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Find the Vector Equation of a Plane Which is at a Distance of 5 Units from the Origin and Its Normal Vector is 2 ^ I − 3 ^ J + 6 ^ K . - Mathematics

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Question

Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is \[2 \hat{i} - 3 \hat{j} + 6 \hat{k} \] .

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Solution

Given: 

\[\text{Normal vector } , \hat{n} = 2 \hat{i} - 3 \hat{j} + 6 \hat{k}  \]
\[\text{ Perpendicular distance, d = 5 units } \]

The vector equation of a plane that is at a distance of 5 units from the origin and has its normal vector \[\hat{n} = 2 \hat{i}  - 3 \hat{j}  + 6 \hat{k} \] is as follows:

\[\vec{r .} \hat{n}  = d\]         
  \[\vec{r .} (2 \hat{i} - 3 \hat{j} + 6 \hat{k}  ) = 5\]

 

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Chapter 29: The Plane - Very Short Answers [Page 84]

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RD Sharma Mathematics [English] Class 12
Chapter 29 The Plane
Very Short Answers | Q 22 | Page 84

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