Find the co-ordinates of the point where the line passing through $\left( 3,-4,-5 \right)$ and $\left( 2,-3,1 \right)$ crosses the plane, passing through the points $\left( 2,2,1 \right)$, $\left( 3,0,1 \right)$ and $\left( 4,-1,0 \right)$.
Answer
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Hint: We start solving the problem by finding the equation of the line passing through the points $\left( 3,-4,-5 \right)$ and $\left( 2,-3,1 \right)$. We find the direction ratios of that line. We then find the collinearity of the points $\left( 2,2,1 \right)$, $\left( 3,0,1 \right)$ and $\left( 4,-1,0 \right)$. After finding the collinearity of the points we find the equation of the plane. After finding the equation of the plane, we substitute the direction ratios in the plane to get the required point.
Complete step-by-step answer:
Given that we had a line passing through the points $\left( 3,-4,-5 \right)$ and $\left( 2,-3,1 \right)$. We need to find the point at which this line crosses the plane, which is passing through the points $\left( 2,2,1 \right)$, $\left( 3,0,1 \right)$ and $\left( 4,-1,0 \right)$.
Let us find the equation of the line passing through the points $\left( 3,-4,-5 \right)$ and $\left( 2,-3,1 \right)$.
We know that the equation of the line passing through the points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is $\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}=r$.
So, the equation of the line passing through the points $\left( 3,-4,-5 \right)$ and $\left( 2,-3,1 \right)$ is $\dfrac{x-3}{2-3}=\dfrac{y+4}{-3+4}=\dfrac{z+5}{1+5}=r$.
$\Rightarrow \dfrac{x-3}{-1}=\dfrac{y+4}{1}=\dfrac{z+5}{6}=r$.
Let us find the direction ratios of this line.
$\Rightarrow \dfrac{x-3}{-1}=r;\dfrac{y+4}{1}=r;\dfrac{z+5}{6}=r$.
$\Rightarrow x-3=-r;y+4=r;z+5=6r$.
$\Rightarrow x=-r+3;y=r-4;z=6r-5$.
Any point on the line is represented by these direction ratios of the line $\left( -r+3,r-4,6r-5 \right)$ -(1).
We now find the equation of the plane passing through $\left( 2,2,1 \right)$, $\left( 3,0,1 \right)$ and $\left( 4,-1,0 \right)$.
To find the equation of the plane containing three points, we first need to make that the three points are not collinear.
We know if three points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)$ are said to be collinear, then the condition to be satisfied is \[\left| \begin{matrix}
{{x}_{1}} & {{y}_{1}} & {{z}_{1}} \\
{{x}_{2}} & {{y}_{2}} & {{z}_{2}} \\
{{x}_{3}} & {{y}_{3}} & {{z}_{3}} \\
\end{matrix} \right|=0\].
Let us find the value of the determinant \[\left| \begin{matrix}
2 & 2 & 1 \\
3 & 0 & 1 \\
4 & -1 & 0 \\
\end{matrix} \right|\].
We know that the determinant of a $3\times 3$ matrix is defined as $\left| \begin{matrix}
a & b & c \\
d & e & f \\
g & h & i \\
\end{matrix} \right|=a\times \left| \begin{matrix}
e & f \\
h & i \\
\end{matrix} \right|-b\times \left| \begin{matrix}
d & f \\
g & i \\
\end{matrix} \right|+c\times \left| \begin{matrix}
d & e \\
g & h \\
\end{matrix} \right|$.
$\Rightarrow \left| \begin{matrix}
2 & 2 & 1 \\
3 & 0 & 1 \\
4 & -1 & 0 \\
\end{matrix} \right|=2\times \left| \begin{matrix}
0 & 1 \\
-1 & 0 \\
\end{matrix} \right|-2\times \left| \begin{matrix}
3 & 1 \\
4 & 0 \\
\end{matrix} \right|+1\times \left| \begin{matrix}
3 & 0 \\
4 & -1 \\
\end{matrix} \right|$.
