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JEE Main 20248 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2024 · 8 April · Evening Shift · Mathematics · Three Dimensional Geometry

Let P(α, β, γ) be the image of the point Q(1,6,4) in the line...

Let P(α,β,γ)\mathrm{P}(\alpha, \beta, \gamma) be the image of the point Q(1,6,4)\mathrm{Q}(1,6,4) in the line x1=y−12=z−23\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}. Then 2α+β+γ2 \alpha+\beta+\gamma is equal to_______Let P(α, β, γ) be the image of the point Q(1,6,4) in the line (x)/(1)=(y-1)/(2)=(z-2)/(3) . Then 2 α+β+γ is equal

Correct answer

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Step-by-step solution

Let P(α, β, γ) be the image of the point Q(1,6,4) in the line (x)/(1)=(y-1)/(2)=(z-2)/(3) . Then 2 α+β+γ is equal
A(t,2t+1,3t+2)QA→=(t−1)i^+(2t−5)j^+(3t−2)k^QA→⋅b→=0(t−1)+2(2t−5)+3(3t−2)=014t=17α=2014β=1214γ=102142α+β+γ=15414=11\begin{aligned} & \mathrm{A}(\mathrm{t}, 2 \mathrm{t}+1,3 \mathrm{t}+2) \\ & \overrightarrow{\mathrm{QA}}=(\mathrm{t}-1) \hat{\mathrm{i}}+(2 \mathrm{t}-5) \hat{\mathrm{j}}+(3 \mathrm{t}-2) \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{QA}} \cdot \overrightarrow{\mathrm{b}}=0 \\ & (\mathrm{t}-1)+2(2 \mathrm{t}-5)+3(3 \mathrm{t}-2)=0 \\ & 14 \mathrm{t}=17 \\ & \alpha=\frac{20}{14} \quad \beta=\frac{12}{14} \quad \gamma=\frac{102}{14} \\ & 2 \alpha+\beta+\gamma=\frac{154}{14}=11\end{aligned}
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