A vector perpendicular to any vector that lies on the plane defined by x+y+z=5, is

Solution:

Let, $\textrm{Ø}=x+y+z-5=0$

We know that the gradient of a level surface function is perpendicular to the level surface at every point.

So, just taking the gradient of the function (level surface) $\textrm{Ø}$ will give us a vector perpendicular to the plane.

$\vec{\nabla}Ø = \frac{\partial\textrm{Ø}}{\partial x}\hat{i} + \frac{\partial\textrm{Ø}}{\partial y}\hat{j} + \frac{\partial\textrm{Ø}}{\partial z}\hat{k}$

Substituting, $\textrm{Ø}=x+y+z-5=0$,

$\vec{\nabla}Ø = \frac{\partial(x+y+z-5)}{\partial x}\hat{i} + \frac{\partial(x+y+z-5)}{\partial y}\hat{j} + \frac{\partial(x+y+z-5)}{\partial z}\hat{k}$

Now, taking partial derivative of $\textrm{Ø}$ of  with respect to x,y and z respectively in the above equation,

We’ll get,

$\vec{\nabla}Ø =\hat{i} + \hat{j} + \hat{k}$

Hence, the vector perpendicular to any vector that lies on the plane defined by x+y+z=5, is:

$\hat{i} + \hat{j} + \hat{k}$

Share