Question

Let the image of the point $P(1, 6, a)$ in the line $L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-a+1}{b}, b>0$, be $(\frac{a}{3}, 0, a+c)$. If $S(\alpha, \beta, \gamma), \alpha>0$, is the point on $L$ such that the distance of $S$ from the foot of perpendicular from the point $P$ on $L$ is $2\sqrt{14}$, then $\alpha + \beta + \gamma$ is equal to:

• A. 19
• B. 20
• C. 21
• D. 22

Hint:

Use the property that the midpoint of a point and its image lies on the line and the segment joining them is perpendicular to the line. Then use the distance formula for points on a line.

Answer 1

The Correct Option is C

This problem involves 3D geometry concepts, specifically the image of a point and distances on a line. Let's define the point $P(1, 6, a)$ and the line $L$ with direction vector $\vec{d} = (1, 2, b)$. The image point is $Q(\frac{a}{3}, 0, a+c)$.

Step 1: Determine $a$. The midpoint of $PQ$, $M = (\frac{1+a/3}{2}, 3, \frac{2a+c}{2})$, lies on $L$.
The $y$-coordinate of $M$ is 3. Substituting this into the $y$-part of the line equation:
$$\frac{y-1}{2} = \frac{3-1}{2} = 1$$
For the $x$-coordinate: $\frac{x}{1} = 1 \implies x = 1$. Thus, $\frac{1+a/3}{2} = 1 \implies 1 + \frac{a}{3} = 2 \implies a = 3$.

Step 2: Determine $b$ and $c$. The vector $\vec{PQ}$ is orthogonal to the line $L$.
$\vec{PQ} = (\frac{3}{3}-1, 0-6, 3+c-3) = (0, -6, c)$.
Direction of $L$ is $(1, 2, b)$. Orthogonality gives:
$$0(1) + (-6)(2) + c(b) = 0 \implies bc = 12 \dots (i)$$
Substituting $a=3$ into the $z$-part of the line equation for point $M$:
$$\frac{z-3+1}{b} = 1 \implies \frac{\frac{6+c}{2} - 2}{b} = 1 \implies \frac{6+c-4}{2} = b \implies c+2 = 2b \dots (ii)$$
Solving (i) and (ii): $b(2b-2) = 12 \implies b^2-b-6=0 \implies b=3$ (as $b>0$). Then $c=4$.
The line $L$ is $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$ and the foot of perpendicular $M$ is $(1, 3, 5)$.

Step 3: Find point $S(\alpha, \beta, \gamma)$. Let $S$ be $(k, 2k+1, 3k+2)$. Foot $M$ is at $k=1$, i.e., $M(1, 3, 5)$.
The distance $MS = \sqrt{(k-1)^2 + (2k-2)^2 + (3k-3)^2} = \sqrt{14(k-1)^2} = |k-1|\sqrt{14}$.
Given $|k-1|\sqrt{14} = 2\sqrt{14} \implies |k-1| = 2$.
Since $\alpha>0$ and $\alpha = k$, we take $k-1 = 2 \implies k = 3$.
Thus $S = (3, 7, 11)$.
Sum $\alpha + \beta + \gamma = 3 + 7 + 11 = 21$.