Question

2:
A vector normal to the
surface xy3z2 = 4 at (-1,-1,2).
0 -41 - 12J + 4K
-41 + 12J + 4K
0 -41 - 12J - 4K
0 -41 - 6J + 4K
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Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

A vector normal to the surface xy³z² = 4 at (-1,-1,2).

• - 4i - 12j + 4k

• - 4i + 12j + 4k

• - 4i - 12j - 4k

• - 4i - 6j + 4k

EVALUATION

Here the given given surface is xy³z² = 4

Now we have to find the vector normal to the surface xy³z² = 4 at (-1,-1,2).

The vector normal to the surface xy³z² = 4

$= \nabla (x {y}^{3} {z}^{2} )$

$\displaystyle =\hat{i} \frac{ \partial}{ \partial x} (x {y}^{3} {z}^{2} ) + \hat{j} \frac{ \partial}{ \partial y} (x {y}^{3} {z}^{2} ) + \hat{k} \frac{ \partial}{ \partial z} (x {y}^{3} {z}^{2} )$

$\displaystyle = {y}^{3} {z}^{2} \hat{i} + 3x {y}^{2} {z}^{2} \hat{j}+ 2x {y}^{3} z \hat{k}$

Now

$\nabla (x {y}^{3} {z}^{2} ) \bigg|_{(-1,-1,2)}$

$\displaystyle = - 4 \hat{i} - 12 \hat{j}+ 4 \hat{k}$

So the vector normal to the surface xy³z² = 4 at (-1,-1,2)

$\displaystyle = - 4 \hat{i} - 12 \hat{j}+ 4 \hat{k}$

FINAL ANSWER

Hence the correct option is

$\displaystyle = - 4 \hat{i} - 12 \hat{j}+ 4 \hat{k}$

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