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 , then is equal to : [2025]

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Solution

Q.
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 $Q (α, β, γ)$ , then $α + β + γ$ is equal to : [2025]

1 18

2 $13$

3 $\frac{47}{3}$

4 $\frac{46}{3}$

Ans.
(4)

Equation of line AB is given by

$\frac{x - 4}{- 1} = \frac{y - 7}{- 2} = \frac{z - 1}{2} = λ (say)$

Any point on line AB is given by $(- λ + 4, - 2 λ + 7, 2 λ + 1)$

Now, mid-point of PQ lies on AB i.e.,

$(\frac{1 + α}{2}, \frac{β}{2}, \frac{3 + γ}{2})$ is a point on AB

$∴ \frac{1 + α}{2} = - λ + 4, \frac{β}{2} = - 2 λ + 7$ and $\frac{3 + γ}{2} = 2 λ + 1$

$\Rightarrow α = - 2 λ + 7, β = 14 - 4 λ and γ = 4 λ - 1$ ... (i)

Now, $\vec{P Q}$ is perpendicular to $\vec{A B}$

$∴ (α - 1) (- 1) + β (- 2) + (γ - 3) (2) = 0$

$\Rightarrow - α - 2 β + 2 γ = 5$

Now, substituting the values of $α$ , $β$ and $γ$ from equation (i) we get

$- (- 2 λ + 7) - 2 (14 - 4 λ) + 2 (4 λ - 1) = 5$

$\Rightarrow 2 λ - 7 - 28 + 8 λ + 8 λ - 2 = 5$

$\Rightarrow 18 λ = 42 \Rightarrow λ = \frac{7}{3}$

$∴ α = 7 - 2 \times \frac{7}{3} = \frac{7}{3}, β = 14 - 4 \times \frac{7}{3} = \frac{14}{3}$

and $γ = 4 \times \frac{7}{3} - 1 = \frac{25}{3}$

$∴ α + β + γ = \frac{7}{3} + \frac{14}{3} + \frac{25}{3} = \frac{46}{3}$ .

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