Vector Fields — Question 4

Question 4 – Solution

Strategy. Compute the distance from the origin first; it controls both the denominator and the inverse-square magnitude.

Step 1: Evaluate at the point At $P=(2,-1,2)$, $$r=\sqrt{2^2+(-1)^2+2^2}=3,\qquad r^3=27.$$ Therefore $$\mathbf G(P)=-\frac{12}{27}\langle 2,-1,2\rangle =\boxed{\left\langle-\frac 89,\frac 49,-\frac 89\right\rangle}.$$

Step 2: Magnitude and unit direction In general, $|\mathbf G|=12r/r^3=12/r^2$. Thus $$|\mathbf G(P)|=\boxed{\frac 43},\qquad \widehat{\mathbf G(P)} =\boxed{\left\langle-\frac 23,\frac 13,-\frac 23\right\rangle}.$$

Step 3: Interpretation The negative sign reverses the outward position direction, so gravity points toward the origin. The denominator vanishes only at the origin; hence the domain is $$\boxed{\mathbb R^3\setminus\{(0,0,0)\}}.$$

Verification The displayed unit vector has length $1$, and multiplying it by $4/3$ reproduces $\mathbf G(P)$.