The 3-D Coordinate System — Question 1

Question 1

A point $P=(x,y,z)$ is known only through its octant and its distances to the coordinate axes.

Given $$x<0,\qquad y>0,\qquad z<0,$$ $$\begin{aligned} d(P,\text{$x$-axis})&=\sqrt{41},\\ d(P,\text{$y$-axis})&=\sqrt{34},\\ d(P,\text{$z$-axis})&=5. \end{aligned}$$

Tasks

1. Determine the coordinates of $P$.

2. Verify your result using all three axis distances.

3. Explain why only two of the three distance measurements would not determine $P$ uniquely.

Question 1 – Solution

Strategy Squared distance to an axis is the sum of the squares of the other two coordinates. Solve first for $x^2,y^2,z^2$, then use the octant to determine signs.

Equations The data give $$y^2+z^2=41,\qquad x^2+z^2=34,\qquad x^2+y^2=25.$$ Adding the last two equations and subtracting the first yields $2x^2=18$, so $x^2=9$. Similarly, $$2y^2=41+25-34=32,\qquad 2z^2=41+34-25=50.$$ Thus $|x|=3$, $|y|=4$, and $|z|=5$. The stated octant fixes the signs: $$\boxed{P=(-3,4,-5)}.$$

Verification The resulting axis distances are $$\sqrt{4^2+(-5)^2}=\sqrt{41},\quad \sqrt{(-3)^2+(-5)^2}=\sqrt{34},\quad \sqrt{(-3)^2+4^2}=5.$$ With only two distances, one squared coordinate would remain free, so the third is essential.