Question

Find the coordinates of the point where the line through the point (3, -4, -5) and (2, -3, 1) crosses the plane 2x + y + z = 7.

Answer 1

Hint: To solve this problem we need to know that when two points $\left( {{a}_{1}},{{b}_{1}},{{c}_{1}} \right)\,and\,\left( {{a}_{2}},{{b}_{2}},{{c}_{2}} \right)$ in 3D are given then the equation of the line passing through both of them is given by $\dfrac{x-{{a}_{1}}}{{{a}_{1}}-{{a}_{2}}}=\dfrac{y-{{b}_{1}}}{{{b}_{2}}-{{b}_{1}}}=\dfrac{z-{{c}_{1}}}{{{c}_{1}}-{{c}_{2}}}$ after finding the equation of the line we will equate the equation of the line with some variable say k, and then represent any point on the line in its general form and will put that point in the equation of the given plant to find the value of k.

Complete step by step answer:
We know that when two points $\left( {{a}_{1}},{{b}_{1}},{{c}_{1}} \right)\,and\,\left( {{a}_{2}},{{b}_{2}},{{c}_{2}} \right)$ in 3D are given then the equation of the line passing through both of them is given by $\dfrac{x-{{a}_{1}}}{{{a}_{1}}-{{a}_{2}}}=\dfrac{y-{{b}_{1}}}{{{b}_{1}}-{{b}_{2}}}=\dfrac{z-{{c}_{1}}}{{{c}_{1}}-{{c}_{2}}}$,
Points that are given to us are point (3, -4, -5) and (2, -3, 1) so equation of the plane we get as,
$\begin{align}
  & \dfrac{x-3}{3-2}=\dfrac{y+4}{-4+3}=\dfrac{z+5}{-5-1} \\
 & \dfrac{x-3}{1}=\dfrac{y+4}{-1}=\dfrac{z+5}{-6} \\
\end{align}$
Now we will find any general point on the above line and for that we will equate the equation of the above lint to a random variable let’s say k,
$\dfrac{x-3}{1}=\dfrac{y+4}{-1}=\dfrac{z+5}{-6}=k$
And any general point we get as,
$\left( k+3,-4-k,-5-6k \right)$
Now if we put the above coordinate in the given plane to find the value of k we will find the coordinates of the intersection point of the line and the plane.
Equation of the plane is 2x + y + z = 7.
Now, Putting$\left( x,y,z \right)=\left( k+3,-4-k,-5-6k \right)$, we get
$\begin{align}
  & 2\left( k+3 \right)+\left( -4-k \right)+\left( -5-6k \right)=7 \\
 & 2k+6-4-k-5-6k=7 \\
 & -5k-3=7 \\
 & k=-\dfrac{10}{5} \\
 & k=-2 \\
\end{align}$
Hence point which passes through the line made by the points (3, -4, -5) and (2, -3, 1) as well as crosses the plane 2x + y + z = 7 is,
$\begin{align}
  & \left( k+3,-4-k,-5-6k \right) \\
 & \left( -2+3,-4+2,-5+12 \right)(k=-2) \\
 & \left( 1,-2,7 \right) \\
\end{align}$

Note:
While solving the questions of 3D vectors we also need to imagine the problem given in 3D, from that we can easily understand what steps we have to follow in order to solve the question.
In the above problem we could also have written the equation of the line as $\dfrac{x-{{a}_{1}}}{{{a}_{2}}-{{a}_{1}}}=\dfrac{y-{{b}_{1}}}{{{b}_{2}}-{{b}_{1}}}=\dfrac{z-{{c}_{1}}}{{{c}_{2}}-{{c}_{1}}}$ in this expression denominator part of each term is reversed but it really does not matter as it is changed for the every term so only general expression will change but we will get the same answer.