question_answer

Find the equation of the plane passing through the point (1, 1, 1) and containing the line \[\vec{r}=(-\,3\hat{i}+\hat{j}+5\hat{k})+\lambda (3\hat{i}-\hat{j}-5\hat{k}).\]

Answer:

Given equation of line is             \[\vec{r}=(-\,3\hat{i}\,+\hat{j}\,+5\hat{k})+\lambda (3\hat{i}\,-\hat{j}\,-5\hat{k})\]                      ? (i) On comparing with \[\vec{r}={{\vec{a}}_{1}}+\lambda {{\vec{b}}_{1}},\] we get \[{{\vec{a}}_{1}}=-\,3\hat{i}\,+\hat{j}\,+5\hat{k}\] and \[{{\vec{b}}_{1}}=3\hat{i}-\hat{j}\,-5\hat{k}\] Let the position vector of given point A (1, 1, 1) be \[{{\vec{a}}_{2}}=\hat{i}+\hat{j}+\hat{k}.\] Now, the vector equation of the plane passing through \[{{\vec{a}}_{2}}=\hat{i}+\hat{j}\,+\hat{k}\] is \[(\vec{r}-{{\vec{a}}_{2}})\cdot \vec{n}=0\]                 ? (ii) The plane which contains the given line must pass through the point \[{{\vec{a}}_{1}}=-\,3\hat{i}+\hat{j}\,+5\hat{k}\] and must be parallel to the line having direction of \[{{\vec{b}}_{1}}=3\hat{i}-\hat{j}\,-5\hat{k}.\] \[\because \] The plane (ii) passes through \[{{\vec{a}}_{1}}=-\,3\hat{i}+\hat{j}\,+5\hat{k},\]therefore \[({{\vec{a}}_{1}}-{{\vec{a}}_{2}})\cdot \vec{n}=0\]                                                                     ? (iii) Also, the plane (ii) is parallel to the line having direction of \[{{\vec{b}}_{1}}=-\,3\hat{i}\,+\hat{j}\,+5\hat{k},\] therefore \[{{\vec{b}}_{1}}\cdot \vec{n}=0\]                                            ? (iv) From Eqs. (iii) and (iv), we conclude that \[\vec{n}\] is of the form \[({{\vec{a}}_{1}}-{{\vec{a}}_{2}})\times {{\vec{b}}_{1}}.\]  \[[\because \,\,\,({{\vec{a}}_{1}}-{{\vec{a}}_{2}})\]and \[\vec{b}\]lies in same plane and \[\vec{n}\] is perpendicular to both of them] Now, \[{{\vec{a}}_{1}}-{{\vec{a}}_{2}}=(-\,3\hat{i}+\hat{j}\,+5\hat{k})-(\hat{i}+\hat{j}\,+\hat{k})\]             \[=-\,4\hat{i}+0\hat{j}+4\hat{k}\] \[\therefore \]      \[\vec{n}=({{\vec{a}}_{1}}-{{\vec{a}}_{2}})\times {{\vec{b}}_{1}}=\left| \begin{matrix}    {\hat{i}} & {\hat{j}} & {\hat{k}}  \\    -\,4 & 0 & 4  \\    3 & -\,1 & -\,5  \\ \end{matrix} \right|\] \[\Rightarrow \]   \[\vec{n}=[\hat{i}\,(0+4)-j(20-12)\,+\hat{k}\,(4)]\]            \[\Rightarrow \]   \[\vec{n}=4\hat{i}-8j\,+4\hat{k}\] \[\Rightarrow \]   \[\vec{n}=4(\hat{i}-2j\,+\hat{k})\] Thus, the repuired equation of the plane passing through \[{{\vec{a}}_{2}}\] and containing the given line is \[(\vec{r}-{{\vec{a}}_{2}})\cdot \vec{n}=0.\] \[\Rightarrow \] \[[\vec{r}-(\hat{i}+\hat{j}\,+\hat{k})]\cdot 4(\hat{i}\,-2\hat{j}\,+\hat{k})=0\] \[\Rightarrow \] \[\vec{r}\cdot (\hat{i}\,-2\hat{j}+\hat{k})-(\hat{i}\,+\hat{j}\,+\hat{k})\cdot (\hat{i}-2\hat{j}\,+\hat{k})=0\] \[\Rightarrow \] \[\vec{r}\cdot (\hat{i}-2\hat{j}\,+\hat{k})-[(1)(1)+(1)(-\,2)+(1)(1)]=0\] \[\Rightarrow \]               \[\vec{r}\cdot (\hat{i}-2\hat{j}\,+\hat{k})-(1-\,2+1)=0\] \[\therefore \]                  \[\vec{r}\cdot (\hat{i}-2\hat{j}\,+\hat{k})=0\]