We know that determinant of $2\times 2$ matrix is defined as $\left| \begin{matrix}
p & q \\
r & s \\
\end{matrix} \right|=\left( p\times s \right)-\left( q\times r \right)$.
$\Rightarrow \left| \begin{matrix}
2 & 2 & 1 \\
3 & 0 & 1 \\
4 & -1 & 0 \\
\end{matrix} \right|=2\times \left( \left( 0\times 0 \right)-\left( -1\times 1 \right) \right)-2\times \left( \left( 3\times 0 \right)-\left( 1\times 4 \right) \right)+1\times \left( \left( 3\times -1 \right)-\left( 4\times 0 \right) \right)$.
$\Rightarrow \left| \begin{matrix}
2 & 2 & 1 \\
3 & 0 & 1 \\
4 & -1 & 0 \\
\end{matrix} \right|=2\times \left( 0+1 \right)-2\times \left( 0-4 \right)+1\times \left( -3-0 \right)$.
$\Rightarrow \left| \begin{matrix}
2 & 2 & 1 \\
3 & 0 & 1 \\
4 & -1 & 0 \\
\end{matrix} \right|=2\times \left( 1 \right)-2\times \left( -4 \right)+1\times \left( -3 \right)$.
$\Rightarrow \left| \begin{matrix}
2 & 2 & 1 \\
3 & 0 & 1 \\
4 & -1 & 0 \\
\end{matrix} \right|=2+8-3$.
$\Rightarrow \left| \begin{matrix}
2 & 2 & 1 \\
3 & 0 & 1 \\
4 & -1 & 0 \\
\end{matrix} \right|=7$.
So, the points $\left( 2,2,1 \right)$, $\left( 3,0,1 \right)$ and $\left( 4,-1,0 \right)$ are not collinear.
We know that the equation of the plane containing three non- collinear points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)$ is \[\left| \begin{matrix}
x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\
{{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\
{{x}_{3}}-{{x}_{1}} & {{y}_{3}}-{{y}_{1}} & {{z}_{3}}-{{z}_{1}} \\
\end{matrix} \right|=0\].
Now, we find the equation of the plane containing the points $\left( 2,2,1 \right)$, $\left( 3,0,1 \right)$ and $\left( 4,-1,0 \right)$.
The equation of the required plane is \[\left| \begin{matrix}
x-2 & y-2 & z-1 \\
3-2 & 0-2 & 1-1 \\
4-2 & -1-2 & 0-1 \\
\end{matrix} \right|=0\].
\[\Rightarrow \left| \begin{matrix}
x-2 & y-2 & z-1 \\
1 & -2 & 0 \\
2 & -3 & -1 \\
\end{matrix} \right|=0\].
\[\Rightarrow \left( x-2 \right)\times \left| \begin{matrix}
-2 & 0 \\
-3 & -1 \\
\end{matrix} \right|-\left( y-2 \right)\times \left| \begin{matrix}
1 & 0 \\
2 & -1 \\
\end{matrix} \right|+\left( z-1 \right)\times \left| \begin{matrix}
1 & -2 \\
2 & -3 \\
\end{matrix} \right|=0\].
\[\Rightarrow \left( x-2 \right)\times \left( \left( -2\times -1 \right)-\left( 0\times -3 \right) \right)-\left( y-2 \right)\times \left( \left( 1\times -1 \right)-\left( 2\times 0 \right) \right)+\left( z-1 \right)\times \left( \left( 1\times -3 \right)-\left( -2\times 2 \right) \right)=0\].
\[\Rightarrow \left( x-2 \right)\times \left( \left( 2 \right)-\left( 0 \right) \right)-\left( y-2 \right)\times \left( \left( -1 \right)-\left( 0 \right) \right)+\left( z-1 \right)\times \left( \left( -3 \right)-\left( -4 \right) \right)=0\].
