Question

Let the point, on the line passing through the points \( P(1, -2, 3) \) and \( Q(5, -4, 7) \), farther from the origin and at a distance of 9 units from the point \( P \), be \( (\alpha, \beta, \gamma) \). Then \( \alpha^2 + \beta^2 + \gamma^2 \) is equal to:

• A. 155
• B. 150
• C. 160
• D. 165

Answer 1

The Correct Option is A

Find the Direction Ratios of Line \(PQ\): 
The direction ratios of the line passing through \(P(1, -2, 3)\) and \(Q(5, -4, 7)\) are:
\[ PQ = (5 - 1, -4 - (-2), 7 - 3) = (4, -2, 4) \] So, the direction ratios are \(4, -2, 4\).

Parametric Form of the Line: 
The parametric form of the line \(PQ\), with point \(P\) as the reference, is:
\[ (x, y, z) = (1, -2, 3) + t(4, -2, 4) \] Expanding each component, we get: \[ x = 1 + 4t, \quad y = -2 - 2t, \quad z = 3 + 4t \]
Calculate the Distance from \(P\) to a Point on the Line: 
The distance from \(P(1, -2, 3)\) to a point on the line parameterized by \(t\) is:
\[ \text{Distance} = \sqrt{(4t)^2 + (-2t)^2 + (4t)^2} = \sqrt{16t^2 + 4t^2 + 16t^2} = \sqrt{36t^2} = 6|t| \]
Given that this distance is 9 units, we set \(6|t| = 9\): \[ |t| = \frac{9}{6} = \frac{3}{2} \]
Since we are looking for the point farther from the origin, we take \(t = \frac{3}{2}\).

Coordinates of the Point \((\alpha, \beta, \gamma)\): 
Substitute \(t = \frac{3}{2}\) into the parametric equations:
\[ \alpha = 1 + 4 \times \frac{3}{2} = 1 + 6 = 7 \]
\[ \beta = -2 - 2 \times \frac{3}{2} = -2 - 3 = -5 \]
\[ \gamma = 3 + 4 \times \frac{3}{2} = 3 + 6 = 9 \]

Thus, the coordinates of the point are \((7, -5, 9)\). 

Calculate \(\alpha^2 + \beta^2 + \gamma^2\): 
\[ \alpha^2 + \beta^2 + \gamma^2 = 7^2 + (-5)^2 + 9^2 = 49 + 25 + 81 = 155 \]