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

Answer: (A)

155

Answer 2

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