Question

Find the distance of the point (2, 12, 5) from the point of intersection of the line \[\vec{r} = 2 \hat{i}  - 4 \hat{j}+ 2 \hat{k}  + \lambda\left( 3 \hat{i}  + 4 \hat{j}  + 2 \hat{k} \right)\] and \[\vec{r} . \left( \hat{i}  - 2 \hat{j}  + \hat{k}  \right) = 0\]

  

Answer 1

The equation of the given line is

\[\vec{r} = 2 \hat{i} - 4 \hat{j}  + 2 \hat{k}  + \lambda\left( 3 \hat{i}  + 4 \hat{j}  + 2 \hat{k} \right)\]

The position vector of any point on the given line is \[\vec{r} = \left( 2 + 3\lambda \right) \hat{i}  + \left( - 4 + 4\lambda \right) \hat{j}  + \left( 2 + 2\lambda \right) \hat{k} \]       .........(1)

If this lies on the plane 

\[\vec{r} . \left( \hat{i}  - 2 \hat{j}  + \hat{k}  \right) = 0\]     then 

\[\left[ \left( 2 + 3\lambda \right) \hat{i}  + \left( - 4 + 4\lambda \right) \hat{j} + \left( 2 + 2\lambda \right) \hat{k} \right] . \left( \hat{i} - 2 \hat{j} + \hat{k} \right) = 0\]

\[ \Rightarrow \left( 2 + 3\lambda \right) - 2\left( - 4 + 4\lambda \right) + \left( 2 + 2\lambda \right) = 0\]

\[ \Rightarrow 2 + 3\lambda + 8 - 8\lambda + 2 + 2\lambda = 0\]

\[ \Rightarrow 3\lambda = 12\]

\[ \Rightarrow \lambda = 4\]

Putting

\[\lambda = 4\]  in (1), we get

\[\left( 2 + 3 \times 4 \right) \hat{i}  + \left( - 4 + 4 \times 4 \right) \hat{j}  + \left( 2 + 2 \times 4 \right) \hat{k} \] or

\[14 \hat{i}  + 12 \hat{j}  + 10 \hat{k} \]

as the coordinate of the point of intersection of the given line and the plane.
The position vector of the given point is 

\[2 \hat{i}  + 12 \hat{j}  + 5 \hat{k} \]

∴ Required distance = Distance between

\[14 \hat{i}  + 12 \hat{j}  + 10 \hat{k} \] and 

\[2 \hat{i}  + 12 \hat{j} + 5 \hat{k} \] 

\[= \left| \left( 14 \hat{i}  + 12 \hat{j}  + 10 \hat{k}  \right) - \left( 2 \hat{i}  + 12 \hat{j}  + 5 \hat{k}  \right) \right|\]
\[ = \left| 12 \hat{i}  + 5 \hat{k}  \right|\]
\[ = \sqrt{{12}^2 + 0^2 + 5^2}\]
\[ = \sqrt{169}\]
\[ = 13 \text{ units } \]