Solveeit Logo
Login

Question

Question: Find the coordinates of the point where line through the points \(A=\left( 3,4,1 \right)\) and \(B=\...

Find the coordinates of the point where line through the points A=(3,4,1)A=\left( 3,4,1 \right) and B=(5,1,6)B=\left( 5,1,6 \right) crosses the xy-plane?

Explanation

Solution

We start solving the problem by recalling the definition of the equation of the line passing through two points in a 3-dimensional system. We then find the equation of the required line and get the general form of the point that represents all the points lying on that line. We then use the fact that the equation of xy-plane is z=0z=0 to find the required point.

Complete step-by-step solution:
According to the problem, we have a line passing through the points A=(3,4,1)A=\left( 3,4,1 \right) and B=(5,1,6)B=\left( 5,1,6 \right). We need to find the coordinates of the point where this line crosses the xy-plane.
Let us find the equation of the line first. We know that the equation of the line passing through the points (x1,y1,z1)\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right) and (x2,y2,z2)\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right) is x−x1x2−x1=y−y1y2−y1=z−z1z2−z1\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}. We use this to find the equation of our line which is passing through the points (3,4,1)\left( 3,4,1 \right) and (5,1,6)\left( 5,1,6 \right).
So, we get the equation of the line as x−35−3=y−41−4=z−16−1\dfrac{x-3}{5-3}=\dfrac{y-4}{1-4}=\dfrac{z-1}{6-1}.
⇒x−32=y−4−3=z−15\Rightarrow \dfrac{x-3}{2}=\dfrac{y-4}{-3}=\dfrac{z-1}{5}.
Let us assume x−32=y−4−3=z−15=r\dfrac{x-3}{2}=\dfrac{y-4}{-3}=\dfrac{z-1}{5}=r(say).
So, we get x−32=r\dfrac{x-3}{2}=r, y−4−3=r\dfrac{y-4}{-3}=r, z−15=r\dfrac{z-1}{5}=r.
⇒x−3=2r\Rightarrow x-3=2r, y−4=−3ry-4=-3r, z−1=5rz-1=5r.
⇒x=2r+3\Rightarrow x=2r+3, y=−3r+4y=-3r+4, z=5r+1z=5r+1.
We have found that (2r+3,−3r+4,5r+1)\left( 2r+3,-3r+4,5r+1 \right) represents every point on the line x−32=y−4−3=z−15\dfrac{x-3}{2}=\dfrac{y-4}{-3}=\dfrac{z-1}{5}. We need to find the point at which it crosses the xy-plane.
We know that the equation of the xy-plane is z=0z=0. So, the z-coordinate of the point will be 0 when the lines cross xy-plane.
So, we get 5r+1=05r+1=0.
⇒5r=−1\Rightarrow 5r=-1.
⇒r=−15\Rightarrow r=\dfrac{-1}{5}.
Let us substitute the value of r in the general form of the point (2r+3,−3r+4,5r+1)\left( 2r+3,-3r+4,5r+1 \right).
So, we the point as (2(−15)+3,−3(−15)+4,5(−15)+1)\left( 2\left( \dfrac{-1}{5} \right)+3,-3\left( \dfrac{-1}{5} \right)+4,5\left( \dfrac{-1}{5} \right)+1 \right).
⇒(−25+3,35+4,−55+1)\Rightarrow \left( \dfrac{-2}{5}+3,\dfrac{3}{5}+4,\dfrac{-5}{5}+1 \right).
⇒(−2+155,3+205,−5+55)\Rightarrow \left( \dfrac{-2+15}{5},\dfrac{3+20}{5},\dfrac{-5+5}{5} \right).
⇒(135,235,05)\Rightarrow \left( \dfrac{13}{5},\dfrac{23}{5},\dfrac{0}{5} \right).
⇒(135,235,0)\Rightarrow \left( \dfrac{13}{5},\dfrac{23}{5},0 \right).
We have found the point as (135,235,0)\left( \dfrac{13}{5},\dfrac{23}{5},0 \right) when the line x−32=y−4−3=z−15\dfrac{x-3}{2}=\dfrac{y-4}{-3}=\dfrac{z-1}{5} crosses xy-plane.
∴\therefore The coordinates of the point where line through the points A=(3,4,1)A=\left( 3,4,1 \right) and B=(5,1,6)B=\left( 5,1,6 \right) crosses the xy-plane is (135,235,0)\left( \dfrac{13}{5},\dfrac{23}{5},0 \right).

Note: We should not confuse with the equation of lines and planes in a 3-dimensional system. Whenever we get this type of problem, we should start by finding the equation of the line. We can also verify whether the obtained point lies on the line by checking the collinearity condition. We can also expect problems to find the area of the triangle that is formed with the origin and the two given points.