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Question-116845




Question Number 116845 by mohammad17 last updated on 07/Oct/20
Commented by mohammad17 last updated on 07/Oct/20
help me sir
$${help}\:{me}\:{sir}\: \\ $$
Answered by TANMAY PANACEA last updated on 07/Oct/20
st line AB ((x−2)/(6−2))=((y−4)/(8−4))→((x−2)/1)=((y−4)/1)  direction ratio (1,1)  r=2i+4j+α(i+j)  l_1    is  r_1 =6i+4j+λ(i+j)  ((x−6)/1)=((y−4)/1) direction ratio of l_1 is (1,1)  so AB  ∥ to l_1   l_2  eqn is ((x−6)/(6−4))=((z−0)/(0−2))  vector eqn l_2 is   r=6i+0k+λ(2i−2k)  or  ((x−4)/(6−4))=((z−2)/(0−2))→r=4i+2k+λ(2i−2k)
$${st}\:{line}\:{AB}\:\frac{{x}−\mathrm{2}}{\mathrm{6}−\mathrm{2}}=\frac{{y}−\mathrm{4}}{\mathrm{8}−\mathrm{4}}\rightarrow\frac{{x}−\mathrm{2}}{\mathrm{1}}=\frac{{y}−\mathrm{4}}{\mathrm{1}} \\ $$$${direction}\:{ratio}\:\left(\mathrm{1},\mathrm{1}\right) \\ $$$${r}=\mathrm{2}{i}+\mathrm{4}{j}+\alpha\left({i}+{j}\right) \\ $$$${l}_{\mathrm{1}} \:\:\:{is}\:\:{r}_{\mathrm{1}} =\mathrm{6}{i}+\mathrm{4}{j}+\lambda\left({i}+{j}\right) \\ $$$$\frac{{x}−\mathrm{6}}{\mathrm{1}}=\frac{{y}−\mathrm{4}}{\mathrm{1}}\:{direction}\:{ratio}\:{of}\:{l}_{\mathrm{1}} {is}\:\left(\mathrm{1},\mathrm{1}\right) \\ $$$${so}\:{AB}\:\:\parallel\:{to}\:{l}_{\mathrm{1}} \\ $$$${l}_{\mathrm{2}} \:{eqn}\:{is}\:\frac{{x}−\mathrm{6}}{\mathrm{6}−\mathrm{4}}=\frac{{z}−\mathrm{0}}{\mathrm{0}−\mathrm{2}} \\ $$$${vector}\:{eqn}\:{l}_{\mathrm{2}} {is}\:\:\:{r}=\mathrm{6}{i}+\mathrm{0}{k}+\lambda\left(\mathrm{2}{i}−\mathrm{2}{k}\right) \\ $$$${or}\:\:\frac{{x}−\mathrm{4}}{\mathrm{6}−\mathrm{4}}=\frac{{z}−\mathrm{2}}{\mathrm{0}−\mathrm{2}}\rightarrow{r}=\mathrm{4}{i}+\mathrm{2}{k}+\lambda\left(\mathrm{2}{i}−\mathrm{2}{k}\right) \\ $$$$ \\ $$

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