On the line ell: 5x+3y-1=0, 3x+tfrac92 z+tfrac152=0 find the…

On the line $\ell:\ 5x+3y-1=0,\quad 3x+\tfrac92 z+\tfrac{15}2=0$ find the point equidistant from $\pi_1:\ 3x+3y-2=0$ and $\pi_2:\ 4x+y+z+4=0$.

Explanation of Method

Points of a line at equal distance from two planes, found as intersections with the two bisector planes.

Step 1: Parametrize

The second equation is $2x+3z+5=0$. With $x=t$: $M=\left(t,\ \tfrac{1-5t}3,\ \tfrac{-5-2t}3\right),\qquad \vec d\parallel\{3,-5,-2\}$.

Step 2: Equal distances

Both normals have length $\sqrt{18}=3\sqrt2$. Substituting $M$: $3x+3y-2=-2t-1,\qquad 4x+y+z+4=\tfrac{5t+8}3$.

\[ 3|2t+1|=|5t+8|\ \Longrightarrow\ t=5\ \text{ or }\ t=-1. \]

Step 3: Result

\[P_1(5,-8,-5),\ \ d=\tfrac{11}{3\sqrt2};\qquad P_2(-1,2,-1),\ \ d=\tfrac1{3\sqrt2}\]

Checks: $P_1$ gives $\pi_1\to-11$ and $\pi_2\to11$. $P_2$ gives $\pi_1\to1$ and $\pi_2\to1$. Both points satisfy the two equations of $\ell$. $\checkmark$

$P_1$ lies on the bisector plane $7x+4y+z+2=0$ (opposite signs). $P_2$ lies on $x-2y+z+6=0$ (same signs).

$P_1$ $P_2$

$3\sqrt2\,d_1=|2t+1|$
$3\sqrt2\,d_2=\tfrac13|5t+8|$

Distances to the two planes along $\ell$ (vertical scale shrunk). They are equal exactly at $t=-1$ ($P_2$) and $t=5$ ($P_1$).

Result

\[\boxed{\,P_1(5,-8,-5),\ \ d=\tfrac{11}{3\sqrt2};\qquad P_2(-1,2,-1),\ \ d=\tfrac1{3\sqrt2}\,}\]

Remark

Correction: the source says that at $P_1$ the two expressions have the same sign. In fact they are $-11$ and $+11$, with opposite signs. It is $P_2$ where the signs agree.