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3D Geometry question

2022 · 28 Jun · Shift 2 · Q41
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3D Geometry question

2022 · 28 Jun · Shift 2 · Q41

JEE MainMathematics3D GeometryNumerical+4 / −1
Let the image of the point P(1, 2, 3) in the line L:x−63=y−12=z−23L:{{x - 6} \over 3} = {{y - 1} \over 2} = {{z - 2} \over 3}L:3x−6​=2y−1​=3z−2​ be Q. Let R (α\alphaα, β\betaβ, γ\gammaγ) be a point that divides internally the line segment PQ in the ratio 1 : 3. Then the value of 22 (α\alphaα+β\betaβ+γ\gammaγ) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 125

  1. Interpret “image of point in a line”
    The image of point P(1,2,3)P(1,2,3)P(1,2,3) in the line LLL means the reflection of PPP about the line LLL.
    So if MMM is the foot of perpendicular from PPP to the line, then MMM is the midpoint of PQPQPQ.

  2. Write the line in parametric form
    Given x−63=y−12=z−23=t\frac{x-6}{3}=\frac{y-1}{2}=\frac{z-2}{3}=t3x−6​=2y−1​=3z−2​=t so a general point on LLL is M(6+3t, 1+2t, 2+3t).M(6+3t,\ 1+2t,\ 2+3t).M(6+3t, 1+2t, 2+3t). The direction vector of the line is d⃗=(3,2,3).\vec d=(3,2,3).d=(3,2,3).

  3. Use perpendicularity condition to find the foot MMM
    Since MMM is the foot of the perpendicular from PPP to LLL, we must have PM→⋅d⃗=0.\overrightarrow{PM}\cdot \vec d=0.PM⋅d=0. Now PM→=(6+3t−1, 1+2t−2, 2+3t−3)=(5+3t, −1+2t, −1+3t).\overrightarrow{PM}=(6+3t-1,\ 1+2t-2,\ 2+3t-3)=(5+3t,\ -1+2t,\ -1+3t).PM=(6+3t−1, 1+2t−2, 2+3t−3)=(5+3t, −1+2t, −1+3t). Thus, (5+3t)3+(−1+2t)2+(−1+3t)3=0.(5+3t)3+(-1+2t)2+(-1+3t)3=0.(5+3t)3+(−1+2t)2+(−1+3t)3=0. Compute: 15+9t−2+4t−3+9t=015+9t-2+4t-3+9t=015+9t−2+4t−3+9t=0 10+22t=010+22t=010+22t=0 t=−511.t=-\frac{5}{11}.t=−115​.

Hence M=(6+3(−511), 1+2(−511), 2+3(−511))M=\left(6+3\left(-\frac{5}{11}\right),\ 1+2\left(-\frac{5}{11}\right),\ 2+3\left(-\frac{5}{11}\right)\right)M=(6+3(−115​), 1+2(−115​), 2+3(−115​)) =(5111, 111, 711).=\left(\frac{51}{11},\ \frac{1}{11},\ \frac{7}{11}\right).=(1151​, 111​, 117​).

  1. Find the reflected point QQQ
    Since MMM is midpoint of PQPQPQ, Q=2M−P.Q=2M-P.Q=2M−P. So Q=(2⋅5111−1, 2⋅111−2, 2⋅711−3)Q=\left(2\cdot \frac{51}{11}-1,\ 2\cdot \frac{1}{11}-2,\ 2\cdot \frac{7}{11}-3\right)Q=(2⋅1151​−1, 2⋅111​−2, 2⋅117​−3) =(10211−1111, 211−2211, 1411−3311)=\left(\frac{102}{11}-\frac{11}{11},\ \frac{2}{11}-\frac{22}{11},\ \frac{14}{11}-\frac{33}{11}\right)=(11102​−1111​, 112​−1122​, 1114​−1133​) =(9111, −2011, −1911).=\left(\frac{91}{11},\ -\frac{20}{11},\ -\frac{19}{11}\right).=(1191​, −1120​, −1119​).

  2. Point R(α,β,γ)R(\alpha,\beta,\gamma)R(α,β,γ) divides PQPQPQ internally in ratio 1:31:31:3
    If RRR divides PQPQPQ internally in the ratio 1:31:31:3, then PR:RQ=1:3.PR:RQ=1:3.PR:RQ=1:3. Using section formula, R=(1⋅xQ+3⋅xP1+3, 1⋅yQ+3⋅yP1+3, 1⋅zQ+3⋅zP1+3).R=\left(\frac{1\cdot x_Q+3\cdot x_P}{1+3},\ \frac{1\cdot y_Q+3\cdot y_P}{1+3},\ \frac{1\cdot z_Q+3\cdot z_P}{1+3}\right).R=(1+31⋅xQ​+3⋅xP​​, 1+31⋅yQ​+3⋅yP​​, 1+31⋅zQ​+3⋅zP​​). Therefore, α=9111+34=9111+33114=12444=3111,\alpha=\frac{\frac{91}{11}+3}{4}=\frac{\frac{91}{11}+\frac{33}{11}}{4}=\frac{124}{44}=\frac{31}{11},α=41191​+3​=41191​+1133​​=44124​=1131​, β=−2011+64=−2011+66114=4644=2322,\beta=\frac{-\frac{20}{11}+6}{4}=\frac{-\frac{20}{11}+\frac{66}{11}}{4}=\frac{46}{44}=\frac{23}{22},β=4−1120​+6​=4−1120​+1166​​=4446​=2223​, γ=−1911+94=−1911+99114=8044=2011.\gamma=\frac{-\frac{19}{11}+9}{4}=\frac{-\frac{19}{11}+\frac{99}{11}}{4}=\frac{80}{44}=\frac{20}{11}.γ=4−1119​+9​=4−1119​+1199​​=4480​=1120​.

So, α+β+γ=3111+2322+2011\alpha+\beta+\gamma=\frac{31}{11}+\frac{23}{22}+\frac{20}{11}α+β+γ=1131​+2223​+1120​ =62+23+4022=12522.=\frac{62+23+40}{22}=\frac{125}{22}.=2262+23+40​=22125​.

  1. Compute the required value 22(α+β+γ)=22⋅12522=125.22(\alpha+\beta+\gamma)=22\cdot \frac{125}{22}=125.22(α+β+γ)=22⋅22125​=125.

Therefore, the required integer is 125.\boxed{125}.125​.

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