A) \[\frac{x-\frac{6}{13}}{2}=\frac{y-\frac{5}{13}}{7}=\frac{z}{-13}\]
B) \[\frac{x-\frac{7}{7}}{2}=\frac{y}{-7}=\frac{z+\frac{5}{7}}{13}\]
C) \[\frac{x-\frac{6}{13}}{2}=\frac{y-\frac{5}{13}}{-7}=\frac{z}{-13}\]
D) \[\frac{x-\frac{4}{7}}{-2}=\frac{y}{7}=\frac{z-\frac{5}{7}}{13}\]
Correct Answer: C
Solution :
\[\vec{n}={{\vec{n}}_{1}}\times {{\vec{n}}_{1}}\] \[\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 3 & -1 & 1 \\ 1 & 4 & -2 \\ \end{matrix} \right|\] \[=\hat{i}(-2)-\hat{j}(-7)+\hat{k}(13)\] \[\overline{n}=2\hat{i}+7\hat{j}+13\hat{k}\] Now \[3x-y+z=1\] \[x+4y-2z=2\] But z = 0 \[3x-y=1\times 4\] \[x+4y=2\] \[13x=6\] \[x=6/13\] \[y=5/13\]?..is \[\] or \[\]
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