Finding the Foot of Perpendicular and Calculating $a^2 + b^2 + \alpha^2$

Problem Setup:

• Given point $P(a, b, 0)$.
• Given line $L: \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - \alpha}{3}$. Let the direction vector be $\vec{d} = (2, 1, 3)$.
• Point A is the foot of the perpendicular from P to L.
• Midpoint M of PA is $(0, \frac{3}{4}, \frac{-1}{4})$.
• Objective: Calculate $a^2 + b^2 + \alpha^2$.

Determine Coordinates of Point A

Let A be a general point on the line L. We can represent A using a parameter $t$:

A = $(1 + 2t, 2 + t, \alpha + 3t)$

Apply Midpoint Condition

The midpoint M of the line segment PA is given by:

$M = (\frac{a + (1 + 2t)}{2}, \frac{b + (2 + t)}{2}, \frac{0 + (\alpha + 3t)}{2})$

Equating this with the given midpoint $(0, \frac{3}{4}, \frac{-1}{4})$:

1. X-coordinate: $\frac{a + 1 + 2t}{2} = 0 \implies a + 1 + 2t = 0 \implies 2t = -a - 1$
2. Y-coordinate: $\frac{b + 2 + t}{2} = \frac{3}{4} \implies b + 2 + t = \frac{3}{2} \implies t = \frac{3}{2} - 2 - b \implies t = -\frac{1}{2} - b$
3. Z-coordinate: $\frac{\alpha + 3t}{2} = \frac{-1}{4} \implies \alpha + 3t = -\frac{1}{2}$

Solve for $a, b, \alpha$

Equating the expressions for $t$ from steps 1 and 2:

$\frac{-a - 1}{2} = -\frac{1}{2} - b$

$-a - 1 = -1 - 2b \implies a = 2b$

Substitute $t = -\frac{1}{2} - b$ into the Z-coordinate equation:

$\alpha + 3(-\frac{1}{2} - b) = -\frac{1}{2}$

$\alpha - \frac{3}{2} - 3b = -\frac{1}{2}$

$\alpha - 3b = 1 \implies \alpha = 3b + 1$

Now, find the specific value of $t$ using the perpendicularity condition. The vector $\vec{PA} = A - P = (1 + 2t - a, 2 + t - b, \alpha + 3t)$.

Since PA is perpendicular to the line L, $\vec{PA} \cdot \vec{d} = 0$:

$(1 + 2t - a) \cdot 2 + (2 + t - b) \cdot 1 + (\alpha + 3t) \cdot 3 = 0$

$2 + 4t - 2a + 2 + t - b + 3\alpha + 9t = 0$

$14t + 4 - 2a - b + 3\alpha = 0$

Substitute $a = 2b$ and $\alpha = 3b + 1$:

$14t + 4 - 2(2b) - b + 3(3b + 1) = 0$

$14t + 4 - 4b - b + 9b + 3 = 0$

$14t + 7 + 4b = 0 \implies 14t = -4b - 7$

We also know $t = -\frac{1}{2} - b$. Substitute this:

$14(-\frac{1}{2} - b) = -4b - 7$

$-7 - 14b = -4b - 7$

$-14b = -4b \implies -10b = 0 \implies b = 0$

Now find $a$ and $\alpha$:

$a = 2b = 2(0) = 0$

$\alpha = 3b + 1 = 3(0) + 1 = 1$

Final Calculation

Calculate $a^2 + b^2 + \alpha^2$:

$a^2 + b^2 + \alpha^2 = 0^2 + 0^2 + 1^2 = 0 + 0 + 1 = 1$