{0}}&{−\mathrm{2}}&{{k}}\\{\mathrm{6}}&{\mathrm{0}}&{\mathrm{2}}\\{\mathrm{3}}&{−\mathrm{1}}&{−\mathrm{5}}\end{vmatrix}=\mathrm{0} \\ $$ $$\begin{vmatrix}{\mathrm{0}}&{−\mathrm{2}}&{{k}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{6}}\\{\mathrm{3}}&{−\mathrm{1}}&{−\mathrm{5}}\end{vmatrix}=\mathrm{0} \\ $$ $${k}+\mathrm{2}×\mathrm{6}=\mathrm{0} \\ $$ $$\Rightarrow{k}=−\mathrm{12} \\ $$

Commented byi jagooll last updated on 16/May/20

it does mean u^→  × (v^→ ×w^→ ) sir?

$$\mathrm{it}\:\mathrm{does}\:\mathrm{mean}\:\overset{\rightarrow} {\mathrm{u}}\:×\:\left(\overset{\rightarrow} {\mathrm{v}}×\overset{\rightarrow} {\mathrm{w}}\right)\:\mathrm{sir}? \\ $$

Commented bymr W last updated on 16/May/20

this is my own understanding and  method which i didn′t learn in school  or read in a book. but i know it is  correct.  say  vector V_1 =(a_1 ,b_1 ,c_1 )  vector V_2 =(a_2 ,b_2 ,c_2 )  vector V_3 =(a_3 ,b_3 ,c_3 )    let′s look the eqn. system  a_1 x+b_1 y+c_1 z=0   ...(i)  a_2 x+b_2 y+c_2 z=0   ...(ii)  a_3 x+b_3 y+c_3 z=0  ...(iii)  we know, if (iii) can be obtained  from (i) and (ii) by adding them  after multiplicated by some values,  then the eqn. system is indeterminate,  i.e.  has infinitely many solutions,  in tbis case we have:   determinant ((a_1 ,b_1 ,c_1 ),(a_2 ,b_2 ,c_2 ),(a_3 ,b_3 ,c_3 ))=0.  but this means the same as that the  vector V_3  is a linear combination of   the vectors V_1  and V_2 .  therefore, if vector (a_3 ,b_3 ,c_3 ) is a  linear combination of vectors  (a_1 ,b_1 ,c_1 ) and (a_2 ,b_2 ,c_2 ) then   determinant ((a_1 ,b_1 ,c_1 ),(a_2 ,b_2 ,c_2 ),(a_3 ,b_3 ,c_3 ))=0.

$${this}\:{is}\:{my}\:{own}\:{understanding}\:{and} \\ $$ $${method}\:{which}\:{i}\:{didn}'{t}\:{learn}\:{in}\:{school} \\ $$ $${or}\:{read}\:{in}\:{a}\:{book}.\:{but}\:{i}\:{know}\:{it}\:{is} \\ $$ $${correct}. \\ $$ $${say} \\ $$ $${vector}\:{V}_{\mathrm{1}} =\left({a}_{\mathrm{1}} ,{b}_{\mathrm{1}} ,{c}_{\mathrm{1}} \right) \\ $$ $${vector}\:{V}_{\mathrm{2}} =\left({a}_{\mathrm{2}} ,{b}_{\mathrm{2}} ,{c}_{\mathrm{2}} \right) \\ $$ $${vector}\:{V}_{\mathrm{3}} =\left({a}_{\mathrm{3}} ,{b}_{\mathrm{3}} ,{c}_{\mathrm{3}} \right) \\ $$ $$ \\ $$ $${let}'{s}\:{look}\:{the}\:{eqn}.\:{system} \\ $$ $${a}_{\mathrm{1}} {x}+{b}_{\mathrm{1}} {y}+{c}_{\mathrm{1}} {z}=\mathrm{0}\:\:\:...\left({i}\right) \\ $$ $${a}_{\mathrm{2}} {x}+{b}_{\mathrm{2}} {y}+{c}_{\mathrm{2}} {z}=\mathrm{0}\:\:\:...\left({ii}\right) \\ $$ $${a}_{\mathrm{3}} {x}+{b}_{\mathrm{3}} {y}+{c}_{\mathrm{3}} {z}=\mathrm{0}\:\:...\left({iii}\right) \\ $$ $${we}\:{know},\:{if}\:\left({iii}\right)\:{can}\:{be}\:{obtained} \\ $$ $${from}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{by}\:{adding}\:{them} \\ $$ $${after}\:{multiplicated}\:{by}\:{some}\:{values}, \\ $$ $${then}\:{the}\:{eqn}.\:{system}\:{is}\:{indeterminate}, \\ $$ $${i}.{e}.\:\:{has}\:{infinitely}\:{many}\:{solutions}, \\ $$ $${in}\:{tbis}\:{case}\:{we}\:{have}: \\ $$ $$\begin{vmatrix}{{a}_{\mathrm{1}} }&{{b}_{\mathrm{1}} }&{{c}_{\mathrm{1}} }\\{{a}_{\mathrm{2}} }&{{b}_{\mathrm{2}} }&{{c}_{\mathrm{2}} }\\{{a}_{\mathrm{3}} }&{{b}_{\mathrm{3}} }&{{c}_{\mathrm{3}} }\end{vmatrix}=\mathrm{0}. \\ $$ $${but}\:{this}\:{means}\:{the}\:{same}\:{as}\:{that}\:{the} \\ $$ $${vector}\:{V}_{\mathrm{3}} \:{is}\:{a}\:{linear}\:{combination}\:{of}\: \\ $$ $${the}\:{vectors}\:{V}_{\mathrm{1}} \:{and}\:{V}_{\mathrm{2}} . \\ $$ $${therefore},\:{if}\:{vector}\:\left({a}_{\mathrm{3}} ,{b}_{\mathrm{3}} ,{c}_{\mathrm{3}} \right)\:{is}\:{a} \\ $$ $${linear}\:{combination}\:{of}\:{vectors} \\ $$ $$\left({a}_{\mathrm{1}} ,{b}_{\mathrm{1}} ,{c}_{\mathrm{1}} \right)\:{and}\:\left({a}_{\mathrm{2}} ,{b}_{\mathrm{2}} ,{c}_{\mathrm{2}} \right)\:{then} \\ $$ $$\begin{vmatrix}{{a}_{\mathrm{1}} }&{{b}_{\mathrm{1}} }&{{c}_{\mathrm{1}} }\\{{a}_{\mathrm{2}} }&{{b}_{\mathrm{2}} }&{{c}_{\mathrm{2}} }\\{{a}_{\mathrm{3}} }&{{b}_{\mathrm{3}} }&{{c}_{\mathrm{3}} }\end{vmatrix}=\mathrm{0}. \\ $$

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′.

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

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