If the image of the point P(a,2,a) in the line (x)/(2)=(y+a)/(1)=(z)/(1) is Q and the imag…

Question

If the image of the point $P(a,2,a)$ in the line $\frac{x}{2}=\frac{y+a}{1}=\frac{z}{1}$ is Q and the image

of Q in the line $\frac{x-2b}{2}=\frac{y-a}{1}=\frac{z+2b}{-5}$ is P , then $a+b$ is equal to $____$ .

Answer: 3

Accepted Answer

For line 1 :

$\frac{x}{2}=\frac{y+a}{1}=\frac{z}{1}=\lambda $

general point $M$ on line is $(2\lambda ,\lambda -a,\lambda )$

$\begin{gathered} P(a,2,a) \ \overset{	o \leftarrow }{PM}=(2\lambda -a,\lambda -a-2,\lambda -a) \end{gathered}$

since $PM$ is perpendicular to line 1 , their dot product is 0

$\begin{gathered} \overset{	o \leftarrow }{PM}\cdot (2,1,1)=0 \ 2(2\lambda -a)+1(\lambda -a-2)+1(\lambda -a)=0 \ 6\lambda -4a-2=0\Rightarrow \lambda =\frac{2a+1}{3} \ M=(\frac{4a+2}{3},\frac{1-a}{3},\frac{2a+1}{3}) \end{gathered}$

for line 2:

$\frac{x-2b}{2}=\frac{y-a}{1}=\frac{z+2b}{-5}$

since $P$ is the image of $Q$, midpoint $M$ lies on line 2

$\begin{gathered} \frac{\frac{4a+2}{3}-2b}{2}=\frac{\frac{1-a}{3}-a}{1}=\frac{\frac{2a+1}{3}+2b}{-5} \ \frac{4a+2-6b}{6}=\frac{1-4a}{3}=\frac{2a+1+6b}{-15} \end{gathered}$

from first two parts:

$\begin{gathered} 4a+2-6b=2(1-4a) \ 4a+2-6b=2-8a\Rightarrow 12a=6b\Rightarrow b=2a \end{gathered}$

from last two parts:

$\begin{gathered} -5(1-4a)=2a+1+6b \ -5+20a=2a+1+6(2a) \ 20a-5=14a+1 \ 6a=6\Rightarrow a=1 \ b=2(1)=2 \ a+b=1+2=3 \end{gathered}$

the value of $a+b$ is 3 .

Solutions · 1