Question

For the spherical surface x² + y² + z² – 1, the unit outward normal vector at the point \(\left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }},0} \right)\) is given by
1. \(\frac{1}{{\sqrt 2 }}\hat i + \frac{1}{{\sqrt 2 }}\hat j\;\)
2. \(\frac{1}{{\sqrt 2 }}\hat i - \frac{1}{{\sqrt 2 }}\hat j\)
3. \(\hat k\)
4. \(\frac{1}{{\sqrt 3 }}\hat i + \frac{1}{{\sqrt 3 }}\hat j + \frac{1}{{\sqrt 3 }}\hat k\)

Answer 1

Correct Answer - Option 1 : \(\frac{1}{{\sqrt 2 }}\hat i + \frac{1}{{\sqrt 2 }}\hat j\;\)

Concept:

The gradient of a scalar function f(x,y,z) is given by:

\({\rm{grad\;f}} = \frac{{\partial {\rm{f}}}}{{\partial {\rm{x}}}}{\rm{\hat i}} + {\rm{\;}}\frac{{\partial {\rm{f}}}}{{\partial {\rm{y}}}}{\rm{\hat j}} + \frac{{\partial {\rm{f}}}}{{\partial {\rm{x}}}}{\rm{\hat k}}\)

Unit outward normal vector of a function f at a point P is given by:

\({\bf{\vec n}} = \frac{{{{\left( {{\bf{grad}}\;{\bf{f}}} \right)}_{{\bf{at}}\;{\bf{p}}}}}}{{\left| {{\bf{grad}}\;{\bf{f}}} \right|}}\)

Calculation:

Given:

Spherical surface x² + y² + z² = 1, point P is \(\left( {\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }},0} \right)\)

F(x,y,z) = x² + y² + z² – 1 = 0

\({\rm{grad\;f}} = \frac{{\partial {\rm{f}}}}{{\partial {\rm{x}}}}{\rm{\hat i}} + {\rm{\;}}\frac{{\partial {\rm{f}}}}{{\partial {\rm{y}}}}{\rm{\hat j}} + \frac{{\partial {\rm{f}}}}{{\partial {\rm{x}}}}{\rm{\hat k}} = 2{\bf{x\hat i}} + 2{\bf{y\hat j}} + 2{\bf{z\hat k}}\)

(grad f)_{at P} = \(\frac{2}{{\sqrt 2 }}{\rm{\hat i}} + \frac{2}{{\sqrt 2 }}{\rm{\hat j}} + 0{\rm{\hat k}} = \sqrt 2 {\rm{\hat i}} + \sqrt 2 {\rm{\hat j}} + 0{\rm{\hat k}}\)

\(\left| {{\rm{grad\;f}}} \right| = \sqrt {2 + 2} = \sqrt 4 = 2\)

Unit outward normal vector of function f at point P is given by:

\(\vec n = \frac{{{{\left( {{\rm{grad\;f}}} \right)}_{{\rm{at\;p}}}}}}{{\left| {{\rm{grad\;f}}} \right|}} = \frac{{\sqrt 2 {\rm{\hat i}} + \sqrt 2 {\rm{\hat j}}}}{2} = \frac{1}{{\sqrt 2 }}{\bf{\hat i}} + \frac{1}{{\sqrt 2 }}{\bf{\hat j}}\)