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Vector equation of plane passing through point with position vector \hat{i}+\hat{j}+\hat{k} and normal to  a given vector 2\hat{i}-\hat{j}+\hat{k} is

Option: 1

\vec r (2 \hat i - \hat j + \hat k ) = 2


Option: 2

\vec r (\hat i - 2\hat j + \hat k ) = 2


Option: 3

\vec r (\hat i + \hat j + \hat k ) = \sqrt 6


Option: 4

\vec r ( \hat i - \hat j - \hat k ) = 1


Answers (1)

As we have learned

Normal form (vector form ) -

\vec{r}\cdot \hat{n}= d ,where d is the distance from origin.

 

- wherein

Let a plane be at a distance of d units from origin and normal unit vector is \hat{n}.Let O be the origin and ON be the perpendicular from origin to given plane then ON=d\hat{n}.

Let P\hat{r} be the general point on the plane NP will be perpendicular to \hat{n}\left (\vec{r}-d\hat{n} \right )\cdot \hat{n}= 0

\vec{r}\cdot \hat{n}= d

 

 

 

Conversion of equation in normal form (vector form ) -

The equation \vec{r}\cdot \vec{n}= D   is converted in normal by divding it by \left | \vec{n} \right |

\frac{\vec{r}\cdot \vec{n}}{\left | \vec{n} \right |}= \frac{D}{\left | \vec{n} \right |}    

we get     \vec{r}\cdot \hat{n}= d

 

-

 

 \vec r (2 \hat i - \hat j + \hat k ) = d

2 - 1+1 = d 

d = 2 

So, \vec r (2 \hat i - \hat j + \hat k ) = 2

 

 

Posted by

Sumit Saini

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