
If a plane passes through the intersection of the plane $2x - y - 4 = 0$ and $y + 2z - 4 = 0$ and also $\left( {1,1,0} \right)$. Then the equation of the plane is:
A. $x - y - z = 0$
B. $2x - z = 0$
C. $x + 2z - 1 = 0$
D. $x - z - 1 = 0$
Answer
162.9k+ views
Hint: The question requires us to determine the equation of the plane that passes through the specified points, and we have already provided the equation at the point where the plane intersects. We will simply put the given equation into the formula to solve it, find the value of $\lambda $ by simplifying it, and then substitute this value to obtain the equation of the plane.
Formula Used: Equation of the plane passes through the intersection point of two planes:
$P_3 = P_1 + \lambda P_2$.
Complete step-by-step solution:
Let the first plane be $P_1 = 2x - y - 4 = 0$ and the second plane be $P_2 = y + 2z - 4 = 0$.
And if the plane passes through the intersection the basic equation of the plane is formed $P_3 = P_1 + \lambda P_2$.
Now substituting the values of both the plane in $P_3 = P_1 + \lambda P_2$.
$P_3 = 2x - y - 4 + \lambda \left( {y + 2z - 4} \right) = 0$.
Also given that the plane passes with $\left( {1,1,0} \right)$ which means the above points will satisfy this equation of the third plane.
Now substituting the values $x,y{\rm{ and }}z$ in the above equation we will get,
$\begin{array}{I}
2 - 1 - 4 + \lambda \left( {1 + 0 - 4} \right) = 0\\
- 3 - 3\lambda = 0\\
3\lambda = - 3\\
\lambda = - 1
\end{array}$
Now substituting the value of $\lambda $ in $P_3 = 2x - y - 4 + \lambda \left( {y + 2z - 4} \right) = 0$ we will get the required equation of the plane.
$\begin{array}{I}
P_3 = 2x - y - 4 - 1\left( {y + 2z - 4} \right) = 0\\
\Rightarrow P_3 = 2x - y - 4 - y - 2z + 4 = 0\\
\Rightarrow P_3 = 2x - 2y - 2z = 0\\
\Rightarrow P_3 = x - y - z = 0
\end{array}$
Hence the required equation of the plane is $P_3 = x - y - z = 0$.
So, option A is correct.
Note:One could mistakenly assume in this problem that the intersection of two planes creates a plane. But no, the intersection of two planes is always a line because if two points are in the same plane, then the entire line that connects them must also be in the same plane.
Formula Used: Equation of the plane passes through the intersection point of two planes:
$P_3 = P_1 + \lambda P_2$.
Complete step-by-step solution:
Let the first plane be $P_1 = 2x - y - 4 = 0$ and the second plane be $P_2 = y + 2z - 4 = 0$.
And if the plane passes through the intersection the basic equation of the plane is formed $P_3 = P_1 + \lambda P_2$.
Now substituting the values of both the plane in $P_3 = P_1 + \lambda P_2$.
$P_3 = 2x - y - 4 + \lambda \left( {y + 2z - 4} \right) = 0$.
Also given that the plane passes with $\left( {1,1,0} \right)$ which means the above points will satisfy this equation of the third plane.
Now substituting the values $x,y{\rm{ and }}z$ in the above equation we will get,
$\begin{array}{I}
2 - 1 - 4 + \lambda \left( {1 + 0 - 4} \right) = 0\\
- 3 - 3\lambda = 0\\
3\lambda = - 3\\
\lambda = - 1
\end{array}$
Now substituting the value of $\lambda $ in $P_3 = 2x - y - 4 + \lambda \left( {y + 2z - 4} \right) = 0$ we will get the required equation of the plane.
$\begin{array}{I}
P_3 = 2x - y - 4 - 1\left( {y + 2z - 4} \right) = 0\\
\Rightarrow P_3 = 2x - y - 4 - y - 2z + 4 = 0\\
\Rightarrow P_3 = 2x - 2y - 2z = 0\\
\Rightarrow P_3 = x - y - z = 0
\end{array}$
Hence the required equation of the plane is $P_3 = x - y - z = 0$.
So, option A is correct.
Note:One could mistakenly assume in this problem that the intersection of two planes creates a plane. But no, the intersection of two planes is always a line because if two points are in the same plane, then the entire line that connects them must also be in the same plane.
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