Question

Find the coordinates of the point where the line through the points $\left( 3,-4,-5 \right)\And \left( 2,-3,1 \right)$ crosses the plane $3x+2y+z+14=0$.

Answer 1

Hint: First of all we will write the equation of a line passing through the two given points. We know that the equation of a line passing through two points $P\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\And Q\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is equal to $\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}=\lambda $. Now, using this formula for the equation of the line we can write the equation of a line passing through two given points. Now, form the equation for each x, y and z expression to $\lambda $ and will get the x, y and z in terms of $\lambda $. Now, the coordinates of x, y and z that we got will be on the plane $3x+2y+z+14=0$ as well because the line intersects the plane so these values of x, y and z will satisfy this plane equation. From the solution of this equation, we get the value of $\lambda $ and substituting this value of $\lambda $ in x, y and z values we get the coordinates of the crossing point.

Complete step by step answer:
We have two points $\left( 3,-4,-5 \right)\And \left( 2,-3,1 \right)$ and an equation of a plane and we have to find the coordinates where the line from these two points crosses.
We know that, the line drawn from two points say $P\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\And Q\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is written in the following form:
$\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}=\lambda $
Using this relation, we can also write the equation of a line passing through two given points $\left( 3,-4,-5 \right)\And \left( 2,-3,1 \right)$ so let us assume first point as P and second point as Q then writing the equation of a line,
$\dfrac{x-3}{2-3}=\dfrac{y-\left( -4 \right)}{-3-\left( -4 \right)}=\dfrac{z-\left( -5 \right)}{1-\left( -5 \right)}=\lambda $
$\begin{align}
  & \Rightarrow \dfrac{x-3}{-1}=\dfrac{y+4}{-3+4}=\dfrac{z+5}{1+5}=\lambda \\
 & \Rightarrow \dfrac{x-3}{-1}=\dfrac{y+4}{1}=\dfrac{z+5}{6}=\lambda \\
\end{align}$
Equating each fraction to $\lambda $ we get,
$\dfrac{x-3}{-1}=\lambda $
On cross multiplying the above equation we get,
$\begin{align}
  & x-3=-\lambda \\
 & \Rightarrow x=3-\lambda \\
\end{align}$
$\begin{align}
  & \dfrac{y+4}{1}=\lambda \\
 & \Rightarrow y=\lambda -4 \\
\end{align}$
$\dfrac{z+5}{6}=\lambda $
On cross multiplication of the above equation we get,
$\begin{align}
  & z+5=6\lambda \\
 & \Rightarrow z=6\lambda -5 \\
\end{align}$
Now, we have got the coordinates of the point lying on the line as:
$\begin{align}
  & x=3-\lambda ; \\
 & y=\lambda -4; \\
 & z=6\lambda -5 \\
\end{align}$
Now, let us assume that this is the point which lies on the plane $3x+2y+z+14=0$ so the above coordinates of x, y and z will satisfy the equation of a plane. Substituting these values of x, y and z in the equation of a plane we get,
$\begin{align}
  & 3\left( 3-\lambda \right)+2\left( \lambda -4 \right)+\left( 6\lambda -5 \right)+14=0 \\
 & \Rightarrow 9-3\lambda +2\lambda -8+6\lambda -5+14=0 \\
\end{align}$
Separately adding the terms which contain $\lambda $ and which do not we get,
$5\lambda +10=0$
Subtracting 10 on both the sides we get,
$5\lambda =-10$
Dividing 5 on both the sides we get,
$\lambda =-\dfrac{10}{5}=-2$
Hence, we have got the value of $\lambda $ so putting this value in the coordinates which contain $\lambda $ we get,
$\begin{align}
  & x=3-\left( -2 \right) \\
 & \Rightarrow x=3+2=5 \\
 & y=\left( -2 \right)-4 \\
 & \Rightarrow y=-6 \\
 & z=6\left( -2 \right)-5 \\
 & \Rightarrow z=-12-5=-17 \\
\end{align}$

From the above, we have got the coordinates of the point where line through the points crosses the plane as:
$x=5,y=-6,z=-17$

Note: You can verify the values of x, y and z that you got by substituting these points in the equation of a plane and see whether they are satisfying the equation of a plane or not.
The coordinates of x, y and z that we got is equal to:
$x=5,y=-6,z=-17$
Now, the given equation of plane is:
$3x+2y+z+14=0$
Substituting the coordinates of x, y and z in the above equation we get,
$\begin{align}
  & 3\left( 5 \right)+2\left( -6 \right)+-17+14=0 \\
 & \Rightarrow 15-12-17+14=0 \\
 & \Rightarrow 29-29=0 \\
 & \Rightarrow 0=0 \\
\end{align}$
As we are getting L.H.S equal to R.H.S so the coordinates of x, y and z that we got are satisfying the equation of a plane.
Hence, the coordinates of x, y and z are correct.