Question

Find the ratio in which the join A(2,1,5) and B(3,4,3) is divided by the plane \[2x + 2y - 2z = 1\]. Also find the coordinates of the point of division.

Answer 1

Hint: First of all, consider the required ratio as a variable then find the coordinates of the point and substitute in the given plane. Thus, we will get the ratio and then substitute in the point which is in terms of the variable to get the coordinates of the point of division.

Complete step-by-step answer:
Given points are \[A\left( {2,1,5} \right){\text{ and }}B\left( {3,4,3} \right)\]
The plane is \[2x + 2y - 2z = 1\]
We know that if \[P\left( {{x_1},{y_1},{z_1}} \right){\text{ and }}Q\left( {{x_2},{y_2},{z_2}} \right)\] are two points, then the point \[R\] which divides the line joining \[P\] and \[Q\] internally in the ratio \[k:1\] is given by \[R = \left( {\dfrac{{k{x_2} + {x_1}}}{{k + 1}},\dfrac{{k{y_2} + {y_1}}}{{k + 1}},\dfrac{{k{z_2} + {z_1}}}{{k + 1}}} \right)\].
Let \[C\] be the point that divides the line joining \[A{\text{ and }}B\] in then ratio \[k:1\]. Then we have
\[
  C = \left( {\dfrac{{k\left( 3 \right) + 2}}{{k + 1}},\dfrac{{k\left( 4 \right) + 1}}{{k + 1}},\dfrac{{k\left( 3 \right) + 5}}{{k + 1}}} \right) \\
  C = \left( {\dfrac{{3k + 2}}{{k + 1}},\dfrac{{4k + 1}}{{k + 1}},\dfrac{{3k + 5}}{{k + 1}}} \right) \\
\]
But this point \[C\] lies on the given plane \[2x + 2y - 2z = 1\]. So, we have
\[
   \Rightarrow 2\left( {\dfrac{{3k + 2}}{{k + 1}}} \right) + 2\left( {\dfrac{{4k + 1}}{{k + 1}}} \right) - 2\left( {\dfrac{{3k + 5}}{{k + 1}}} \right) = 1 \\
   \Rightarrow \dfrac{2}{{k + 1}}\left( {3k + 2 + 4k + 1 - 3k - 5} \right) = 1 \\
   \Rightarrow 2\left( {3k + 2 + 4k + 1 - 3k - 5} \right) = k + 1 \\
   \Rightarrow 6k + 8k - 6k + 4 + 2 - 10 = k + 1 \\
   \Rightarrow 8k - 4 = k + 1 \\
   \Rightarrow 8k - k = 1 + 4 \\
   \Rightarrow 7k = 5 \\
  \therefore k = \dfrac{5}{7} \\
\]
So, the required ratio \[k:1 = \dfrac{5}{7}:1 = 5:7\]
By substituting the value of \[k\] in point \[C\], we get
\[
  C = \left( {\dfrac{{3\left( {\dfrac{5}{7}} \right) + 2}}{{\left( {\dfrac{5}{7}} \right) + 1}},\dfrac{{4\left( {\dfrac{5}{7}} \right) + 1}}{{\left( {\dfrac{5}{7}} \right) + 1}},\dfrac{{3\left( {\dfrac{5}{7}} \right) + 5}}{{\left( {\dfrac{5}{7}} \right) + 1}}} \right) \\
  C = \left( {\dfrac{{15 + 14}}{{5 + 7}},\dfrac{{20 + 7}}{{5 + 7}},\dfrac{{15 + 35}}{{5 + 7}}} \right) \\
  \therefore C = \left( {\dfrac{{29}}{{12}},\dfrac{{27}}{{12}},\dfrac{{50}}{{12}}} \right) \\
\]
Thus, the ratio in which join the \[A\left( {2,1,5} \right){\text{ and }}B\left( {3,4,3} \right)\] is divided by the plane \[2x + 2y - 2z = 1\] is \[5:7\]and the coordinates of the point of division is \[\left( {\dfrac{{29}}{{12}},\dfrac{{27}}{{12}},\dfrac{{50}}{{12}}} \right)\].

Note: If \[P\left( {{x_1},{y_1},{z_1}} \right){\text{ and }}Q\left( {{x_2},{y_2},{z_2}} \right)\] are two points, then the point \[R\] which divides the line joining \[P\] and \[Q\] internally in the ratio \[k:1\] is given by \[R = \left( {\dfrac{{k{x_2} + {x_1}}}{{k + 1}},\dfrac{{k{y_2} + {y_1}}}{{k + 1}},\dfrac{{k{z_2} + {z_1}}}{{k + 1}}} \right)\].