\[\Rightarrow \left( x-2 \right)\times \left( 2-0 \right)-\left( y-2 \right)\times \left( -1-0 \right)+\left( z-1 \right)\times \left( -3+4 \right)=0\].
\[\Rightarrow \left( x-2 \right)\times \left( 2 \right)-\left( y-2 \right)\times \left( -1 \right)+\left( z-1 \right)\times \left( 1 \right)=0\].
\[\Rightarrow 2x-4+y-2+z-1=0\].
\[\Rightarrow 2x+y+z-7=0\] -(2).
We have found the equation of plane passing through points $\left( 2,2,1 \right)$, $\left( 3,0,1 \right)$ and $\left( 4,-1,0 \right)$ as \[2x+y+z-7=0\].
Let us substitute the direction cosines of lines we found in equation (1) in the equation of plane (in equation (2)).
We substitute $\left( -r+3,r-4,6r-5 \right)$ in \[2x+y+z-7=0\].
$\Rightarrow 2\left( -r+3 \right)+\left( r-4 \right)+\left( 6r-5 \right)-7=0$.
$\Rightarrow -2r+6+r-4+6r-5-7=0$.
$\Rightarrow 5r-10=0$.
$\Rightarrow 5r=10$.
\[\Rightarrow r=\dfrac{10}{5}\].
\[\Rightarrow r=2\].
Let us substitute the value of ‘r’ in equation (1).
The point is $\left( -2+3,2-4,6\left( 2 \right)-5 \right)$.
$\Rightarrow \left( 1,-2,12-5 \right)$.
$\Rightarrow \left( 1,-2,7 \right)$.
We have found the point at which the line crosses the plane as $\left( 1,-2,7 \right)$.
∴ The co-ordinates of the point where the line passing through $\left( 3,-4,-5 \right)$ and $\left( 2,-3,1 \right)$ crosses the plane, passing through the points $\left( 2,2,1 \right)$, $\left( 3,0,1 \right)$ and $\left( 4,-1,0 \right)$ is $\left( 1,-2,7 \right)$.
Note: We should remember that direction ratios represent all points on the given line. We should find the equation of the plane without checking the collinearity of the given points. We should know that at the point of intersection of two lines, two planes or line and a plane, the values of direction ratios of both lines, planes or line and a plane is equal.
Complete step-by-step answer:
Given that we had a line passing through the points $\left( 3,-4,-5 \right)$ and $\left( 2,-3,1 \right)$. We need to find the point at which this line crosses the plane, which is passing through the points $\left( 2,2,1 \right)$, $\left( 3,0,1 \right)$ and $\left( 4,-1,0 \right)$.
Let us find the equation of the line passing through the points $\left( 3,-4,-5 \right)$ and $\left( 2,-3,1 \right)$.
We know that the equation of the line passing through the points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is $\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}=r$.
So, the equation of the line passing through the points $\left( 3,-4,-5 \right)$ and $\left( 2,-3,1 \right)$ is $\dfrac{x-3}{2-3}=\dfrac{y+4}{-3+4}=\dfrac{z+5}{1+5}=r$.
$\Rightarrow \dfrac{x-3}{-1}=\dfrac{y+4}{1}=\dfrac{z+5}{6}=r$.
Let us find the direction ratios of this line.
$\Rightarrow \dfrac{x-3}{-1}=r;\dfrac{y+4}{1}=r;\dfrac{z+5}{6}=r$.
$\Rightarrow x-3=-r;y+4=r;z+5=6r$.
$\Rightarrow x=-r+3;y=r-4;z=6r-5$.
Any point on the line is represented by these direction ratios of the line $\left( -r+3,r-4,6r-5 \right)$ -(1).
We now find the equation of the plane passing through $\left( 2,2,1 \right)$, $\left( 3,0,1 \right)$ and $\left( 4,-1,0 \right)$.
