मराठी

Find the Vector Equation of a Plane Which is at a Distance of 5 Units from the Origin and Which is Normal to the Vector ^ I − 2 ^ J − 2 ^ K .

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प्रश्न

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

 

बेरीज
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उत्तर

`  \text{ It is given that the normal vector, }\vec{n} = \hat{i   } - 2   \hat{j  }- 2  \hat{k  }`

`  \text{ Now, }  \hat{ n } = \frac{\vec{n}}{| \vec{n} |} = \frac{ \hat{i } - 2  \hat{j  } - 2 \hat{k  }}{\sqrt{1 + 4 + 4}} = \frac{\hat{i } - 2 \hat{j } - 2 \hat{k }}{3} = \frac{1}{3} \hat{i  } - \frac{2}{3} \hat{j  } - \frac{2}{3}  \hat{ k  }`

The equation of a plane in normal form is  

`   \vec{r} . \hat{n } =\text{  d (where d is the distance of the plane from the origin) }  `

`  \text{ Substituting}   \hat{n }=\frac{1}{3} \hat{i  } - \frac{2}{3} \hat{j }- \frac{2}{3} \hat{k } andd= 5` 

Here, 

`  \vec{r} . ( \frac{1}{3} \hat{i }- \frac{2}{3}\hat{ j } - \frac{2}{3} \hat{ k ) = 5 `

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 28: The Plane - Exercise 29.04 [पृष्ठ १९]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 28 The Plane
Exercise 29.04 | Q 2 | पृष्ठ १९
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