Let be the image of the point Q(1, 6, 4) in the line x 1 = y – 1 2 = z – 2 3 . Then is equal to __________. [2024]

Question

Let $P (α, β, γ)$ be the image of the point Q(1, 6, 4) in the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ . Then $2 α + β + γ$ is equal to __________. [2024]

Smart Achievers · Accepted Answer

(11)

We have, $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = t$ (say)

$\vec{Q R} \cdot (\hat{i} + 2 \hat{j} + 3 \hat{k}) = 0$

$\Rightarrow (t - 1) + (2 t - 5) \times 2 + (3 t - 2) \times 3 = 0 \Rightarrow t = \frac{17}{14}$

$\Rightarrow R \equiv (\frac{17}{14}, \frac{48}{14}, \frac{79}{14})$

Since, R is the mid point of PQ

$∴ \frac{α + 1}{2} = \frac{17}{14}, \frac{β + 6}{2} = \frac{48}{14}, \frac{γ + 4}{2} = \frac{79}{14}$

$\Rightarrow α = \frac{10}{7}, β = \frac{6}{7}, γ = \frac{51}{7}$

Hence, $2 α + β + γ = 2 \times \frac{10}{7} + \frac{6}{7} + \frac{51}{7} = \frac{20 + 6 + 51}{7} = \frac{77}{7} = 11$ .

Solution

Q.
Let $P (α, β, γ)$ be the image of the point Q(1, 6, 4) in the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ . Then $2 α + β + γ$ is equal to __________. [2024]

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Solution

Q. Let P(α,β,γ) be the image of the point Q(1, 6, 4) in the line x1=y–12=z–23. Then 2α+β+γ is equal to __________. [2024]

Ans. (11) We have, x1=y–12=z–23=t (say) QR→·(i^+2j^+3k^)=0 ⇒ (t–1)+(2t–5)×2+(3t–2)×3=0 ⇒ t=1714 ⇒ R≡(1714,4814,7914) Since, R is the mid point of PQ ∴ α+12=1714, β+62=4814, γ+42=7914 ⇒ α=107, β=67, γ=517 Hence, 2α+β+γ=2×107+67+517=20+6+517=777=11.

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Q.
Let $P (α, β, γ)$ be the image of the point Q(1, 6, 4) in the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ . Then $2 α + β + γ$ is equal to __________. [2024]