A) \[\frac{234}{7}units\]
B) \[\frac{288}{21}units\]
C) \[\frac{221}{3}units\]
D) \[\frac{234}{21}units\]
Correct Answer: B
Solution :
Given, lines are \[\frac{x-7}{3}=\frac{y+4}{-16}=\frac{z-6}{7}\]and\[\frac{x-10}{3}=\frac{y-30}{8}=\frac{4-z}{5}\] The vector form of given equation are\[r=7\hat{i}-4\hat{j}+6\hat{k}+\lambda (3\hat{i}-16\hat{j}+7\hat{k})\] and \[r=10\hat{i}+30\hat{j}+4\hat{k}+\mu (3\hat{i}+8\hat{j}-5\hat{k})\] On comparing these equation with \[r={{a}_{1}}+\lambda {{b}_{1}}\]and\[r={{a}_{2}}+\mu {{b}_{2}},\]we get \[{{a}_{1}}=7\hat{i}-4\hat{j}+6\hat{k}\] \[{{a}_{2}}=10\hat{i}+30\hat{j}+4\hat{k}\] \[{{b}_{1}}=3\hat{i}-16\hat{j}+7\hat{k}\]and\[{{b}_{2}}=3\hat{i}+8\hat{j}-5\hat{k}\] Now,\[({{a}_{2}}-{{a}_{1}})\] \[=10\hat{i}+30\hat{j}+4\hat{k})-(7\hat{i}-4\hat{j}+6\hat{k})\] \[=3\hat{i}+34\hat{j}-2\hat{k}\] \[{{b}_{1}}\times {{b}_{2}}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \\ \end{matrix} \right|\] \[=(80-56)\hat{i}+(21+15)\hat{j}+(24+48)\hat{k}\] \[=24\hat{i}+36\hat{j}+72\hat{k}\] Then\[|{{b}_{1}}\times {{b}_{2}}|=\sqrt{{{(24)}^{2}}+{{(36)}^{2}}+{{(72)}^{2}}}\] \[=\sqrt{576+1296+5184}\] \[=\sqrt{7056}=84\] \[\therefore \]Shortest distance\[=\frac{100000}{99}\approx 1010\] \[\Rightarrow \] \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\left( \text{g} \right)\text{=}{{\text{p}}_{\text{1}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
