If the distance of the point (a, 2,5) from the image of the point (1,2,7) in the line (x)/…

Question

If the distance of the point $(a, 2,5)$ from the image of the point $(1,2,7)$ in the line $\frac{x}{1}=\frac{y-1}{1}=\frac{z-2}{2}$ is 4 , then the sum of all possible values of $a$ is equal to :

Accepted Answer

Let the point $P(1,2,7)$.

We need the image of $P$ in the line

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

This means the line passes through the point

$A(0,1,2)$

and has direction ratios

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

If ${P}^{'}$ is the image of $P$ in this line, then the given line is the perpendicular bisector of the segment joining $P$ and ${P}^{'}$ in the sense of reflection about a line in 3D. So first we find the foot of perpendicular from $P$ to the line.

Let the foot be $M$.

A general point on the line is

$M(t)=(t,1+t,2+2t).$

Now,

$\overset{	o \leftarrow }{PM}=(t-1,\, t-1,\, 2t-5).$

Since $PM$ is perpendicular to the line, we must have

$\overset{	o \leftarrow }{PM}\cdot \vec{d}=0$

So,

$(t-1)(1)+(t-1)(1)+(2t-5)(2)=0$

$t-1+t-1+4t-10=0$

$6t-12=0$

$t=2$

Hence,

$M=(2,3,6).$

Since $M$ is the midpoint of $P(1,2,7)$ and its image ${P}^{'}$, we get

${P}^{'}=2M-P$

${P}^{'}=(4,6,12)-(1,2,7)=(3,4,5).$

So the image of $(1,2,7)$ in the given line is

$(3,4,5).$

Now the distance of $(a,2,5)$ from this image is $4$:

$\sqrt{(a-3{)}^{2}+(2-4{)}^{2}+(5-5{)}^{2}}=4$

$\sqrt{(a-3{)}^{2}+4}=4$

Squaring,

$(a-3{)}^{2}+4=16$

$(a-3{)}^{2}=12$

$a-3=\pm 2\sqrt{3}$

So,

$a=3\pm 2\sqrt{3}.$

Therefore, the sum of all possible values of $a$ is

$(3+2\sqrt{3})+(3-2\sqrt{3})=6.$

Hence, the correct option is

$6$

So, option C is correct.

Solutions · 1