Let the image of the point P(1,6,a) in the line L:(x)/(1)=(y-1)/(2)=(z-a+1)/(b),b0, be ((…

Question

Let the image of the point $P(1,6,a)$ in the line $L:\frac{x}{1}=\frac{y-1}{2}=\frac{z-a+1}{b},b>0$, be $(\frac{a}{3},0,a+c)$. If $S(\alpha ,\beta ,\gamma ),\alpha >0$, is the point on L such that the distance of S from the foot of perpendicular from the point P on L is $2\sqrt{14}$, then $\alpha +\beta +\gamma $ is equal to:

Accepted Answer

Let the line be written in parametric form as

$L:\frac{x}{1}=\frac{y-1}{2}=\frac{z-a+1}{b}=t$

So a general point on $L$ is

$Q(t)=(t,1+2t,a-1+bt).$

We are given:

Point $P=(1,6,a)$

Its image in the line $L$ is $(\frac{a}{3},0,a+c)$

For reflection in a line in 3D, the line is the perpendicular bisector of the segment joining the point and its image.

So if image of $P$ is ${P}^{'}=(\frac{a}{3},0,a+c)$, then:

Midpoint of $P{P}^{'}$ lies on $L$

The line $L$ is perpendicular to $\overset{	o \leftarrow }{P{P}^{'}}$

1. Midpoint of $P$ and ${P}^{'}$
Midpoint $M$ is

$M=(\frac{1+\frac{a}{3}}{2},\frac{6+0}{2},\frac{a+(a+c)}{2})=(\frac{a+3}{6},3,\frac{2a+c}{2})$

Since $M$ lies on $L$, for some parameter $t$,

$(t,1+2t,a-1+bt)=(\frac{a+3}{6},3,\frac{2a+c}{2})$

From the $y$-coordinate,

$1+2t=3\, \implies \, 2t=2\, \implies \, t=1$

Then from the $x$-coordinate,

$t=\frac{a+3}{6}$

So

$1=\frac{a+3}{6}\, \implies \, a+3=6\, \implies \, a=3$

Now from the $z$-coordinate,

$a-1+b=\frac{2a+c}{2}$

Substitute $a=3$:

$3-1+b=\frac{6+c}{2}$

$2+b=\frac{6+c}{2}$

$4+2b=6+c$

$c=2b-2$

2. Perpendicularity condition
Direction vector of line $L$ is

$\vec{d}=(1,2,b)$

Now

$\overset{	o \leftarrow }{P{P}^{'}}=(\frac{a}{3}-1,0-6,a+c-a)=(\frac{a}{3}-1,-6,c)$

Since $a=3$,

$\overset{	o \leftarrow }{P{P}^{'}}=(0,-6,c)$

This must be perpendicular to $\vec{d}=(1,2,b)$, so

$(0,-6,c)\cdot (1,2,b)=0$

$0-12+bc=0$

$bc=12$

But we already have

$c=2b-2$

So

$b(2b-2)=12$

$2{b}^{2}-2b-12=0$

${b}^{2}-b-6=0$

$(b-3)(b+2)=0$

Given $b>0$, so

$b=3$

Then

$c=2b-2=6-2=4$

Thus the line is

$L:\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$

and the image point is

${P}^{'}=(1,0,7).$

3. Foot of perpendicular from $P$ to $L$
Let the foot be $F$. Since midpoint of $P$ and its image lies on the mirror line, $F$ is exactly the midpoint of $P$ and ${P}^{'}$:

$F=(\frac{1+1}{2},\frac{6+0}{2},\frac{3+7}{2})=(1,3,5)$

Check that this lies on $L$:

For $t=1$,

$Q(1)=(1,3,5)$

Correct.

4. Point $S(\alpha ,\beta ,\gamma )$ on $L$ such that $SF=2\sqrt{14}$
A general point on $L$ is

$S(t)=(t,1+2t,2+3t)$

Since $F=(1,3,5)$ corresponds to $t=1$, we get

$\overset{	o \leftarrow }{FS}=(t-1,2t-2,3t-3)=(t-1)(1,2,3)$

Hence

$SF=\sqrt{(t-1{)}^{2}+(2t-2{)}^{2}+(3t-3{)}^{2}}$

$=|t-1|\sqrt{1+4+9}=|t-1|\sqrt{14}$

Given

$SF=2\sqrt{14}$

So

$|t-1|=2$

Thus

$t=3\, \, t=-1$

Since $\alpha >0$ and $\alpha =x=t$, we need

$t>0\, \implies \, t=3$

Therefore

$S=(3,7,11)$

So

$\alpha +\beta +\gamma =3+7+11=21$

Hence the correct answer is

$21$

So, Option C is correct.

Solutions · 1