Question

A line passes through two points A(2, -3, -1) and B(8, -1, 2). The coordinates of a point on this line at a distance of 14 units from A are
A) (14, 1, 5)
B) (-10, -7, 7)
C) (86, 25, 41)
D) (0, 0, 0)

Answer 1

Hint: If the three-dimensional coordinates of the points A and B are given as $\left( {{x_1},{y_1},{z_1}} \right)$and $\left( {{x_2},{y_2},{z_2}} \right)$ then considering the rectangular coordinates of point R as \[\left( {x,{\text{ }}y,{\text{ }}z} \right)\]
The Cartesian equation of a line is as follows:
$\dfrac{{x - {x_1}}}{{{x_2} - {x_1}}} = \dfrac{{y - {y_1}}}{{{y_2} - {y_1}}} = \dfrac{{z - {z_1}}}{{{z_2} - {z_1}}}$
The distance formula is an algebraic expression used to determine the distance between two points using the coordinates of the points.
Distance between two points A$\left( {{x_1},{y_1},{z_1}} \right)$ and B$\left( {{x_1},{y_1},{z_1}} \right)$ be ‘r’
${\left( {{x_1} - {x_2}} \right)^2} + {\left( {{y_1} - {y_2}} \right)^2} + {\left( {{z_1} - {z_2}} \right)^2} = {r^2}$
In this question, assume the point is P(x, y, z) by using the equation of line passing through two points and equating it to constant $\lambda $ we can find out the value of the coordinates of the point on the line. We will get the value of coordinates in terms of$\lambda $. With the help of distance formula, applying for point A and point p as the distance between the points is given, with this we will get the value of$\lambda $. Finally, we have to put the value of $\lambda $ in the value of coordinates we get earlier. Hence, we will get the coordinates of the point P.

Complete answer:
According to the question,
The coordinates of point A = $\left( {{x_1},{y_1},{z_1}} \right) = \left( {2, - 3, - 1} \right)$
The coordinates of point B = $\left( {{x_2},{y_2},{z_2}} \right) = \left( {8, - 1,2} \right)$
Let the point on the line be P having the coordinate as (x, y, z)
On substituting the values,
Equation of line passing through two points
$\begin{gathered}
  \dfrac{{x - {x_1}}}{{{x_2} - {x_1}}} = \dfrac{{y - {y_1}}}{{{y_2} - {y_1}}} = \dfrac{{z - {z_1}}}{{{z_2} - {z_1}}} = \lambda \\
  \dfrac{{x - 2}}{{8 - 2}} = \dfrac{{y - ( - 3)}}{{ - 1 - ( - 3)}} = \dfrac{{z - ( - 1)}}{{2 - ( - 1)}} = \lambda \\
  \dfrac{{x - 2}}{6} = \dfrac{{y + 3}}{2} = \dfrac{{z + 1}}{3} = \lambda \\
\end{gathered} $
Hence we get,
\[x - 2 = 6\lambda \]
\[\begin{gathered}
  x = 6\lambda + 2 \ldots \left( 1 \right) \\
  y + 3 = 2\lambda \\
  y = 2\lambda - 3 \ldots \left( 2 \right) \\
  z + 1 = 3\lambda \\
  z = 3\lambda - 1 \ldots \left( 3 \right) \\
\end{gathered} \]
Distance between A $\left( {2, - 3, - 1} \right)$ and general point P(x, y, z) of line is 14 unit.
From the equations (1), (2) and (3)
$\begin{gathered}
  {\left( {6\lambda + 2 - 2} \right)^2} + {\left( {2\lambda - 3 - \left( { - 3} \right)} \right)^2} + {\left( {3\lambda - 1 - \left( { - 1} \right)} \right)^2} = {14^2} \\
  {\left( {6\lambda } \right)^2} + {\left( {2\lambda } \right)^2} + {\left( {3\lambda } \right)^2} = 196 \\
  36{\lambda ^2} + 4{\lambda ^2} + 9{\lambda ^2} = 196 \\
  49{\lambda ^2} = 196 \\
  {\lambda ^2} = 4 \\
  \lambda = \pm 2 \\
\end{gathered} $
Now, by putting the value of $\lambda $
When \[\lambda = 2\]
x=14, y=1, z=5
When \[\lambda = - 2\]
x=(-10), y=(-7), z=(-7)
So, the coordinates of the point is (14, 1, 5)
∴Option (A) is correct

Note: If the endpoints of a line segment is (x1, y1, z1) and (x2, y2, z2) then the midpoint of the line segment has the coordinates:
$\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3},\dfrac{{{z_1} + {z_2} + {z_3}}}{3}} \right)$