Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector \(3\hat i+5\hat j-6\hat k.\)
Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector \(3\hat i+5\hat j-6\hat k.\)
Solution and Explanation
The normal vector is,
\(\overrightarrow n= 3\hat i+5\hat j-6\hat k.\)
∴\(\hat n=\frac{\overrightarrow n}{\mid \overrightarrow n\mid}\)
=\(\frac{3\hat i+5\hat j-6\hat k.}{\sqrt{(3)^2+(5)^2+(6)^2}}\)
=\(\frac{3\hat i+5\hat j-6\hat k.}{\sqrt 70}\)
It is known that the equation of the plane with position vector \(\overrightarrow r\) is given by,
\(\overrightarrow r.\hat n=d\)
\(\Rightarrow \hat r.\bigg(\frac{3\hat i+5\hat j-6\hat k.}{\sqrt {70}}\bigg)=7\)
This is the vector equation of the required plane.
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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.