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

To solve this problem, we start by utilizing the properties of the image of a point with respect to a line. The line is given by $L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-a+1}{b}$. Let the point be $P(1, 6, a)$ and its image be $Q(\frac{a}{3}, 0, a+c)$.

The midpoint of $PQ$, let's call it $M$, must lie on the line $L$. The coordinates of $M$ are:
$$M = \left( \frac{1 + \frac{a}{3}}{2}, \frac{6 + 0}{2}, \frac{a + a + c}{2} \right) = \left( \frac{a+3}{6}, 3, \frac{2a+c}{2} \right)$$
Substituting $M$ into the equation of the line $L$:
$$\frac{\frac{a+3}{6}}{1} = \frac{3-1}{2} = \frac{\frac{2a+c}{2} - a + 1}{b}$$
From the first two terms: $\frac{a+3}{6} = 1 \implies a+3 = 6 \implies a = 3$.
Now substitute $a=3$ into the third term:
$$\frac{\frac{6+c}{2} - 3 + 1}{b} = 1 \implies \frac{6+c-6+2}{2b} = 1 \implies \frac{c+2}{2b} = 1 \implies c + 2 = 2b \dots (1)$$

Additionally, the vector $\vec{PQ}$ must be perpendicular to the direction of the line $L$, which is $\vec{v} = (1, 2, b)$.
$\vec{PQ} = (Q_x - P_x, Q_y - P_y, Q_z - P_z) = (\frac{3}{3}-1, 0-6, 3+c-3) = (0, -6, c)$.
Since $\vec{PQ} \perp \vec{v}$, their dot product is zero:
$$0(1) + (-6)(2) + c(b) = 0 \implies -12 + bc = 0 \implies bc = 12 \dots (2)$$
Substitute $c = 2b-2$ from (1) into (2):
$$b(2b-2) = 12 \implies 2b^2 - 2b - 12 = 0 \implies b^2 - b - 6 = 0$$
Solving this quadratic equation: $(b-3)(b+2) = 0$. Given $b>0$, we have $b=3$.
Then $c = 2(3) - 2 = 4$.
So the line $L$ is $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$.
The foot of the perpendicular $M$ is the midpoint of $PQ$: $M = (1, 3, 5)$.

A general point $S$ on line $L$ can be represented as $(k, 2k+1, 3k+2)$.
The distance $MS = 2\sqrt{14}$:
$$\sqrt{(k-1)^2 + (2k+1-3)^2 + (3k+2-5)^2} = 2\sqrt{14}$$
$$(k-1)^2 + (2k-2)^2 + (3k-3)^2 = (2\sqrt{14})^2 = 56$$
$$(k-1)^2 + 4(k-1)^2 + 9(k-1)^2 = 56 \implies 14(k-1)^2 = 56 \implies (k-1)^2 = 4$$
$$k-1 = \pm 2 \implies k = 3 \text{ or } k = -1$$
For $k=3$, $S = (3, 7, 11)$. Here $\alpha = 3>0$, which fits the condition.
For $k=-1$, $S = (-1, -1, -1)$, which has $\alpha<0$.
Thus, $S = (3, 7, 11)$, so $\alpha=3, \beta=7, \gamma=11$.
$\alpha + \beta + \gamma = 3 + 7 + 11 = 21$.