A plane passes through negative two, negative two, three and has normal negative four, one, negative four. Give its equation in vector form.
Okay, in this scenario, we have a plane that is oriented with respect to an 𝑥𝑦𝑧-coordinate frame. We’re told that this plane has a normal vector, we’ll call it 𝐧, with components negative four, one, negative four. We also know that our plane passes through a point, we’ll call 𝑃 zero, with coordinates negative two, negative two, three.
Knowing all this, we want to give the equation of this plane in its vector form. We can recall that the vector form of a plane equation is given by this expression. If we take the dot product of a vector normal to the plane and a vector to a general point in the plane, that is equal to the dot product of that same normal vector and a vector to a known point in the plane. In our case, that known point is 𝑃 zero, and the vector to it, what we call 𝐫 zero, looks like this. The vector 𝐫 zero has components that are equal to the coordinates of point 𝑃 zero. And so we know the components of a normal vector to our plane as well as the components of a vector to a known point in the plane. We can now apply this knowledge to the vector form of the plane’s equation.
In this equation, we replace 𝐧 with the vector negative four, one, negative four. We leave the vector 𝐫 which, as we said, points to an arbitrary point in the plane. And then for 𝐫 zero, we use the vector negative two, negative two, three. We can simplify this result a bit by carrying out this dot product. Recall that that involves multiplying the respective components of these two vectors together. That gives us eight minus two minus 12 or negative six. And now we’ve simplified as far as we can. The vector form of our plane’s equation is the vector negative four, one, negative four dotted with 𝐫 being equal to negative six.