Question

Let the point \( A \) be the foot of perpendicular drawn from the point \( P(a, b, 0) \) on the line} \[ \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - \alpha}{3}. \] If the midpoint of the line segment \( PA \) is \( \left( \frac{3}{4}, \frac{4}{3}, -\frac{1}{4} \right) \), then the value of \( a^2 + b^2 + \alpha^2 \) is:

• A. 21
• B. 76
• C. 62
• D. 9

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We are given a point \(P\) and the midpoint of the segment connecting \(P\) to its foot of perpendicular \(A\) on a line.
Using the midpoint formula, we can express the coordinates of \(A\) in terms of the unknowns \(a\) and \(b\).
Since \(A\) is the foot of the perpendicular, it must lie on the given line, and the vector \(\vec{PA}\) must be orthogonal to the line's direction vector.
Step 2: Key Formula or Approach:
The midpoint \(M\) of points \(P(x_1, y_1, z_1)\) and \(A(x_2, y_2, z_2)\) is \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)\).
The direction vector of the line is \(\vec{d} = 2\hat{i} + \hat{j} + 3\hat{k}\).
The orthogonality condition is \(\vec{PA} \cdot \vec{d} = 0\).
Step 3: Detailed Explanation:
Let the coordinates of point \(A\) be \((x_2, y_2, z_2)\).
Using the given midpoint \(\left(0, \frac{3}{4}, -\frac{1}{4}\right)\), we can set up equations.
\[ \frac{a + x_2}{2} = 0 \implies x_2 = -a \] \[ \frac{b + y_2}{2} = \frac{3}{4} \implies y_2 = \frac{3}{2} - b \] \[ \frac{0 + z_2}{2} = -\frac{1}{4} \implies z_2 = -\frac{1}{2} \] So, point \(A\) is \(\left(-a, \frac{3}{2} - b, -\frac{1}{2}\right)\).
Since \(A\) lies on the line \(\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - \alpha}{3}\), we substitute its coordinates.
\[ \frac{-a - 1}{2} = \frac{\frac{3}{2} - b - 2}{1} = \frac{-\frac{1}{2} - \alpha}{3} = k \] Simplify the middle term.
\[ \frac{-a - 1}{2} = -\frac{1}{2} - b \implies -a - 1 = -1 - 2b \implies a = 2b \] Now, form the vector \(\vec{PA}\).
\[ \vec{PA} = (-a - a)\hat{i} + \left(\frac{3}{2} - b - b\right)\hat{j} + \left(-\frac{1}{2} - 0\right)\hat{k} = -2a\hat{i} + \left(\frac{3}{2} - 2b\right)\hat{j} - \frac{1}{2}\hat{k} \] Apply the orthogonality condition \(\vec{PA} \cdot \vec{d} = 0\).
\[ 2(-2a) + 1\left(\frac{3}{2} - 2b\right) + 3\left(-\frac{1}{2}\right) = 0 \] \[ -4a + \frac{3}{2} - 2b - \frac{3}{2} = 0 \] \[ -4a - 2b = 0 \implies 2a + b = 0 \implies b = -2a \] We now have a system: \(a = 2b\) and \(b = -2a\).
Substitute \(b\) into the first equation: \(a = 2(-2a) \implies a = -4a \implies 5a = 0 \implies a = 0\).
Since \(a = 0\), it follows that \(b = 0\).
Now substitute \(a\) and \(b\) back into the line equation to find \(\alpha\).
\[ \frac{0 - 1}{2} = \frac{-\frac{1}{2} - \alpha}{3} \] \[ -\frac{1}{2} = \frac{-1 - 2\alpha}{6} \] \[ -3 = -1 - 2\alpha \implies 2\alpha = 2 \implies \alpha = 1 \] Finally, compute the required sum.
\[ a^2 + b^2 + \alpha^2 = 0^2 + 0^2 + 1^2 = 1 \] Step 4: Final Answer:
The value of \(a^2 + b^2 + \alpha^2\) is \(1\).