Question

Find the equation of the plane that bisects the line segment joining the points (1, 2, 3) and (3, 4, 5) and is at right angle to it.

 

Answer 1

\[\text{ The normal is passing through the points A(1, 2, 3) and B(3, 4, 5). So } ,\]
\[ \vec{n} = \vec{AB} = \vec{OB} - \vec{OA} =\left( \text{ 3 }\hat{i}  + \text{  4 }\hat{j}  + \text{ 5  }\hat{k}  \right) - \left( \hat{i}  + \text{ 2 }\hat{j}  + \text{ 3 }\hat{k}  \right) = \text{ 2 } \hat{i}  + \text{ 2 }\hat{j}  + \text{ 2  }\hat{k} \]
\[\text{ Mid-point of AB } =\left( \frac{1 + 3}{2}, \frac{2 + 4}{2}, \frac{3 + 5}{2} \right)=\left( 2, 3, 4 \right)\]
\[ \text{ Since the plane passes through } \left( 2, 3, 4 \right), \vec{a} =2 \hat{i}  + 3 \hat{j}  + 4 \hat{k}  \]
\[\text{ We know that the vector equation of the plane passing through a point }  \vec{a} \text{ and normal to}  \vec { n } \text { is } \]
\[ \vec{r} . \vec{n} = \vec{a} . \vec{n} \]
\[\text{ Substituting } \vec{a} = \hat{i} - \hat{j}  + \hat{ k} \text{ and }  \vec{n} = \text{ 4 } \hat{i} + \text{  2 } \hat{j} - \text{ 3 }\hat{k}  , \text{ we get } \]
\[ \vec{r} . \left( \text{ 2 } \hat{i}  + \text{  2 }\hat{j} + \text{ 2 }\hat{k}  \right) = \left( \text{ 2 }\hat{i} + \text{ 3 } \hat{j}  + \text{ 4 } \hat{k}  \right) . \left( \text{ 2 } \hat{i} + \text{  2  }\hat{j}  + \text{ 2 }\hat{k}  \right)\]
\[ \Rightarrow \vec{r} . \left( \text{ 2 } \hat{i}  + \text{ 2 } \hat{j}  + \text{  2 } \hat{k}  \right) = 4 + 6 + 8\]
\[ \Rightarrow \vec{r} . \left( \text{  2 }\hat{i}  + \text{ 2 } \hat{j}  + \text{ 2 } \hat{k}  \right) = 18\]
\[ \Rightarrow \vec{r} . \left[ \text{ 2 }\left( \hat{i} + \hat{j}  + \hat{k}  \right) \right] = 18\] 
\[ \Rightarrow \vec{r} . \left( \hat{i}  + \hat{j}  + \hat{k} \right) = 9\]
\[\text{ Substituting } \vec{r} = \text{ x }\hat{i}  + \text{ y }\hat{j}  + z \hat{k}  \text{ in the vector equation, we get } \]
\[\left( \text{ x }\hat{i}  + \text{ y } \hat{j}  + z \hat{k}  \right) . \left( \hat{i}  + \hat{j}  + \hat{k} \right) = 9\]
\[ \Rightarrow x + y + z = 9\]