Question

find the equation of the plane passing through the point (1, 2, 1) and perpendicular to the line joining the points (1, 4, 2) and (2, 3, 5). Find also the perpendicular distance of the origin from this plane

Answer 1

\[ \text { The normal is passing through the points A(1, 4, 2) and B (2, 3, 5).  }\]
\[\text{ So } , \vec{n} = \vec{AB} = \vec{OB} - \vec{OA} =\left(\text{  2 } \hat{i} +\text{  3 } \hat{j} + \text{  5  }\hat{k} \right) - \left( \hat{i} + \text{  4 }\hat{j} +\text{  2 }  \hat{k}  \right) = \hat{i}  - \hat{j}  + \text{  3 }\hat{k}  \]
\[ \text{ We know that the vector equation of the plane passing through a point } (1, 2, 1) ( \vec{a} ) \text{ and normal to }  \vec{n} \text{ is } \]
\[ \vec{r} . \vec{n} = \vec{a} . \vec{n} \]
\[ \text{ Substituting } \vec{a} = \hat{i}  +\text{  2 } \hat{j}  + \hat{k}  \text{ and }  \vec{n} = \hat{i}  - \hat{j}  + \text{  3 } \hat{k}  , \text{ we get } \]
\[ \vec{r} . \left( \hat{i} - \hat{j} +\text{   3 } \hat{k}  \right) = \left( \hat{i} +\text{  2 } \hat{j} + \hat{k}  \right) . \left( \hat{i}  - \hat{j}  + \text{  3 } \hat{k}  \right)\]
\[ \Rightarrow \vec{r} . \left( \hat{i} - \hat{j}  +\text{   3 } \hat{k}  \right) = 1 - 2 + 3\]
\[ \Rightarrow \vec{r} . \left( \hat{i}  - \hat{j} + \text{  3 }\hat{k} \right) = 2 . . . \left( 1 \right)\]
\[ \text{ To find the perpendicular distance of this plane from the origin, we have to reduce this to normal form } .\]
\[ \text{ We have } \vec{n} = \hat{i}  - \hat{j}  + \text{  3 } \hat{k}  ; \left| \vec{n} \right| = \sqrt{1 + 1 + 9} = \sqrt{11}\]
\[ \text{ Dividing (1) by }  \sqrt{11}, \text{ we get } \]
\[ \vec{r} . \left( \frac{1}{\sqrt{11}} \hat{i}  - \frac{1}{\sqrt{11}} \hat{j}  + \frac{3}{\sqrt{11}} \hat{k}  \right) = \frac{2}{\sqrt{11}},  \text{ which is the normal form of plane } (1).\]
\[ \text{ So, the perpendicular distance of plane (1) from the origin } =\frac{2}{\sqrt{11}}\]