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

Hint:

If two lines are axes of symmetry for the same pair of points (P, Q), they must intersect at the midpoint of PQ.

Answer 1

Correct Answer: 14

Let the first line be $L_1$ and the second be $L_2$.
Since Q is the image of P in $L_1$ and P is the image of Q in $L_2$, both lines are perpendicular bisectors of the segment PQ.
Thus, $L_1$ and $L_2$ must intersect at the midpoint M of PQ.
Find the intersection of $L_1$ and $L_2$.
General point on $L_1$: $M_1 = (2\lambda, \lambda-a, \lambda)$.
General point on $L_2$: $M_2 = (2\mu+2b, \mu+a, -5\mu-2b)$.
Equating coordinates: $2\lambda = 2\mu+2b \implies \lambda = \mu+b$.
$\lambda-a = \mu+a \implies \lambda = \mu+2a$.
Comparing $\lambda$, we get $b=2a$.
Substitute $\lambda = \mu+2a$ and $b=2a$ into z-coordinate: $\mu+2a = -5\mu-4a$.
$6\mu = -6a \implies \mu = -a$.
Then $\lambda = a$. The intersection point M is $(2a, 0, a)$.
Since M is the midpoint, vector $\vec{MP}$ is perpendicular to $L_1$.
$P(a, 2, a)$, $M(2a, 0, a)$. $\vec{MP} = (-a, 2, 0)$.
Direction of $L_1$ is $\vec{d}_1 = (2, 1, 1)$.
Dot product: $\vec{MP} \cdot \vec{d}_1 = -2a + 2 + 0 = 0 \implies a = 1$.
Since $b = 2a$, we have $b = 2$.
$a + b = 1 + 2 = 3$.