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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 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?
$$\mathrm{mr}\:\mathrm{W}\:\mathrm{and}\:\mathrm{Pritwish}.\:\mathrm{my}\:\mathrm{way}\:\mathrm{is} \\ $$$$\mathrm{correct}? \\ $$
Commented by PRITHWISH SEN 2 last updated on 16/May/20
yes sir.
$$\mathrm{yes}\:\mathrm{sir}. \\ $$
Commented by i 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 by i 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 by mr 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 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′.
$$\mathrm{Actually}\:\mathrm{if}\:\mathrm{three}\:\mathrm{vectors}\:\mathrm{form}\:\mathrm{a}\:\mathrm{linear}\:\mathrm{combination} \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{parallelepiped}\:\mathrm{so}\:\mathrm{formed} \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{three}\:\mathrm{vectors}\:\mathrm{must}\:\mathrm{be}\:\mathrm{zero}. \\ $$$$\mathrm{And}\:\mathrm{since}\:\mathrm{the}\:\mathrm{determinant}\:\mathrm{means}\:\mathrm{the}\:\mathrm{volume} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{parallelepiped}\:\mathrm{formed}\:\mathrm{by}\:\mathrm{the}\:\mathrm{three}\:\mathrm{vectors} \\ $$$$\mathrm{then}\:\mathrm{it}\:\mathrm{must}\:\mathrm{be}\:'\boldsymbol{\mathrm{zero}}'. \\ $$

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