Question

2. The magnitude of the vector drawn in a direction perpendicular to the surface
x2 + 2y2 + z = 7 at the point (1,-1, 2) is
.
3
(iv) 6
(ii)
2.
Summer
(1
(iii) 3
WIN
(AMIETE​

Answer 1

Given:

Surface given is x²+2y²+z=7

Point given = (1,-1,2)

To Find:

Find the magnitude of the perpendicular vector to the surface x²+2y²+z=7.

Solution:

f= x²+2y²+z-7

To find the perpendicular vector to any surface we have to find the gradient of the surface. It is given as:

∇f=(i∂/∂x + j∂/∂y + k∂/∂z) (f)

    =(i∂/∂x + j∂/∂y + k∂/∂z) (x²+2y²+z-7)

    =i∂/∂x(x²+2y²+z-7) +j∂/∂y(x²+2y²+z-7) +k∂/∂z(x²+2y²+z-7)

    =i.2x + j4y +k

Now put the values of x=1, y=-1, z=2,

    =i×2×1 +j×4×(-1) +k×1

    =2i -4j +k

Now the magnitude of ∇f is given as :

I∇f I = $\sqrt{2^2+{-4}^2+1^2}$

       = $\sqrt{21}$

Hence the magnitude of a vector perpendicular to the surface x²+2y²+z=7 is  $\sqrt{21}$.