Find the locus of the point of intersection of three mutually perpendicular tangent planes to the paraboloid $ax^2+by^2=2cz.$

Question

Find the locus of the point of intersection of three mutually perpendicular tangent planes to the paraboloid $ax^2+by^2=2cz.$

The solution given in a regional handout(or extract) is hereby attached:

However, I don't get how are they concluding $l_r^2+m_r^2+n_r^2=1(\forall r\in \{1,2,3\})$ unless $l_r,m_r,n_r$ are direction cosines. This relation is only true if they are direction cosines and hence when they assume equation of the plane as $lx+my+nz=p$, they essentially considered, $l,m,n$ as direction cosines and $p$ as the length of perpendicular from the origin, right? But if this is correct, I dont still get how is $l_1^2+l_2^2+l_3^2=0$ ? When I tried adding equations of all planes, it didn't simplify to the required form, but still how are they getting all those relations about normals of planes?

Answer 1

I had the same doubt but in different concept - Director Sphere to Ellipsoid. You can go through it for a bit more clarity as to what author is doing in your given concept.

After going through it you'll see that the doubt you have is same in both (which you have written in paragraph).

Now coming to your doubt - it's a concept called 3 Mutually Orthogonal Vectors || Orthogonal Matrix ∈ Vector Spaces.

We get it by putting mutually orthogonal vectors in Matrix A and using relation

A . A^T = A^T . A = I

• Do it in terms of < l1, m1, n1 >, <l2, m2, n2>, <l3, m3, n3>.

• You'll get the desired result which author has assumed in your case (i.e. core of your doubt).

• I'll take 2-3 pages to actually get why that results comes.

• Do it once and you'll be set for life. Since it's a trivial concept.

• To shorten it you can take ∑ forms in your matrix product.

You can find it in initial chapters of any book on vector spaces. You can refer Schaum's Outline. You can find your doubt in literal form of a ques in many books on vector spaces.

Answer 2

Three planes be $l_ix+m_iy+n_iz=p_i\tag*{}$ where $i=1,2,3$.

Putting it in matrix form, we have: $\underbrace{\begin{pmatrix}l_1&m_1&n_1\\ l_2&m_2&n_2\\ l_3&m_3&n_3\end{pmatrix}}_{=\mathcal O}\begin{pmatrix}x\\ y \\ z\end{pmatrix}=\begin{pmatrix}p_1\\p_2\\p_3\end{pmatrix}\tag*{}$

If they are mutually perpendicular, WLOG, we can assume that: $\begin{align}&\Omega(i,j)\\&:=\left(l_i\mathbf{\hat{i}}+m_i\mathbf{\hat{j}}+n_i\mathbf{\hat k}\right)\mathbf{\cdot}\left(l_j\mathbf{\hat{i}}+m_j\mathbf{\hat{j}}+n_j\mathbf{\hat k}\right)\\&=l_il_j+m_im_j+n_in_j\\&=\begin{cases}0 & \forall \ i\neq j\\ 1 & \forall \ i=j \end{cases}\end{align}\tag*{}$ i.e., we have normal unit vectors for equation of each plane.

The assumption above is equivalent to assuming $\mathcal{OO}^\mathrm T=\mathrm I_3$. Notice that:

$$\begin{align}\mathcal{OO}^\mathrm T&=\begin{pmatrix}\Omega(1,1) & & \\ \Omega(1,2) & \Omega(2,2)& \\ \Omega(1,3) & \Omega(2,3) & \Omega(3,3)\end{pmatrix}\\&=\mathrm I_3\end{align}$$

N.B.: $\mathcal {OO}^\mathrm T$ is symmetric so I have omitted repeated entries.

Since $\mathcal {OO}^\mathrm T=\mathrm I_3$, we have that $\mathcal O$ is orthogonal. It follows that $\mathcal{O}^\mathrm T\mathcal O =\mathcal{OO}^\mathrm T =\mathrm I_3$.

$$\therefore \begin{align}\mathcal {O}^\mathrm T\mathcal{O} &=\begin{pmatrix}\Sigma l_i^2 & &\\ \Sigma l_im_i &\Sigma m_i^2&\\ \Sigma l_i n_i & \Sigma m_i n_i & \Sigma n_i^2\end{pmatrix}=\mathrm I_3\end{align}$$

Comparing corresponding entries, we have: $$\begin{cases}\sum l_i^2=\sum m_i^2=\sum n_i^2=1\\ \\ \sum l_im_i=\sum m_in_i=\sum n_il_i=0\end{cases}$$ which was to be shown. $\blacksquare$

For a system of three mutually perpendicular planes, there's always a $3\times 3$ orthogonal matrix associated with it. :)

The fact about orthogonal matrices: if you pick two rows, multiply the corresponding elements and sum it up, it's

• either $0$ (if you picked different rows)
• or $1$ (if you picked the same row twice).

This result holds for columns too.