Question

The distance of the point $\left( {1,2,3} \right)$ to the line $x = 2 + t,y = 2 - 3t,z = 5t$
A) $3\dfrac{4}{7}$
B) $3\dfrac{2}{7}$
C) $4\dfrac{4}{7}$
D) $3\dfrac{4}{9}$

Answer 1

Hint:
Let us assume the line perpendicular to the given line, that is passing through the point $\left( {1,2,3} \right)$ meets the given line at a point $A$. Find the coordinates of $A$ using the property that the line vector \[\overrightarrow {PA} \] will be perpendicular to the given line. Find the distance between point $A$ and point $P$ using the distance formula.

Complete step by step solution:
Let us assume the given line $x = 2 + t,y = 2 - 3t,z = 5t$ represented by $l$.
Let us assume the line perpendicular to the given line $l$, that is passing through the point $\left( {1,2,3} \right)$ meets the given line $l$ at a point $A$.
The positional vector \[\overrightarrow {PA} \]can be given by positional vector of $A$$ - $positional vector of $P$.
The positional vector of $A$is given by the equation of the given line $l$ as $\left( {2 + t} \right)\hat i + \left( {2 - 3t} \right)\hat j + 5t\hat k$.
The positional vector of $P$is given by the point $\left( {1,2,3} \right)$ as $\hat i + 2\hat j + 3\hat k$.
The positional vector of \[\overrightarrow {PA} \] becomes
$
  \left( {2 + t} \right)\hat i + \left( {2 - 3t} \right)\hat j + 5t\hat k - \left( {\hat i + 2\hat j + 3\hat k} \right) \\
   = \left( {1 + t} \right)\hat i - 3t\hat j + \left( {5t - 3} \right)\hat k \\
$
The direction ratio of the given line $l$ can be found by the coefficient of $t$in the equation of the line $x = 2 + t,y = 2 - 3t,z = 5t$.
The direction ratio of $l$are $\left( {1, - 3,5} \right)$, and can be represented in the vector form as
$\vec l = \hat i - 3\hat j + 5\hat k$
The line \[\overrightarrow {PA} \] is perpendicular to the given line $l$, therefore the dot product of the two vectors must be zero.
$
  \overrightarrow {PA} .\vec l = 0 \\
  \left( {\left( {1 + t} \right)\hat i - 3t\hat j + \left( {5t - 3} \right)\hat k} \right).\left( {\hat i - 3\hat j + 5\hat k} \right) = 0 \\
$
On solving, we get the value of $t$
$
  1 + t + 9t + 25t - 15 = 0 \\
  35t = 14 \\
  t = \dfrac{2}{5} \\
$
Thus coordinates of the point $A$can be calculated by substituting the value \[\dfrac{2}{5}\] for $t$in the equation of the line$x = 2 + t,y = 2 - 3t,z = 5t$
 $
  x = 2 + \dfrac{2}{5},y = 2 - 3\left( {\dfrac{2}{5}} \right),z = 5\left( {\dfrac{2}{5}} \right) \\
  x = \dfrac{{12}}{5},y = \dfrac{4}{5},z = 2 \\
 $
The distance between the point $A$$\left( {\dfrac{{12}}{5},\dfrac{4}{5},2} \right)$and point $P$$\left( {1,2,3} \right)$can be calculated by the distance formula.
$
  \left| {PA} \right| = \sqrt {{{\left( {\dfrac{{12}}{5} - 1} \right)}^2} + {{\left( {\dfrac{4}{5} - 2} \right)}^2} + {{\left( {2 - 3} \right)}^2}} \\
   = \sqrt {\dfrac{{49}}{{25}} + \dfrac{{36}}{{25}} + 1} \\
   = \sqrt {\dfrac{{110}}{{25}}} \\
   = \dfrac{{\sqrt {110} }}{5} \\
$

Thus, none of the given options are correct.

Note:
The dot product of two perpendicular vectors is always zero. The perpendicular distance between two points $\left( {{x_1},{y_1},{z_1}} \right)$ and $\left( {{x_2},{y_2},{z_2}} \right)$ in the coordinate geometry is given by $\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2} + {{\left( {{z_1} - {z_2}} \right)}^2}} $.