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Question Number 141129 by bobhans last updated on 16/May/21
For what values of λ are the  vectors λi^�  + 2j^�  + k^�  , 3i^�  +4j^�  +λk^�    , j^�  + k^�   coplanar
$${For}\:{what}\:{values}\:{of}\:\lambda\:{are}\:{the} \\ $$$${vectors}\:\lambda\hat {{i}}\:+\:\mathrm{2}\hat {{j}}\:+\:\hat {{k}}\:,\:\mathrm{3}\hat {{i}}\:+\mathrm{4}\hat {{j}}\:+\lambda\hat {{k}}\: \\ $$$$,\:\hat {{j}}\:+\:\hat {{k}}\:\:{coplanar}\: \\ $$
Answered by EDWIN88 last updated on 16/May/21
  ((3),(4),(λ) ) = m  ((λ),(2),(1) ) + n ((0),(1),(1) )  ⇒ mλ = 3 ; λ=(3/m)  ⇒ { ((2m+n=4)),((m+n=λ)) :} ⇒m=4−λ  ⇒ m=4−(3/m) ; m^2 −4m+3 = 0  ⇒(m−1)(m−3)=0  { ((m=1 ⇒λ=3)),((m=3⇒λ= 1)) :}   λ = { 1, 3 } ⋇
$$\:\begin{pmatrix}{\mathrm{3}}\\{\mathrm{4}}\\{\lambda}\end{pmatrix}\:=\:\mathrm{m}\:\begin{pmatrix}{\lambda}\\{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}\:+\:\mathrm{n}\begin{pmatrix}{\mathrm{0}}\\{\mathrm{1}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$\Rightarrow\:\mathrm{m}\lambda\:=\:\mathrm{3}\:;\:\lambda=\frac{\mathrm{3}}{\mathrm{m}} \\ $$$$\Rightarrow\begin{cases}{\mathrm{2m}+\mathrm{n}=\mathrm{4}}\\{\mathrm{m}+\mathrm{n}=\lambda}\end{cases}\:\Rightarrow\mathrm{m}=\mathrm{4}−\lambda \\ $$$$\Rightarrow\:\mathrm{m}=\mathrm{4}−\frac{\mathrm{3}}{\mathrm{m}}\:;\:\mathrm{m}^{\mathrm{2}} −\mathrm{4m}+\mathrm{3}\:=\:\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{m}−\mathrm{1}\right)\left(\mathrm{m}−\mathrm{3}\right)=\mathrm{0}\:\begin{cases}{\mathrm{m}=\mathrm{1}\:\Rightarrow\lambda=\mathrm{3}}\\{\mathrm{m}=\mathrm{3}\Rightarrow\lambda=\:\mathrm{1}}\end{cases} \\ $$$$\:\lambda\:=\:\left\{\:\mathrm{1},\:\mathrm{3}\:\right\}\:\divideontimes\:\: \\ $$

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