A) \[-\frac{16}{19}\]
B) \[-\frac{19}{16}\]
C) \[\frac{19}{32}\]
D) \[\frac{32}{19}\]
Correct Answer: D
Solution :
\[\frac{x-1}{\alpha }=\frac{y+1}{-1}=\frac{z-0}{1}={{L}_{1}}\] \[x+y+z+1=0\] \[2x-y+7+3=0={{L}_{2}}\] so point P(0,1,-2) [on line \[{{L}_{2}}\]] point on other line [on line \[{{L}_{1}}\]] Q (1 , -1 , 0) [on line \[{{L}_{2}}\]] \[Q(1,-1,0)\] \[\overrightarrow{PQ}=\hat{i}-2\hat{j}+2\hat{k}\]vector II to line \[{{L}_{2}}\] \[\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & 1 & 1 \\ 2 & -1 & 1 \\ \end{matrix} \right|=2\hat{i}+\hat{j}-3\hat{k}\]vector \[{{\bot }_{r}}\] to \[48{{L}_{2}}\] \[\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ \alpha & -1 & 1 \\ 2 & -1 & -3 \\ \end{matrix} \right|\]\[\vec{n}=2\hat{i}+\hat{j}(3\alpha +2)+\hat{k}(\alpha +2)\] distance \[=\overrightarrow{PQ}.\hat{m}\] \[=\frac{2-2\left| 3\alpha +2 \right|+2\left( \alpha +2 \right)}{\sqrt{4+{{\left( 3\alpha +2 \right)}^{2}}+{{\left( \alpha +2 \right)}^{2}}}}=\frac{1}{\sqrt{3}}\] \[\Rightarrow \]\[3{{\left( 2-4\alpha \right)}^{2}}=10{{\alpha }^{2}}+16\alpha +12\] \[\Rightarrow \]\[6\left[ 1+4{{\alpha }^{2}}-4\alpha \right]=5{{\alpha }^{2}}+5\alpha +6\] \[\Rightarrow \]\[19{{\alpha }^{2}}-32\alpha =0\]\[\Rightarrow \]\[\alpha =\frac{32}{19}\]
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