Question

If the image of the point $ P(1, 0, 3) $ in the line joining the points $ A(4, 7, 1) $ and $ B(3, 5, 3) $ is $ Q(\alpha, \beta, \gamma) $, then $ \alpha + \beta + \gamma $ is equal to:

• A. 47/3
• B. 46/3
• C. 18
• D. 13

Hint:

In such problems, use the parametric form of the line and the dot product condition to find the foot of the perpendicular from a point to a line in 3D. Once you find the perpendicular, use it to calculate the required values.

Answer 1

The Correct Option is B

To determine the reflection of a point \( P \) across a line, the initial step involves locating the foot of the perpendicular from \( P \) to the line. This foot of the perpendicular acts as the midpoint between \( P \) and its reflection, denoted as \( Q \).

Concept Used:

The solution employs the following principles from 3D geometry:

1. Equation of a line in 3D: A line passing through a point with position vector \( \vec{a} \) and parallel to a vector \( \vec{d} \) can be represented by the vector equation \( \vec{r} = \vec{a} + \lambda \vec{d} \), where \( \lambda \) is a scalar parameter.
2. Perpendicular Vectors: Two vectors are orthogonal if their scalar product equals zero.
3. Midpoint Formula: For a line segment PQ with midpoint M, the position vector of M is \( \vec{m} = \frac{\vec{p} + \vec{q}}{2} \). Rearranging this yields \( \vec{q} = 2\vec{m} - \vec{p} \).

Step-by-Step Solution:

Step 1: Define the equation of the line passing through points A(4, 7, 1) and B(3, 5, 3).

The direction vector \( \vec{d} \) for the line is computed as the vector \( \vec{AB} \).

\[\vec{d} = \vec{B} - \vec{A} = (3-4)\hat{i} + (5-7)\hat{j} + (3-1)\hat{k} = -\hat{i} - 2\hat{j} + 2\hat{k}\]

The vector equation of the line originating from point A(4, 7, 1) is formulated as:

\[\vec{r} = (4\hat{i} + 7\hat{j} + \hat{k}) + \lambda(-\hat{i} - 2\hat{j} + 2\hat{k})\]

Any point M on this line will have coordinates of the form \( M(4-\lambda, 7-2\lambda, 1+2\lambda) \).

Step 2: Determine the foot of the perpendicular (M) from point P(1, 0, 3) to the line.

The vector \( \vec{PM} \) is calculated as follows:

\[\vec{PM} = ( (4-\lambda) - 1 )\hat{i} + ( (7-2\lambda) - 0 )\hat{j} + ( (1+2\lambda) - 3 )\hat{k}\]\[\vec{PM} = (3-\lambda)\hat{i} + (7-2\lambda)\hat{j} + (-2+2\lambda)\hat{k}\]

As \( \vec{PM} \) is perpendicular to the line, its dot product with the direction vector \( \vec{d} \) must be zero.

\[\vec{PM} \cdot \vec{d} = 0\]\[( (3-\lambda)(-1) + (7-2\lambda)(-2) + (-2+2\lambda)(2) ) = 0\]\[-3 + \lambda - 14 + 4\lambda - 4 + 4\lambda = 0\]\[9\lambda - 21 = 0 \implies \lambda = \frac{21}{9} = \frac{7}{3}\]

Substituting \( \lambda = 7/3 \) into the coordinates of M yields the foot of the perpendicular:

\[x_M = 4 - \frac{7}{3} = \frac{5}{3}\]\[y_M = 7 - 2\left(\frac{7}{3}\right) = 7 - \frac{14}{3} = \frac{7}{3}\]\[z_M = 1 + 2\left(\frac{7}{3}\right) = 1 + \frac{14}{3} = \frac{17}{3}\]

Therefore, the foot of the perpendicular is \( M\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right) \).

Step 3: Calculate the coordinates of the image point \( Q(\alpha, \beta, \gamma) \).

The point M serves as the midpoint of the segment PQ.

Applying the midpoint formula:

\[M = \left( \frac{x_P + x_Q}{2}, \frac{y_P + y_Q}{2}, \frac{z_P + z_Q}{2} \right)\]\[\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right) = \left( \frac{1 + \alpha}{2}, \frac{0 + \beta}{2}, \frac{3 + \gamma}{2} \right)\]

Solving for \( \alpha, \beta, \gamma \):

\[\frac{1 + \alpha}{2} = \frac{5}{3} \implies 3 + 3\alpha = 10 \implies 3\alpha = 7 \implies \alpha = \frac{7}{3}\]\[\frac{\beta}{2} = \frac{7}{3} \implies 3\beta = 14 \implies \beta = \frac{14}{3}\]\[\frac{3 + \gamma}{2} = \frac{17}{3} \implies 9 + 3\gamma = 34 \implies 3\gamma = 25 \implies \gamma = \frac{25}{3}\]

The coordinates of the image point are \( Q\left(\frac{7}{3}, \frac{14}{3}, \frac{25}{3}\right) \).

Final Computation & Result

The objective is to compute the sum \( \alpha + \beta + \gamma \).

\[\alpha + \beta + \gamma = \frac{7}{3} + \frac{14}{3} + \frac{25}{3}\]\[\alpha + \beta + \gamma = \frac{7 + 14 + 25}{3} = \frac{46}{3}\]

Consequently, the value of \( \alpha + \beta + \gamma \) is \( \frac{46}{3} \).