If the image of the point P(a, 2, a) in the line L_1: x2=y+a1=z1 is Q, and the image of Q …

Question

If the image of the point $P(a, 2, a)$ in the line $L_1:\ \dfrac{x}{2}=\dfrac{y+a}{1}=\dfrac{z}{1}$ is $Q$, and the image of $Q$ in the line $L_2:\ \dfrac{x-2b}{2}=\dfrac{y-a}{1}=\dfrac{z+2b}{-5}$ is again the point $P$, then $a+b$ is equal to \_\_\_\_.

Answer: 3

Accepted Answer

Let $M$ be the foot of the perpendicular from $P(a,2,a)$ to $L_1$. A general point on $L_1$ is $(2\lambda,\ \lambda-a,\ \lambda)$, so
$$\vec{PM}=(2\lambda-a,\ \lambda-a-2,\ \lambda-a).$$
Since $\vec{PM}\perp L_1$ (direction $(2,1,1)$):
$$2(2\lambda-a)+1(\lambda-a-2)+1(\lambda-a)=0 \implies 6\lambda-4a-2=0 \implies \lambda=\frac{2a+1}{3}.$$
So
$$M=\left(\frac{4a+2}{3},\ \frac{1-a}{3},\ \frac{2a+1}{3}\right).$$
Since $Q$ is the image of $P$ in $L_1$, $M$ is the midpoint of $PQ$. Since $P$ is also the image of $Q$ in $L_2$, this same midpoint $M$ must lie on $L_2$. Equating $M$'s coordinates with the parametric form of $L_2$:
$$\frac{\frac{4a+2}{3}-2b}{2}=\frac{\frac{1-a}{3}-a}{1}=\frac{\frac{2a+1}{3}+2b}{-5}.$$
From the first two ratios: $4a+2-6b=2(1-4a) \Rightarrow 12a=6b \Rightarrow b=2a.$
From the last two ratios: $-5(1-4a)=2a+1+6b \Rightarrow 20a-5=14a+1 \Rightarrow 6a=6 \Rightarrow a=1.$
Then $b=2$, so $a+b=3$.

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