Consider a line L passing through the points P(1, 2, 1) and Q(2, 1, –1). If the mirror image of the point A(2, 2, 2) in the line L is , then is equal to __________. [2024]

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Solution

Q.
Consider a line L passing through the points P(1, 2, 1) and Q(2, 1, –1). If the mirror image of the point A(2, 2, 2) in the line L is $(α, β, γ)$ , then $α + β + 6 γ$ is equal to __________. [2024]

Ans.
(6)

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

Any point on the line PQ is of the form

$B (λ + 1, - λ + 2, - 2 λ + 1)$

$\vec{A B} = (λ + 1 - 2) \hat{i} + (- λ + 2 - 2) \hat{j} + (- 2 λ + 1 - 2) \hat{k}$

$\Rightarrow \vec{A B} = (λ - 1) \hat{i} - λ \hat{j} + (- 2 λ - 1) \hat{k} and \vec{P Q} = 1 \hat{i} - \hat{j} - 2 \hat{k}$

$\vec{A B} \cdot \vec{P Q} = 0$

$\Rightarrow (λ - 1) + λ + 2 (2 λ + 1) = 0$

$\Rightarrow λ - 1 + λ + 4 λ + 2 = 0$

$\Rightarrow 6 λ = - 1 \Rightarrow λ = \frac{- 1}{6}$

$∴ B (\frac{- 1}{6} + 1, \frac{1}{6} + 2, - 2 (\frac{- 1}{6}) + 1) = B (\frac{5}{6}, \frac{13}{6}, \frac{8}{6})$

Now B is mid point of AA'

$\Rightarrow (\frac{5}{6}, \frac{13}{6}, \frac{8}{6}) = (\frac{α + 2}{2}, \frac{β + 2}{2}, \frac{γ + 2}{2})$

$\Rightarrow α + 2 = \frac{10}{6}, β + 2 = \frac{26}{6}, γ + 2 = \frac{16}{6}$

$\Rightarrow α = \frac{- 1}{3}, β = \frac{7}{3}, γ = \frac{2}{3}$

Hence, $α + β + 6 γ = \frac{- 1}{3} + \frac{7}{3} + 6 \times \frac{2}{3} = \frac{18}{3} = 6$ .

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