The 3-D Coordinate System — Question 7

Question 7 – Solution

Strategy Distances to the $yz$- and $xz$-planes determine $|a|$ and $|b|$. The octant fixes signs, and the distance to the origin determines $c$.

Known coordinates We have $|a|=2$ and $|b|=3$, so $a=-2$ and $b=3$.

Use the radial distance Since $a^2+b^2+c^2=49$, $$4+9+c^2=49\quad\Longrightarrow\quad c^2=36.$$ The octant condition $z<0$ selects $c=-6$. Hence $\boxed{P=(-2,3,-6)}$.

Verification Its distance from the origin is $\sqrt{4+9+36}=7$, and its plane distances are 2 and 3 as required.