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:
Show Hint
- 47/3
- 46/3
- 18
- 13
The Correct Option is B
Solution and Explanation
To find the image of a point \( P \) in a line, we first need to find the foot of the perpendicular from the point \( P \) to the line. This foot of the perpendicular will then serve as the midpoint between the point \( P \) and its image \( Q \).
Concept Used:
The solution uses the following concepts from 3D geometry:
- Equation of a line in 3D: The vector equation of a line passing through a point with position vector \( \vec{a} \) and parallel to a vector \( \vec{d} \) is given by \( \vec{r} = \vec{a} + \lambda \vec{d} \), where \( \lambda \) is a scalar parameter.
- Perpendicular Vectors: Two vectors are perpendicular if their dot product is zero.
- Midpoint Formula: If M is the midpoint of the line segment PQ, then the position vector of M is given by \( \vec{m} = \frac{\vec{p} + \vec{q}}{2} \). This implies \( \vec{q} = 2\vec{m} - \vec{p} \).
Step-by-Step Solution:
Step 1: Determine the equation of the line passing through points A(4, 7, 1) and B(3, 5, 3).
The direction vector \( \vec{d} \) of the line is given by 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 passing through point A(4, 7, 1) can be written 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 has coordinates \( M(4-\lambda, 7-2\lambda, 1+2\lambda) \).
Step 2: Find the foot of the perpendicular (M) from point P(1, 0, 3) to the line.
The vector \( \vec{PM} \) is given by:
\[ \vec{PM} = ( (4-\lambda) - 1 )\hat{i} + ( (7-2\lambda) - 0 )\hat{j} + ( (1+2\lambda) - 3 )\hat{k} \] \[ \vec{PM} = (3-\lambda)\hat{i} + (7-2\lambda)\hat{j} + (-2+2\lambda)\hat{k} \]Since \( \vec{PM} \) is perpendicular to the line, its dot product with the direction vector \( \vec{d} \) must be zero.
\[ \vec{PM} \cdot \vec{d} = 0 \] \[ ( (3-\lambda)(-1) + (7-2\lambda)(-2) + (-2+2\lambda)(2) ) = 0 \] \[ -3 + \lambda - 14 + 4\lambda - 4 + 4\lambda = 0 \] \[ 9\lambda - 21 = 0 \implies \lambda = \frac{21}{9} = \frac{7}{3} \]Now, substitute \( \lambda = 7/3 \) into the coordinates of M to find the foot of the perpendicular:
\[ x_M = 4 - \frac{7}{3} = \frac{5}{3} \] \[ y_M = 7 - 2\left(\frac{7}{3}\right) = 7 - \frac{14}{3} = \frac{7}{3} \] \[ z_M = 1 + 2\left(\frac{7}{3}\right) = 1 + \frac{14}{3} = \frac{17}{3} \]So, the foot of the perpendicular is \( M\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right) \).
Step 3: Find the coordinates of the image point \( Q(\alpha, \beta, \gamma) \).
The point M is the midpoint of the segment PQ.
Using 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 image point is \( Q\left(\frac{7}{3}, \frac{14}{3}, \frac{25}{3}\right) \).
Final Computation & Result
The problem asks for the sum \( \alpha + \beta + \gamma \).
\[ \alpha + \beta + \gamma = \frac{7}{3} + \frac{14}{3} + \frac{25}{3} \] \[ \alpha + \beta + \gamma = \frac{7 + 14 + 25}{3} = \frac{46}{3} \]Thus, the value of \( \alpha + \beta + \gamma \) is \( \frac{46}{3} \).
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