# Vertical Planes

#### edenbrown

##### New member
I have been given the equation of the graph f(x,y)=4y^2+6yx^2+17
this goes through the point (2,1,45) and I have also calculated the tangent plane's cartesian description: z - 24x - 32y = -35
Also knowing that the vector 4i+3j+0k is a direction vector of the vertical plane that contains the point (2,1,45) how would you work out the cartesian description of the vertical plane?

#### pka

##### Elite Member
I have been given the equation of the graph f(x,y)=4y^2+6yx^2+17
this goes through the point (2,1,45) and I have also calculated the tangent plane's cartesian description: z - 24x - 32y = -35
Also knowing that the vector 4i+3j+0k is a direction vector of the vertical plane that contains the point (2,1,45) how would you work out the cartesian description of the vertical plane?
Frankly I am puzzled by the terminology in this question.
First planes have normal vectors. If it wants the plane with normal $$\displaystyle 4\vec{i}+3\vec{j}+0\vec{k}$$ containing point $$\displaystyle (2,1,45)$$ then that plane is $$\displaystyle 4(x-2)+3(y-1)=0$$.

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#### Dr.Peterson

##### Elite Member
Frankly I am puzzled by the terminology in this question.
First planes have normal vectors. If it wants the plane with normal $$\displaystyle 4\vec{i}+3\vec{j}+0\vec{k}$$ containing point $$\displaystyle (2,1,45)$$ then that plane is $$\displaystyle 4(x-2)+3(y-1)=0$$.
However that plane is not vertical! In fact it is parallel to the $$\displaystyle xy\text{-plane}$$.
Thus it appears that you have some concepts confused with one another. Please review and correct.
That normal vector is in the xy plane, but isn't it true that the plane is vertical (that is, contains a vertical line, and is perpendicular to the xy plane)? That agrees with the fact that the equation of the plane does not restrict z.

#### pka

##### Elite Member
That normal vector is in the xy plane, but isn't it true that the plane is vertical (that is, contains a vertical line, and is perpendicular to the xy plane)? That agrees with the fact that the equation of the plane does not restrict z.
Yes my mistake. You are correct. Of course $$\displaystyle (4\vec{i}+3\vec{j})\cdot(\vec{k})=0$$ so the plane is parallel to the $$\displaystyle z\text{-axis}$$