Step 1: Find the Image Q of Point P in Line L1
Given point P = ($a$, 2, $a$) and line $L_1: \frac{x}{2} = \frac{y+a}{1} = \frac{z}{1}$. The direction vector of $L_1$ is $d_1 = <2, 1, 1>$. Let M be the foot of the perpendicular from P to $L_1$. M lies on $L_1$. Parametric form of $L_1$: M($2k$, $-a+k$, $k$). Vector PM = $<2k-a, k-a-2, k-a>$.
Since PM $\perp d_1$, the dot product PM $\cdot d_1 = 0$: $2(2k-a) + 1(k-a-2) + 1(k-a) = 0$ $4k - 2a + k - a - 2 + k - a = 0$ $6k - 4a - 2 = 0 \implies 3k = 2a + 1$
The coordinates of M are: $x_M = 2k = \frac{4a+2}{3}$ $y_M = -a+k = -a + \frac{2a+1}{3} = \frac{-a+1}{3}$ $z_M = k = \frac{2a+1}{3}$
M is the midpoint of P($a$, 2, $a$) and its image Q($x_Q$, $y_Q$, $z_Q$). Using the midpoint formula ($x_M = \frac{a+x_Q}{2}$, etc.): $x_Q = 2x_M - a = 2(\frac{4a+2}{3}) - a = \frac{8a+4-3a}{3} = \frac{5a+4}{3}$ $y_Q = 2y_M - 2 = 2(\frac{-a+1}{3}) - 2 = \frac{-2a+2-6}{3} = \frac{-2a-4}{3}$ $z_Q = 2z_M - a = 2(\frac{2a+1}{3}) - a = \frac{4a+2-3a}{3} = \frac{a+2}{3}$ Thus, Q = $(\frac{5a+4}{3}, \frac{-2a-4}{3}, \frac{a+2}{3})$.
Step 2: Find the Relationship between a and b using Line L2
The image of Q in line $L_2: \frac{x-2b}{2} = \frac{y-a}{1} = \frac{z+2b}{-5}$ is P($a$, 2, $a$). The direction vector of $L_2$ is $d_2 = <2, 1, -5>$. Let N be the midpoint of the segment QP. N must lie on $L_2$. Coordinates of N: $x_N = \frac{a+x_Q}{2} = \frac{a + \frac{5a+4}{3}}{2} = \frac{3a+5a+4}{6} = \frac{8a+4}{6} = \frac{4a+2}{3}$ $y_N = \frac{2+y_Q}{2} = \frac{2 + \frac{-2a-4}{3}}{2} = \frac{6-2a-4}{6} = \frac{-2a+2}{6} = \frac{-a+1}{3}$ $z_N = \frac{a+z_Q}{2} = \frac{a + \frac{a+2}{3}}{2} = \frac{3a+a+2}{6} = \frac{4a+2}{6} = \frac{2a+1}{3}$
N lies on $L_2$. Parametric form of $L_2$ is N($2b+2m$, $a+m$, $-2b-5m$). Equating the y-coordinates: $y_N = a+m \implies \frac{-a+1}{3} = a+m$ $m = \frac{-a+1}{3} - a = \frac{-4a+1}{3}$
Equating the x-coordinates: $x_N = 2b+2m \implies \frac{4a+2}{3} = 2b + 2m$ Substitute $m$: $\frac{4a+2}{3} = 2b + 2(\frac{-4a+1}{3})$ $4a+2 = 6b + 2(-4a+1) = 6b - 8a + 2$ $12a = 6b \implies b = 2a$ (Relation 1)
Equating the z-coordinates: $z_N = -2b-5m \implies \frac{2a+1}{3} = -2b - 5m$ Substitute $m$: $\frac{2a+1}{3} = -2b - 5(\frac{-4a+1}{3})$ $2a+1 = -6b - 5(-4a+1) = -6b + 20a - 5$ $6b = 18a - 6 \implies b = 3a - 1$ (Relation 2)
Step 3: Solve for a and b
We have two relations between $a$ and $b$: 1. $b = 2a$ 2. $b = 3a - 1$ Equating the expressions for $b$: $2a = 3a - 1$ $a = 1$ Substitute $a=1$ into Relation 1: $b = 2(1) = 2$
Step 4: Calculate a + b
With $a=1$ and $b=2$, the required value is: $a+b = 1 + 2 = 3$
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