Q.If the foot of the perpendicular drawn from the origin to the plane is $(4, -2, -3)$, then find the equation of the plane in vector and Cartesian form. OR If $f(1) = 10$ and $f'(x) \geq 2$ for $1 \leq x \leq 4$, how small can $f(4)$ possibly be?

Main part: the foot of the perpendicular gives both the normal direction and a point on the plane, so its position vector directly yields the plane equation. OR alternative: apply the Mean Value Theorem on [1,4] to bound f(4)−f(1) below.

Main part

1. Let $F=(4,-2,-3)$ be the foot of the perpendicular from the origin $O$ to the required plane.
2. Since $OF$ is perpendicular to the plane, the direction of $\overrightarrow{OF}=4\vec i-2\vec j-3\vec k$ is a normal vector $\vec n$ to the plane.
3. The plane passes through $F$, whose position vector is $\vec a=4\vec i-2\vec j-3\vec k$ — the same as $\vec n$ here.
4. Vector equation of a plane through point $\vec a$ with normal $\vec n$: $\vec r\cdot\vec n=\vec a\cdot\vec n$.
5. $\vec a\cdot\vec n=(4)(4)+(-2)(-2)+(-3)(-3)=16+4+9=29$.
6. So the vector equation is $\vec r\cdot(4\vec i-2\vec j-3\vec k)=29$.
7. Writing $\vec r=x\vec i+y\vec j+z\vec k$, the Cartesian form is $4x-2y-3z=29$.

OR — alternative

8. We are given $f(1)=10$ and $f'(x)\geq2$ for all $x\in[1,4]$; $f$ is differentiable (hence continuous) on $[1,4]$. …