Question:
The distance of the origin from the plane through the points $(1, 1, 0)$, $(1, 2, 1)$ and $(-2, 2, -1)$ is
The distance of the origin from the plane through the points $(1, 1, 0)$, $(1, 2, 1)$ and $(-2, 2, -1)$ is
Updated On: Sep 3, 2024
- $\frac{3}{\sqrt{11}}$
- $\frac{5}{\sqrt{22}}$
- $3$
- $\frac{4}{\sqrt{22}}$
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The Correct Option is B
Solution and Explanation
The equation of plane is $\begin{vmatrix}x-1&y-1&z\\ 0&1&1\\ -3&1&-1\end{vmatrix}=0$
$\Rightarrow 2x + 3y - 3z = 5$
Distance of origin $\left(0, 0,0\right)$ from the plane is
$\left|\frac{-5}{\sqrt{2^{2}+3^{2}+3^{2}}}\right|=\frac{5}{\sqrt{22}}$.
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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.