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If the plane $lx+my+nz=p$ where $l^2+m^2+n^2=1$ meets the coordinate axes in X, Y, Z and G is the centroid of the triangle XYZ and if the perpendicular to the plane at G, meets the coordinate planes in A, B, C, then prove that $\frac{1}{GA}+\frac{1}{GB}+\frac{1}{GC}=\frac{3}{p}$

The above problem is not clear to me. Please provide me a diagrammatic representation of the problem.

user3575652
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1 Answers1

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The equation of the plane can be written in the standard form as $$ \frac{x}{p/l} + \frac{y}{p/m} + \frac{z}{p/n} = 1. $$ So the coordinates of the intercepts are $\mathbf X = (p/l, 0, 0)$ (just set $y$ and $z$ equal to zero in the above equation and you'll see), $\mathbf Y = (0, p/m, 0)$, $\mathbf Z = (0, 0, p/n)$.

The coordinates of the centroid is the average of those of $X$, $Y$ and $Z$, and $$ \mathbf G = \left( \frac{p}{3\,l}, \frac{p}{3\,m}, \frac{p}{3\,n}\right). $$

If we draw a perpendicular line to the plane passing through $G$, then the coordinates can be written as $$ \left( \frac{p}{3\,l} - l \,t, \frac{p}{3\,m} - m \, t, \frac{p}{3\,n} - n \, t \right). $$ This is because $(l, m, n)$ is a vector perpendicular to the plane, and any vector parallel it can be written as $(-l\,t, -m\,t, -n\,t$ (the negative sign is chosen for convenience). The distance from $G$ can be computed as $$ \sqrt{(l\,t)^2 + (m\,t)^2 + (n\,t)^2} = \sqrt{l^2 + m^2 + n^2} |t| = |t|. \tag{1}$$

Now when this perpendicular line meets the $xy$-plane, the $z$ coordinate must be zero, which means $$ \frac{p}{3\,n} - n \, t = 0. $$ and $t = p/(3n^2)$. By (1), this value gives the distance from the intersection $C$ to $G$, so $$ GC = \frac{p}{3 \, n^2}. $$ Similarly, \begin{align} GA &= \frac{p}{3 \, l^2}, \\ GB &= \frac{p}{3 \, m^2}. \end{align}

Finally, $$ \frac{1}{GA} + \frac{1}{GB} + \frac{1}{GC} = \frac{3 \, l^2+ 3 \, m^2 + 3 \, n^2}{p} = \frac{3}{p}. $$

hbp
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