Question

Find the coordinates of the point where the line $\dfrac{x+1}{2}=\dfrac{y+2}{3}=\dfrac{z+3}{4}$ meets the plane $x+y+4z=6$.

Answer 1

Hint: In this question, we are given an equation of line and an equation of plane. We have to find the coordinates of the point where the given line meets the given plane. For this, we will first find the general form of coordinate of points on the given line. Since the line will meet the plane at one point, that point will lie on the plane and hence, satisfy the equation of the plane. Using this, we will find the particular point.

Complete step by step answer:
Here, we are given the equation of line as:
$\dfrac{x+1}{2}=\dfrac{y+2}{3}=\dfrac{z+3}{4}$.
Let us suppose it to be equal to k, we get:
\[\dfrac{x+1}{2}=\dfrac{y+2}{3}=\dfrac{z+3}{4}=k\]
Solving for x, y, z we get, for x $\dfrac{x+1}{2}=k$.
Cross multiplying we get:
\[\begin{align}
  & \Rightarrow x+1=2k \\
 & \Rightarrow x=2k-1\cdots \cdots \cdots \left( 1 \right) \\
\end{align}\]
For y $\dfrac{y+2}{3}=k$.
Cross multiplying we get:
\[\begin{align}
  & \Rightarrow y+2=3k \\
 & \Rightarrow y=3k-2\cdots \cdots \cdots \left( 2 \right) \\
\end{align}\]
For z, $\dfrac{z+3}{4}=k$.
Cross multiplying we get:
\[\begin{align}
  & \Rightarrow z+3=4k \\
 & \Rightarrow z=4k-3\cdots \cdots \cdots \left( 3 \right) \\
\end{align}\]
Hence, $\left( x,y,z \right)=\left( 2k-1,3k-2,4k-3 \right)$.
This point represents all points lying on the line, where k = 0, 1, 2. . . .
Now, we have to find one of these points which meets the plane $x+y+4z=6$. Hence, the found point will satisfy the equation of the plane.
Putting value of x, y and z in the equation of plane, we get:
\[\begin{align}
  & \Rightarrow 2k-1+3k-2+4\left( 4k-3 \right)=6 \\
 & \Rightarrow 5k-3+16k-12=6 \\
 & \Rightarrow 21k-15=6 \\
 & \Rightarrow 21k=21 \\
 & \Rightarrow k=1 \\
\end{align}\]
Hence, the point which will be on the plane will be given by $\left( 2k-1,3k-2,4k-3 \right)$ when k = 1.
Putting k = 1, we get point as
\[\begin{align}
  & \Rightarrow \left( 2\left( 1 \right)-1,3\left( 1 \right)-2,4\left( 1 \right)-3 \right) \\
 & \Rightarrow \left( 2-1,3-2,4-3 \right) \\
 & \Rightarrow \left( 1,1,1 \right) \\
\end{align}\]

Hence, required point is (1, 1, 1) where lines $\dfrac{x+1}{2}=\dfrac{y+2}{3}=\dfrac{z+3}{4}$ meets the plane $x+y+4z=6$.

Note: Here, students should take care of the signs while finding the general value of x, y and z. These points represent points on line and only the point lying on the plane too will be an intersecting point so we have to put the value in the equation of the plane. Students can check their answers by putting points in the equation of the plane and the equation of the line.