Mathematics · 3D Geometry

JEE Main 2024 — 8 April, Shift 2 — Question 22

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 \qquad

Answer: 11

Numerical answer — enter this value.

Step-by-step solution

figure

∣ A(1714,4814,7914)x1=y−12=z−23\frac{\left\lvert\, \mathrm{A}\left(\frac{17}{14}, \frac{48}{14}, \frac{79}{14}\right)\right.}{\frac{\mathrm{x}}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-2}{3}}

P(α,β,γ),b→=i^+2j^+3k^\mathrm{P}(\alpha, \beta, \gamma), \quad \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}

A(t,2t+1,3t+2)A(t, 2 t+1,3 t+2)

QA→=(t−1)i^+(2t−5)j^+(3t−2)k^\overrightarrow{\mathrm{QA}}=(\mathrm{t}-1) \hat{\mathrm{i}}+(2 \mathrm{t}-5) \hat{\mathrm{j}}+(3 \mathrm{t}-2) \hat{\mathrm{k}}

QA→⋅b⃗=0\overrightarrow{\mathrm{QA}} \cdot \vec{b}=0

(t−1)+2(2t−5)+3(3t−2)=0(\mathrm{t}-1)+2(2 \mathrm{t}-5)+3(3 \mathrm{t}-2)=0

14t=1714 \mathrm{t}=17

α=2014,β=1214,γ=10214\alpha=\frac{20}{14}, \quad \beta=\frac{12}{14} ,\quad \gamma=\frac{102}{14}

2α+β+γ=15414=112 \alpha+\beta+\gamma=\frac{154}{14}=11

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
3D Geometry
Topic
Vector & Cartesian forms of lines and planes