
The point of intersection of the line joining the points $\left( {3,4,1} \right)$ and $\left( {5,1,6} \right)$ and the xy-plane is:
A. $\left( {13,23,0} \right)$
B. $\left( {\dfrac{{13}}{5},\dfrac{{23}}{5},0} \right)$
C. $\left( { - 13,23,0} \right)$
D. $\left( {\dfrac{{ - 13}}{5},\dfrac{{23}}{5},0} \right)$
Answer
162.9k+ views
Hint: In order to solve this type of question, first assume the ratio as k:1. Then, using the section formula i.e., $\left( {x,y,z} \right) = \left( {\dfrac{{{m_1}{x_2} + {m_2}{x_1}}}{{{m_1} + {m_2}}},\dfrac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_1} + {m_2}}},\dfrac{{{m_1}{z_2} + {m_2}{z_1}}}{{{m_1} + {m_2}}}} \right)$ we will find the ratio. Here, $\left( {{x_1},{y_1},{z_1}} \right) = \left( {3,4,1} \right)$ and $\left( {{x_2},{y_2},{z_2}} \right) = \left( {5,1,6} \right)$. Also, ${m_1} = k$ and ${m_2} = 1.$ Then, find the x and y-coordinate the required point of intersection.
Formula used:
$\left( {x,y,z} \right) = \left( {\dfrac{{{m_1}{x_2} + {m_2}{x_1}}}{{{m_1} + {m_2}}},\dfrac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_1} + {m_2}}},\dfrac{{{m_1}{z_2} + {m_2}{z_1}}}{{{m_1} + {m_2}}}} \right)$
Complete step by step solution:
We are given that,
$A\left( {3,4,1} \right)$ and $B\left( {5,1,6} \right)$
Let xy plane divide the line joining the points in the ratio of $k:1.$
In the xy plane, z-coordinate must be $0.$
Solving for z-coordinate,
Using section formula, compare the z-coordinate to get the required ratio,
$ \Rightarrow \dfrac{{k\left( 6 \right) + 1\left( 1 \right)}}{{k + 1}} = 0$
$6k + 1 = 0$
On solving,
$k = \dfrac{{ - 1}}{6}$
Now, solving for x and y
$x = \dfrac{{5k + 1\left( 3 \right)}}{{k + 1}}$ and $y = \dfrac{{1k + 1\left( 4 \right)}}{{k + 1}}$
Substituting the value of k and solving it, we get
$x = \dfrac{{13}}{5}$ and $y = \dfrac{{23}}{5}$
Thus, the point of intersection is $\left( {\dfrac{{13}}{5},\dfrac{{23}}{5},0} \right)$.
Hence, the correct option is B.
Note: The line segment divides the xy plane, this means that the line joining the given points is parallel to z-axis. So, the z-coordinate is 0. Also, make sure that the coordinates of the point that divides the plane is $\left( {x,y,0} \right)$ and not $\left( {0,0,z} \right)$, otherwise it may lead to incorrect answer.
Formula used:
$\left( {x,y,z} \right) = \left( {\dfrac{{{m_1}{x_2} + {m_2}{x_1}}}{{{m_1} + {m_2}}},\dfrac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_1} + {m_2}}},\dfrac{{{m_1}{z_2} + {m_2}{z_1}}}{{{m_1} + {m_2}}}} \right)$
Complete step by step solution:
We are given that,
$A\left( {3,4,1} \right)$ and $B\left( {5,1,6} \right)$
Let xy plane divide the line joining the points in the ratio of $k:1.$
In the xy plane, z-coordinate must be $0.$
Solving for z-coordinate,
Using section formula, compare the z-coordinate to get the required ratio,
$ \Rightarrow \dfrac{{k\left( 6 \right) + 1\left( 1 \right)}}{{k + 1}} = 0$
$6k + 1 = 0$
On solving,
$k = \dfrac{{ - 1}}{6}$
Now, solving for x and y
$x = \dfrac{{5k + 1\left( 3 \right)}}{{k + 1}}$ and $y = \dfrac{{1k + 1\left( 4 \right)}}{{k + 1}}$
Substituting the value of k and solving it, we get
$x = \dfrac{{13}}{5}$ and $y = \dfrac{{23}}{5}$
Thus, the point of intersection is $\left( {\dfrac{{13}}{5},\dfrac{{23}}{5},0} \right)$.
Hence, the correct option is B.
Note: The line segment divides the xy plane, this means that the line joining the given points is parallel to z-axis. So, the z-coordinate is 0. Also, make sure that the coordinates of the point that divides the plane is $\left( {x,y,0} \right)$ and not $\left( {0,0,z} \right)$, otherwise it may lead to incorrect answer.
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