Find the intercepts cut off by the plane 2x+y-z = 5
Find the intercepts cut off by the plane 2x+y-z = 5
Solution and Explanation
2x+y-z=5...(1)
Dividing both sides of equation(1) by 5, we obtain
\(\frac{2}{5}x+\frac{y}{5}-\frac{z}{5}=1\)
\(\Rightarrow \frac{x}{\frac{5}{2}}+\frac{y}{5}+\frac{z}{-5}=1\)...(2)
It is known that the equation of a plane in intercept form is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\), where a,b,c are the intercepts cut off by the planet at x, y, and z axes respectively.
Therefore, for the given equation, a=\(\frac{5}{2}\), b=5, and c=-5
Thus, the intercepts cut off by the plane are \(\frac{5}{2}\), 5, and -5.
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Concepts Used:
Plane
A surface comprising all the straight lines that join any two points lying on it is called a plane in geometry. A plane is defined through any of the following uniquely:
- Using three non-collinear points
- Using a point and a line not on that line
- Using two distinct intersecting lines
- Using two separate parallel lines
Properties of a Plane:
- In a three-dimensional space, if there are two different planes than they are either parallel to each other or intersecting in a line.
- A line could be parallel to a plane, intersects the plane at a single point or is existing in the plane.
- If there are two different lines that are perpendicular to the same plane then they must be parallel to each other.
- If there are two separate planes which are perpendicular to the same line then they must be parallel to each other.