Question

Findthe angle between the normals to the surface xy=z² at the points(4, 1, 2) and(3,3,-3)​

Answer 1

Given:

Surface xy=z², points (4,1,2) and (3,3,-3).

To find:

The angle between the normals to the surface xy=z² at the points (4, 1, 2) and (3,3,-3)​.

Solution:

We can represent the curve as $F(x, y, z)=c$, where c is a constant.

According to the question,  $F(x, y, z)=x y-z^{2}=0$.

$Now, normal vector to F is:\\ \nabla F=\left(\frac{\partial}{\partial x} i+\frac{\partial}{\partial y} j+\frac{\partial}{\partial z} k\right)\left(x y-z^{2}\right)=y i+x j-2 z k$

$The normal vector at (4,1,2) is i+4 j-4 k , and the normal vector at (3,3,-3) is 3 i+3 j+6 k .$

$Finally, the angle between the two vectors i+4 j-4 k and 3 i+3 j+6 k is given by \begin{aligned}\theta &=\cos ^{-1}\left(\frac{( i+4 j-4 k) \cdot(3 i+3 j+6 k)}{\left|i+4 j-4 k\right||3 i+3 j+6 k \mid}\right) \\&=\cos ^{-1}(-\frac{1}{\sqrt{22} } ) \approx 77.7^{\circ} .\end{aligned}$

Therefore, the angle between the normals to the surface xy=z² at the points (4, 1, 2) and (3,3,-3)​ is 77.7°.

Answer 2

Answer:

Step-by-step explanation: