Question and Answers Forum

All Questions      Topic List

Vector Calculus Questions

Previous in All Question      Next in All Question      

Previous in Vector Calculus      Next in Vector Calculus      

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)) :}

$$\overset{\rightarrow} {\mathrm{b}}\:=\:\mathrm{2}\overset{\rightarrow} {\mathrm{c}}+\lambda\overset{\rightarrow} {\mathrm{a}}\:;\:\mid\overset{\rightarrow} {\mathrm{b}}\mid\:=\:\mid\mathrm{2}\overset{\rightarrow} {\mathrm{c}}+\lambda\overset{\rightarrow} {\mathrm{a}}\mid \\ $$$$\Rightarrow\:\mathrm{4}\:=\:\sqrt{\mid\mathrm{2}\overset{\rightarrow} {\mathrm{c}}\mid^{\mathrm{2}} +\mid\lambda\overset{\rightarrow} {\mathrm{a}}\mid^{\mathrm{2}} +\mathrm{2}\mid\mathrm{2}\overset{\rightarrow} {\mathrm{c}}\mid\mid\lambda\overset{\rightarrow} {\mathrm{a}}\mid\left(\frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$$$\Rightarrow\mathrm{16}\:=\:\mathrm{4}+\lambda^{\mathrm{2}} +\lambda\: \\ $$$$\Rightarrow\lambda^{\mathrm{2}} +\lambda−\mathrm{12}=\mathrm{0} \\ $$$$\Rightarrow\left(\lambda+\mathrm{4}\right)\left(\lambda−\mathrm{3}\right)=\mathrm{0}\:\begin{cases}{\lambda_{\mathrm{1}} =−\mathrm{4}}\\{\lambda_{\mathrm{2}} =\mathrm{3}}\end{cases} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com