A point M starts from M_0(0,1,1) with velocity vec V=\1,0,2\…
A point $M$ starts from $M_0(0,1,1)$ with velocity $\vec V=\{1,0,2\}$. Let $d_1$ and $d_2$ be its distances from the planes $\pi_1:\ x+4y-3z+9=0,\qquad \pi_2:\ 5x+z+8=0$ $1^\circ$ When is $d_1=d_2$? $2^\circ$ When is $d=d_1+d_2$ smallest?
Explanation of Method
When a moving point is equally far from two planes, and when the sum of its distances is smallest.
Step 1: Distances as functions of time
$M=(t,\,1,\,1+2t)$. Both normals have length $\sqrt{26}$, so
\[ d_1=\frac{|10-5t|}{\sqrt{26}},\qquad d_2=\frac{|7t+9|}{\sqrt{26}}. \]
Step 2: $1^\circ$ Equal distances
We need $|10-5t|=|7t+9|$.
\[ 10-5t=7t+9\ \Rightarrow\ t=\tfrac1{12},\quad M\left(\tfrac1{12},1,\tfrac76\right),\quad d_1=d_2=\tfrac{115}{12\sqrt{26}}. \]
\[ 10-5t=-(7t+9)\ \Rightarrow\ t=-\tfrac{19}2,\quad M\left(-\tfrac{19}2,1,-18\right),\quad d_1=d_2=\tfrac{115}{2\sqrt{26}}. \]
Step 3: $2^\circ$ Minimum of the sum
Let $f(t)=|10-5t|+|7t+9|$, so $d=f/\sqrt{26}$. The break points are $t=-\tfrac97$ and $t=2$:
| interval | $f(t)$ | behaviour |
|---|---|---|
| $t\le-\tfrac97$ | $1-12t$ | decreasing |
| $-\tfrac97\le t\le2$ | $19+2t$ | increasing |
| $t\ge2$ | $12t-1$ | increasing |
So the minimum is at $t=-\tfrac97$, where $M$ lies on $\pi_2$ ($d_2=0$): \[t=-\tfrac97,\quad M\left(-\tfrac97,\,1,\,-\tfrac{11}7\right),\quad d_{\min}=\frac{115}{7\sqrt{26}}=\frac{115\sqrt{26}}{182}\approx3.22\]
- $\sqrt{26}\,d_1=|10-5t|$
- $\sqrt{26}\,d_2=|7t+9|$
- $\sqrt{26}\,(d_1+d_2)$
Result
\[\boxed{\,t=-\tfrac97,\quad M\left(-\tfrac97,\,1,\,-\tfrac{11}7\right),\quad d_{\min}=\frac{115}{7\sqrt{26}}=\frac{115\sqrt{26}}{182}\approx3.22\,}\]
Remark
Note on time: $t=-\tfrac97$ and $t=-\tfrac{19}2$ lie before the start ($t=0$). They are points of the whole line of motion. If only $t\ge0$ counts, then $d_1=d_2$ happens only at $t=\tfrac1{12}$. The sum is also smallest at the start, $t=0$, where $d=\tfrac{19}{\sqrt{26}}\approx3.73$.