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Question: Find the ratio in which the join A(2,1,5) and B(3,4,3) is divided by the plane \[2x + 2y - 2z = 1\]....

Find the ratio in which the join A(2,1,5) and B(3,4,3) is divided by the plane 2x+2y2z=12x + 2y - 2z = 1. Also find the coordinates of the point of division.

Explanation

Solution

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(2,1,5) and B(3,4,3)A\left( {2,1,5} \right){\text{ and }}B\left( {3,4,3} \right)
The plane is 2x+2y2z=12x + 2y - 2z = 1
We know that if P(x1,y1,z1) and Q(x2,y2,z2)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 RR which divides the line joining PP and QQ internally in the ratio k:1k:1 is given by R=(kx2+x1k+1,ky2+y1k+1,kz2+z1k+1)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 CC be the point that divides the line joining A and BA{\text{ and }}B in then ratio k:1k:1. Then we have

C=(k(3)+2k+1,k(4)+1k+1,k(3)+5k+1) C=(3k+2k+1,4k+1k+1,3k+5k+1)  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 CC lies on the given plane 2x+2y2z=12x + 2y - 2z = 1. So, we have

2(3k+2k+1)+2(4k+1k+1)2(3k+5k+1)=1 2k+1(3k+2+4k+13k5)=1 2(3k+2+4k+13k5)=k+1 6k+8k6k+4+210=k+1 8k4=k+1 8kk=1+4 7k=5 k=57  \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=57:1=5:7k:1 = \dfrac{5}{7}:1 = 5:7
By substituting the value of kk in point CC, we get

C=(3(57)+2(57)+1,4(57)+1(57)+1,3(57)+5(57)+1) C=(15+145+7,20+75+7,15+355+7) C=(2912,2712,5012)  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(2,1,5) and B(3,4,3)A\left( {2,1,5} \right){\text{ and }}B\left( {3,4,3} \right) is divided by the plane 2x+2y2z=12x + 2y - 2z = 1 is 5:75:7and the coordinates of the point of division is (2912,2712,5012)\left( {\dfrac{{29}}{{12}},\dfrac{{27}}{{12}},\dfrac{{50}}{{12}}} \right).

Note: If P(x1,y1,z1) and Q(x2,y2,z2)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 RR which divides the line joining PP and QQ internally in the ratio k:1k:1 is given by R=(kx2+x1k+1,ky2+y1k+1,kz2+z1k+1)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).