To find the equation of the plane containing three points, we first need to make that the three points are not collinear.
We know if three points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)$ are said to be collinear, then the condition to be satisfied is \[\left| \begin{matrix}
{{x}_{1}} & {{y}_{1}} & {{z}_{1}} \\
{{x}_{2}} & {{y}_{2}} & {{z}_{2}} \\
{{x}_{3}} & {{y}_{3}} & {{z}_{3}} \\
\end{matrix} \right|=0\].
Let us find the value of the determinant \[\left| \begin{matrix}
2 & 2 & 1 \\
3 & 0 & 1 \\
4 & -1 & 0 \\
\end{matrix} \right|\].
We know that the determinant of a $3\times 3$ matrix is defined as $\left| \begin{matrix}
a & b & c \\
d & e & f \\
g & h & i \\
\end{matrix} \right|=a\times \left| \begin{matrix}
e & f \\
h & i \\
\end{matrix} \right|-b\times \left| \begin{matrix}
d & f \\
g & i \\
\end{matrix} \right|+c\times \left| \begin{matrix}
d & e \\
g & h \\
\end{matrix} \right|$.
$\Rightarrow \left| \begin{matrix}
2 & 2 & 1 \\
3 & 0 & 1 \\
4 & -1 & 0 \\
\end{matrix} \right|=2\times \left| \begin{matrix}
0 & 1 \\
-1 & 0 \\
\end{matrix} \right|-2\times \left| \begin{matrix}
3 & 1 \\
4 & 0 \\
\end{matrix} \right|+1\times \left| \begin{matrix}
3 & 0 \\
4 & -1 \\
\end{matrix} \right|$.
We know that determinant of $2\times 2$ matrix is defined as $\left| \begin{matrix}
p & q \\
r & s \\
\end{matrix} \right|=\left( p\times s \right)-\left( q\times r \right)$.
$\Rightarrow \left| \begin{matrix}
2 & 2 & 1 \\
3 & 0 & 1 \\
4 & -1 & 0 \\
\end{matrix} \right|=2\times \left( \left( 0\times 0 \right)-\left( -1\times 1 \right) \right)-2\times \left( \left( 3\times 0 \right)-\left( 1\times 4 \right) \right)+1\times \left( \left( 3\times -1 \right)-\left( 4\times 0 \right) \right)$.
$\Rightarrow \left| \begin{matrix}
2 & 2 & 1 \\
3 & 0 & 1 \\
4 & -1 & 0 \\
\end{matrix} \right|=2\times \left( 0+1 \right)-2\times \left( 0-4 \right)+1\times \left( -3-0 \right)$.
$\Rightarrow \left| \begin{matrix}
2 & 2 & 1 \\
3 & 0 & 1 \\
4 & -1 & 0 \\
\end{matrix} \right|=2\times \left( 1 \right)-2\times \left( -4 \right)+1\times \left( -3 \right)$.
$\Rightarrow \left| \begin{matrix}
2 & 2 & 1 \\
3 & 0 & 1 \\
4 & -1 & 0 \\
\end{matrix} \right|=2+8-3$.
$\Rightarrow \left| \begin{matrix}
2 & 2 & 1 \\
3 & 0 & 1 \\
4 & -1 & 0 \\
\end{matrix} \right|=7$.
So, the points $\left( 2,2,1 \right)$, $\left( 3,0,1 \right)$ and $\left( 4,-1,0 \right)$ are not collinear.
We know that the equation of the plane containing three non- collinear points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)$ is \[\left| \begin{matrix}
x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\
{{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\
{{x}_{3}}-{{x}_{1}} & {{y}_{3}}-{{y}_{1}} & {{z}_{3}}-{{z}_{1}} \\
\end{matrix} \right|=0\].
Now, we find the equation of the plane containing the points $\left( 2,2,1 \right)$, $\left( 3,0,1 \right)$ and $\left( 4,-1,0 \right)$.
