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3D Geometry question

2024 · 1 Feb · Shift 2 · Q47
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3D Geometry question

2024 · 1 Feb · Shift 2 · Q47

JEE MainMathematics3D GeometryMCQ+4 / −1
If the mirror image of the point P(3,4,9)P(3,4,9)P(3,4,9) in the line x−13=y+12=z−21\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}3x−1​=2y+1​=1z−2​ is (α,β,γ)(\alpha, \beta, \gamma)(α,β,γ), then 14 (α+β+γ)(\alpha+\beta+\gamma)(α+β+γ) is :
  1. A
    102
  2. B
    138
  3. C
    132
  4. D
    108
View written solutionFree

Correct answer: D

  1. Interpret the line

The given line is

x−13=y+12=z−21=t\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}=t3x−1​=2y+1​=1z−2​=t

So its parametric form is

x=1+3t,y=−1+2t,z=2+tx=1+3t,\quad y=-1+2t,\quad z=2+tx=1+3t,y=−1+2t,z=2+t

Hence, a point on the line is A(1,−1,2)A(1,-1,2)A(1,−1,2) and its direction vector is d⃗=(3,2,1).\vec d=(3,2,1).d=(3,2,1).

  1. Find the foot of perpendicular from P(3,4,9)P(3,4,9)P(3,4,9) to the line

Let the foot of perpendicular be QQQ on the line. Then

Q=(1+3t,−1+2t,2+t).Q=(1+3t,-1+2t,2+t).Q=(1+3t,−1+2t,2+t).

Since QQQ is the projection of PPP on the line, we must have

PQ→⊥d⃗.\overrightarrow{PQ}\perp \vec d.PQ​⊥d.

Now,

PQ→=Q−P=(1+3t−3,−1+2t−4,2+t−9)=(3t−2,2t−5,t−7).\overrightarrow{PQ}=Q-P=(1+3t-3,-1+2t-4,2+t-9)=(3t-2,2t-5,t-7).PQ​=Q−P=(1+3t−3,−1+2t−4,2+t−9)=(3t−2,2t−5,t−7).

So,

(3t−2,2t−5,t−7)⋅(3,2,1)=0.(3t-2,2t-5,t-7)\cdot(3,2,1)=0.(3t−2,2t−5,t−7)⋅(3,2,1)=0.

Compute:

3(3t−2)+2(2t−5)+(t−7)=03(3t-2)+2(2t-5)+(t-7)=03(3t−2)+2(2t−5)+(t−7)=0 9t−6+4t−10+t−7=09t-6+4t-10+t-7=09t−6+4t−10+t−7=0 14t−23=014t-23=014t−23=0 t=2314.t=\frac{23}{14}.t=1423​.

Therefore,

Q=(1+3⋅2314,−1+2⋅2314,2+2314)Q=\left(1+3\cdot\frac{23}{14},-1+2\cdot\frac{23}{14},2+\frac{23}{14}\right)Q=(1+3⋅1423​,−1+2⋅1423​,2+1423​) Q=(8314,167,5114).Q=\left(\frac{83}{14},\frac{16}{7},\frac{51}{14}\right).Q=(1483​,716​,1451​).
  1. Use midpoint property for mirror image

If (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) is the mirror image of P(3,4,9)P(3,4,9)P(3,4,9) in the line, then the line is the perpendicular bisector in 3D sense of segment joining PPP and its image. Thus QQQ is the midpoint of PPP and image point P′(α,β,γ)P'(\alpha,\beta,\gamma)P′(α,β,γ).

Hence,

α=2Qx−3=2⋅8314−3=16614−4214=12414=627,\alpha=2Q_x-3=2\cdot\frac{83}{14}-3=\frac{166}{14}-\frac{42}{14}=\frac{124}{14}=\frac{62}{7},α=2Qx​−3=2⋅1483​−3=14166​−1442​=14124​=762​, β=2Qy−4=2⋅167−4=327−287=47,\beta=2Q_y-4=2\cdot\frac{16}{7}-4=\frac{32}{7}-\frac{28}{7}=\frac{4}{7},β=2Qy​−4=2⋅716​−4=732​−728​=74​, γ=2Qz−9=2⋅5114−9=10214−12614=−2414=−127.\gamma=2Q_z-9=2\cdot\frac{51}{14}-9=\frac{102}{14}-\frac{126}{14}=-\frac{24}{14}=-\frac{12}{7}.γ=2Qz​−9=2⋅1451​−9=14102​−14126​=−1424​=−712​.

So,

α+β+γ=627+47−127=547.\alpha+\beta+\gamma=\frac{62}{7}+\frac{4}{7}-\frac{12}{7}=\frac{54}{7}.α+β+γ=762​+74​−712​=754​.
  1. Compute the required value
14(α+β+γ)=14⋅547=2⋅54=108.14(\alpha+\beta+\gamma)=14\cdot\frac{54}{7}=2\cdot 54=108.14(α+β+γ)=14⋅754​=2⋅54=108.
  1. Match with options

Thus the correct option is 108\boxed{108}108​ which is Option D.

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