A) \[[{{\vec{n}}_{2}}{{\vec{n}}_{3}}{{\vec{n}}_{4}}](\vec{r}.{{\vec{n}}_{1}}-{{\vec{q}}_{1}})=[{{\vec{n}}_{1}}{{\vec{n}}_{3}}{{\vec{n}}_{4}}](\vec{r}.{{\vec{n}}_{2}}-{{\vec{q}}_{2}})\]
B) \[[{{\vec{n}}_{1}}{{\vec{n}}_{2}}{{\vec{n}}_{4}}](\vec{r}.{{\vec{n}}_{4}}{{q}_{4}})=[{{\vec{n}}_{4}}{{\vec{n}}_{3}}{{\vec{n}}_{1}}](\vec{r}.{{\vec{n}}_{2}}-{{q}_{2}})\]
C) \[[{{\vec{n}}_{4}}{{\vec{n}}_{3}}{{\vec{n}}_{1}}](\vec{r}.{{\vec{n}}_{4}}-{{q}_{4}})=[{{\vec{n}}_{1}}{{\vec{n}}_{2}}\vec{n} 3](\vec{r}.{{\vec{n}}_{2}}={{q}_{2}})\]
D) None of these
Correct Answer: A
Solution :
[a] \[(\vec{r}.{{\vec{n}}_{1}}+\lambda \vec{r}.{{\vec{n}}_{2}}={{q}_{1}}+\lambda {{q}_{2}})\] ??(i) Where \[\lambda \] is a parameter? So, \[{{\vec{n}}_{1}}+\lambda {{\vec{n}}_{2}}\] is normal to plane (i), now, any plane parallel to the line of intersection of the planes \[\vec{r}.{{\vec{n}}_{3}}={{q}_{3}}\] and \[\vec{r}.{{\vec{n}}_{4}}={{q}_{4}}\] is of the form \[\vec{r}.({{\vec{n}}_{3}}\times {{\vec{n}}_{4}})=d.\] Hence, we must have \[[{{\vec{n}}_{1}}+\lambda {{\vec{n}}_{2}}].[{{\vec{n}}_{3}}\times {{\vec{n}}_{4}}]=0\] or \[[{{\vec{n}}_{1}}{{\vec{n}}_{3}}{{\vec{n}}_{4}}]+\lambda [{{\vec{n}}_{2}}{{\vec{n}}_{3}}{{\vec{n}}_{4}}]=0\] or \[\lambda =\frac{-[{{{\vec{n}}}_{1}}{{{\vec{n}}}_{3}}{{{\vec{n}}}_{4}}]}{[{{{\vec{n}}}_{2}}{{{\vec{n}}}_{3}}{{{\vec{n}}}_{4}}]}\] on putting this value in Eq. (i), we have the equation of the required plane as \[\vec{r}.{{\vec{n}}_{1}}-{{q}_{1}}=\frac{[{{{\vec{n}}}_{1}}{{{\vec{n}}}_{2}}{{{\vec{n}}}_{4}}]}{[{{{\vec{n}}}_{2}}{{{\vec{n}}}_{3}}{{{\vec{n}}}_{4}}]}(r.{{\vec{n}}_{2}}-{{q}_{2}})\] or \[[{{\vec{n}}_{2}}{{\vec{n}}_{3}}{{\vec{n}}_{4}}](\vec{r}.{{\vec{n}}_{1}}-{{q}_{1}})=[{{\vec{n}}_{1}}{{\vec{n}}_{3}}{{\vec{n}}_{4}}](\vec{r}.{{\vec{n}}_{2}}-{{q}_{2}})\]
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