The unit normal to the surface x² + y² + z² = 3 at (1,1,1) is A)1/√16 (2i+2j+2k) B)1/√3(i+j+ k)
C) 1/√6(i+2j+3k)
D)1/√3(i+2j+2k)
x2+y2+z2=3f(x,y,z)=x2+y2+z2−3=0x² + y² + z² = 3\\ f ( x , y , z )= x² + y² + z² - 3=0x2+y2+z2=3f(x,y,z)=x2+y2+z2−3=0
The gradient of f(x,y,z)f ( x , y , z )f(x,y,z) at point x,y,zx , y , zx,y,z is a vector normal to the surface at this point.
The gradient is obtained as follows
∇f(x,y,z)=(fx,fy,fz)=2x+2y+2z∇ f ( x , y , z ) = ( f_ x , f_ y , f_ z ) = 2x+2y+2z∇f(x,y,z)=(fx,fy,fz)=2x+2y+2z at point (1,1,1) and the unit vector is
=2,2,22+2+2=28,28,28=12,12,12=\frac{2,2,2}{\sqrt{2+2+2}}\\ =\frac{2}{\sqrt{8}}, \frac{2}{\sqrt{8}},\frac{2}{\sqrt{8}}\\ =\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}=2+2+22,2,2=82,82,82=21,21,21
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments