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 :

• A. 11
• B. 9
• C. 6
• D. 4

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
First, find the image of point $M(1, 2, 7)$ in the given line. Then, use the distance formula between the image and $(a, 2, 5)$ to find possible values of $a$.
: Key Formula or Approach:
Foot of perpendicular $F$ on line $\vec{r} = \vec{a} + \lambda \vec{d}$ is found using $(\vec{MF} \cdot \vec{d}) = 0$.
Image $M'$ satisfies $\vec{F} = \frac{\vec{M} + \vec{M'}}{2}$.
Step 2: Detailed Explanation:
General point on line: $F = (\lambda, \lambda + 1, 2\lambda + 2)$.
Vector $MF = (\lambda - 1, \lambda - 1, 2\lambda - 5)$.
$MF \cdot (1, 1, 2) = 0 \implies (\lambda - 1) + (\lambda - 1) + 2(2\lambda - 5) = 0 \implies 6\lambda = 12 \implies \lambda = 2$.
Foot $F = (2, 3, 6)$.
Image $M' = (2(2)-1, 2(3)-2, 2(6)-7) = (3, 4, 5)$.
Distance from $(a, 2, 5)$ to $(3, 4, 5)$ is 4:
$\sqrt{(a-3)^2 + (2-4)^2 + (5-5)^2} = 4 \implies (a-3)^2 + 4 = 16$.
$(a-3)^2 = 12 \implies a - 3 = \pm 2\sqrt{3}$.
Possible values of $a$: $3 + 2\sqrt{3}$ and $3 - 2\sqrt{3}$.
Sum of values $= (3 + 2\sqrt{3}) + (3 - 2\sqrt{3}) = 6$.
Step 3: Final Answer:
The sum of all possible values of $a$ is 6.