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Question-134949




Question Number 134949 by rexford last updated on 08/Mar/21
Answered by bobhans last updated on 30/Jan/22
b^→  = 2c^→ +λa^→  ; ∣b^→ ∣ = ∣2c^→ +λa^→ ∣  ⇒ 4 = (√(∣2c^→ ∣^2 +∣λa^→ ∣^2 +2∣2c^→ ∣∣λa^→ ∣((1/4))))  ⇒16 = 4+λ^2 +λ   ⇒λ^2 +λ−12=0  ⇒(λ+4)(λ−3)=0  { ((λ_1 =−4)),((λ_2 =3)) :}
$$\overset{\rightarrow} {\mathrm{b}}\:=\:\mathrm{2}\overset{\rightarrow} {\mathrm{c}}+\lambda\overset{\rightarrow} {\mathrm{a}}\:;\:\mid\overset{\rightarrow} {\mathrm{b}}\mid\:=\:\mid\mathrm{2}\overset{\rightarrow} {\mathrm{c}}+\lambda\overset{\rightarrow} {\mathrm{a}}\mid \\ $$$$\Rightarrow\:\mathrm{4}\:=\:\sqrt{\mid\mathrm{2}\overset{\rightarrow} {\mathrm{c}}\mid^{\mathrm{2}} +\mid\lambda\overset{\rightarrow} {\mathrm{a}}\mid^{\mathrm{2}} +\mathrm{2}\mid\mathrm{2}\overset{\rightarrow} {\mathrm{c}}\mid\mid\lambda\overset{\rightarrow} {\mathrm{a}}\mid\left(\frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$$$\Rightarrow\mathrm{16}\:=\:\mathrm{4}+\lambda^{\mathrm{2}} +\lambda\: \\ $$$$\Rightarrow\lambda^{\mathrm{2}} +\lambda−\mathrm{12}=\mathrm{0} \\ $$$$\Rightarrow\left(\lambda+\mathrm{4}\right)\left(\lambda−\mathrm{3}\right)=\mathrm{0}\:\begin{cases}{\lambda_{\mathrm{1}} =−\mathrm{4}}\\{\lambda_{\mathrm{2}} =\mathrm{3}}\end{cases} \\ $$

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