Question-60085

Question Number 60085 by mr W last updated on 17/May/19

Commented by mr W last updated on 17/May/19

$${the}\:{edge}\:{length}\:{of}\:{cube}\:{is}\:{a}. \\ $$$${find}\:{the}\:{minimum}\:{distance}\:{between} \\ $$$${the}\:{blue}\:{lines}. \\ $$

Commented by ajfour last updated on 17/May/19

Commented by ajfour last updated on 17/May/19

back left corner be origin. x axis rightwards, y axis upwards. r_1 =λa(i+j) r_2 =ai+μ(k−i) d_(min) =((ai.(i+j)×(k−i))/(∣(i+j)×(k−i)∣)) =(a/( (√3))) i.(−j+i+k) = (a/( (√3))) .

$$\mathrm{back}\:\mathrm{left}\:\mathrm{corner}\:\mathrm{be}\:\mathrm{origin}. \\ $$$$\mathrm{x}\:\mathrm{axis}\:\mathrm{rightwards},\:\mathrm{y}\:\mathrm{axis}\:\mathrm{upwards}. \\ $$$$\mathrm{r}_{\mathrm{1}} =\lambda\mathrm{a}\left(\mathrm{i}+\mathrm{j}\right) \\ $$$$\mathrm{r}_{\mathrm{2}} =\mathrm{ai}+\mu\left(\mathrm{k}−\mathrm{i}\right) \\ $$$$\mathrm{d}_{\mathrm{min}} =\frac{\mathrm{ai}.\left(\mathrm{i}+\mathrm{j}\right)×\left(\mathrm{k}−\mathrm{i}\right)}{\mid\left(\mathrm{i}+\mathrm{j}\right)×\left(\mathrm{k}−\mathrm{i}\right)\mid} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{a}}{\:\sqrt{\mathrm{3}}}\:\mathrm{i}.\left(−\mathrm{j}+\mathrm{i}+\mathrm{k}\right)\:=\:\frac{\mathrm{a}}{\:\sqrt{\mathrm{3}}}\:. \\ $$

Commented by mr W last updated on 17/May/19

$${clear}\:{and}\:{simple}\:{way}, \\ $$$${thank}\:{you}\:{sir}!\: \\ $$$${let}\:{me}\:{try}\:{without}\:{using}\:{vectors}. \\ $$

Answered by mr W last updated on 18/May/19

Commented by mr W last updated on 18/May/19

A(a,a,0) B(a,0,a) eqn. of line AB: ((x−a)/(a−a))=((y−a)/(0−a))=((z−0)/(a−0))=λ or { ((x=a)),((y=a(1−λ))),((z=aλ)) :} C(a,a,a) D(0,0,a) eqn. of line CD: ((x−a)/(0−a))=((y−a)/(0−a))=((z−a)/(a−a))=μ or { ((x=a(1−μ))),((y=a(1−μ))),((z=a)) :} D=PQ^2 =a^2 (1−μ−1)^2 +a^2 (1−μ−1+λ)^2 +a^2 (1−λ)^2 =a^2 {μ^2 +(λ−μ)^2 +(1−λ)^2 } =a^2 {2λ^2 +2μ^2 −2λμ−2λ+1} (∂D/∂λ)=0⇒4λ−2μ−2=0⇒2λ−μ=1 (∂D/∂μ)=0⇒4μ−2λ=0⇒2μ−λ=0 ⇒λ=(2/3), μ=(1/3) D_(min) =a^2 ((1/9)+(1/9)+(1/9))=(a^2 /3) ⇒PQ_(min) =(√D_(min) )=(a/( (√3)))

$${A}\left({a},{a},\mathrm{0}\right) \\ $$$${B}\left({a},\mathrm{0},{a}\right) \\ $$$${eqn}.\:{of}\:{line}\:{AB}: \\ $$$$\frac{{x}−{a}}{{a}−{a}}=\frac{{y}−{a}}{\mathrm{0}−{a}}=\frac{{z}−\mathrm{0}}{{a}−\mathrm{0}}=\lambda \\ $$$${or} \\ $$$$\begin{cases}{{x}={a}}\\{{y}={a}\left(\mathrm{1}−\lambda\right)}\\{{z}={a}\lambda}\end{cases} \\ $$$$ \\ $$$${C}\left({a},{a},{a}\right) \\ $$$${D}\left(\mathrm{0},\mathrm{0},{a}\right) \\ $$$${eqn}.\:{of}\:{line}\:{CD}: \\ $$$$\frac{{x}−{a}}{\mathrm{0}−{a}}=\frac{{y}−{a}}{\mathrm{0}−{a}}=\frac{{z}−{a}}{{a}−{a}}=\mu \\ $$$${or} \\ $$$$\begin{cases}{{x}={a}\left(\mathrm{1}−\mu\right)}\\{{y}={a}\left(\mathrm{1}−\mu\right)}\\{{z}={a}}\end{cases} \\ $$$$ \\ $$$${D}={PQ}^{\mathrm{2}} ={a}^{\mathrm{2}} \left(\mathrm{1}−\mu−\mathrm{1}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \left(\mathrm{1}−\mu−\mathrm{1}+\lambda\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \left(\mathrm{1}−\lambda\right)^{\mathrm{2}} \\ $$$$={a}^{\mathrm{2}} \left\{\mu^{\mathrm{2}} +\left(\lambda−\mu\right)^{\mathrm{2}} +\left(\mathrm{1}−\lambda\right)^{\mathrm{2}} \right\} \\ $$$$={a}^{\mathrm{2}} \left\{\mathrm{2}\lambda^{\mathrm{2}} +\mathrm{2}\mu^{\mathrm{2}} −\mathrm{2}\lambda\mu−\mathrm{2}\lambda+\mathrm{1}\right\} \\ $$$$\frac{\partial{D}}{\partial\lambda}=\mathrm{0}\Rightarrow\mathrm{4}\lambda−\mathrm{2}\mu−\mathrm{2}=\mathrm{0}\Rightarrow\mathrm{2}\lambda−\mu=\mathrm{1} \\ $$$$\frac{\partial{D}}{\partial\mu}=\mathrm{0}\Rightarrow\mathrm{4}\mu−\mathrm{2}\lambda=\mathrm{0}\Rightarrow\mathrm{2}\mu−\lambda=\mathrm{0} \\ $$$$\Rightarrow\lambda=\frac{\mathrm{2}}{\mathrm{3}},\:\mu=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${D}_{{min}} ={a}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{9}}\right)=\frac{{a}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\Rightarrow{PQ}_{{min}} =\sqrt{{D}_{{min}} }=\frac{{a}}{\:\sqrt{\mathrm{3}}} \\ $$

Commented by ajfour last updated on 18/May/19

$$\boldsymbol{{BOLD}}\:\mathrm{n}\:\mathcal{BEAUTIFUL}\:{way},\:{Sir}! \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply