Let the point A be the foot of perpendicular drawn from the point P(a, b, 0) on the line[(…

Question

Let the point A be the foot of perpendicular drawn from the point P$(a, b, 0)$ on the line$$\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-\alpha}{3}.$$If the midpoint of the line segment PA is $$\left(0, \frac{3}{4}, -\frac{1}{4}\right),$$ then t...

Accepted Answer

Let the given line be written in parametric form by taking

$\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-\alpha }{3}=r.$

Then a general point $A$ on the line is

$A(1+2r,\, 2+r,\, \alpha +3r).$

Given $P(a,b,0)$ and the midpoint of $PA$ is $(0,\frac{3}{4},-\frac{1}{4})$. Using midpoint formula coordinate-wise:

$\begin{gathered} \frac{2r+1+a}{2}=0\Rightarrow 2r+a=-1 \ \frac{r+2+b}{2}=\frac{3}{4}\Rightarrow r+b=-\frac{1}{2} \end{gathered}$

Similarly, from the $z$-coordinate of the midpoint:

$\begin{gathered} \frac{3r+\alpha }{2}=\frac{-1}{4}\Rightarrow 3r+\alpha =\frac{-1}{2} \end{gathered}$

Now, $A$ is the foot of the perpendicular from $P$ to the line, so $\overset{	o \leftarrow }{PA}$ is perpendicular to the direction vector of the line.

The direction ratios of the line are $(2,1,3)$, so we use the dot product condition:

$2({x}_{A}-a)+1({y}_{A}-b)+3({z}_{A}-0)=0.$

Using the midpoint point $(0,\frac{3}{4},-\frac{1}{4})$, we can write $\overset{	o \leftarrow }{PA}=2((0,\frac{3}{4},-\frac{1}{4})-P)$, so its components are proportional to $(-a,\frac{3}{4}-b,-\frac{1}{4})$. Hence the perpendicularity condition becomes:

$\begin{gathered} 2\cdot a+1\cdot (b-3/4)+3\cdot \frac{1}{4}=0\Rightarrow a+b=0 \end{gathered}$

From $2r+a=-1$, we get $a=-1-2r$.

From $r+b=-\frac{1}{2}$, we get $b=-\frac{1}{2}-r$.

Substitute in $a+b=0$:

$\begin{gathered} \Rightarrow 2(-1-2r)+(-r-1/2)=0 \ \Rightarrow -3/2=3r=1\Rightarrow r=-1/2,a=0,b=0,\alpha =1 \ 	herefore {a}^{2}+{b}^{2}+{\alpha }^{2}=1 \end{gathered}$

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