The equation of the required plane is \[\left| \begin{matrix}
x-2 & y-2 & z-1 \\
3-2 & 0-2 & 1-1 \\
4-2 & -1-2 & 0-1 \\
\end{matrix} \right|=0\].
\[\Rightarrow \left| \begin{matrix}
x-2 & y-2 & z-1 \\
1 & -2 & 0 \\
2 & -3 & -1 \\
\end{matrix} \right|=0\].
\[\Rightarrow \left( x-2 \right)\times \left| \begin{matrix}
-2 & 0 \\
-3 & -1 \\
\end{matrix} \right|-\left( y-2 \right)\times \left| \begin{matrix}
1 & 0 \\
2 & -1 \\
\end{matrix} \right|+\left( z-1 \right)\times \left| \begin{matrix}
1 & -2 \\
2 & -3 \\
\end{matrix} \right|=0\].
\[\Rightarrow \left( x-2 \right)\times \left( \left( -2\times -1 \right)-\left( 0\times -3 \right) \right)-\left( y-2 \right)\times \left( \left( 1\times -1 \right)-\left( 2\times 0 \right) \right)+\left( z-1 \right)\times \left( \left( 1\times -3 \right)-\left( -2\times 2 \right) \right)=0\].
\[\Rightarrow \left( x-2 \right)\times \left( \left( 2 \right)-\left( 0 \right) \right)-\left( y-2 \right)\times \left( \left( -1 \right)-\left( 0 \right) \right)+\left( z-1 \right)\times \left( \left( -3 \right)-\left( -4 \right) \right)=0\].
\[\Rightarrow \left( x-2 \right)\times \left( 2-0 \right)-\left( y-2 \right)\times \left( -1-0 \right)+\left( z-1 \right)\times \left( -3+4 \right)=0\].
\[\Rightarrow \left( x-2 \right)\times \left( 2 \right)-\left( y-2 \right)\times \left( -1 \right)+\left( z-1 \right)\times \left( 1 \right)=0\].
\[\Rightarrow 2x-4+y-2+z-1=0\].
\[\Rightarrow 2x+y+z-7=0\] -(2).
We have found the equation of plane passing through points $\left( 2,2,1 \right)$, $\left( 3,0,1 \right)$ and $\left( 4,-1,0 \right)$ as \[2x+y+z-7=0\].
Let us substitute the direction cosines of lines we found in equation (1) in the equation of plane (in equation (2)).
We substitute $\left( -r+3,r-4,6r-5 \right)$ in \[2x+y+z-7=0\].
$\Rightarrow 2\left( -r+3 \right)+\left( r-4 \right)+\left( 6r-5 \right)-7=0$.
$\Rightarrow -2r+6+r-4+6r-5-7=0$.
$\Rightarrow 5r-10=0$.
$\Rightarrow 5r=10$.
\[\Rightarrow r=\dfrac{10}{5}\].
\[\Rightarrow r=2\].
Let us substitute the value of ‘r’ in equation (1).
The point is $\left( -2+3,2-4,6\left( 2 \right)-5 \right)$.
$\Rightarrow \left( 1,-2,12-5 \right)$.
$\Rightarrow \left( 1,-2,7 \right)$.
We have found the point at which the line crosses the plane as $\left( 1,-2,7 \right)$.
∴ The co-ordinates of the point where the line passing through $\left( 3,-4,-5 \right)$ and $\left( 2,-3,1 \right)$ crosses the plane, passing through the points $\left( 2,2,1 \right)$, $\left( 3,0,1 \right)$ and $\left( 4,-1,0 \right)$ is $\left( 1,-2,7 \right)$.
Note: We should remember that direction ratios represent all points on the given line. We should find the equation of the plane without checking the collinearity of the given points. We should know that at the point of intersection of two lines, two planes or line and a plane, the values of direction ratios of both lines, planes or line and a plane is equal.
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