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3D Geometry question

2024 · 1 Feb · Shift 2 · Q45
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3D Geometry question

2024 · 1 Feb · Shift 2 · Q45

JEE MainMathematics3D GeometryMCQ+4 / −1
Let P\mathrm{P}P and Q\mathrm{Q}Q be the points on the line x+38=y−42=z+12\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}8x+3​=2y−4​=2z+1​ which are at a distance of 6 units from the point R(1,2,3)\mathrm{R}(1,2,3)R(1,2,3). If the centroid of the triangle PQR is (α,β,γ)(\alpha, \beta, \gamma)(α,β,γ), then α2+β2+γ2\alpha^2+\beta^2+\gamma^2α2+β2+γ2 is :
  1. A
    18
  2. B
    24
  3. C
    26
  4. D
    36
View written solutionFree

Correct answer: A

  1. Write the line in parametric form

Given

x+38=y−42=z+12=t\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}=t8x+3​=2y−4​=2z+1​=t

So a general point on the line is

P(t)=(−3+8t,  4+2t,  −1+2t).P(t)=(-3+8t,\;4+2t,\;-1+2t).P(t)=(−3+8t,4+2t,−1+2t).
  1. Use the condition that the point is 6 units from R(1,2,3)R(1,2,3)R(1,2,3)

Distance squared from P(t)P(t)P(t) to RRR is:

(−3+8t−1)2+(4+2t−2)2+(−1+2t−3)2=62.(-3+8t-1)^2+(4+2t-2)^2+(-1+2t-3)^2=6^2.(−3+8t−1)2+(4+2t−2)2+(−1+2t−3)2=62.

That is,

(8t−4)2+(2t+2)2+(2t−4)2=36.(8t-4)^2+(2t+2)^2+(2t-4)^2=36.(8t−4)2+(2t+2)2+(2t−4)2=36.

Expand:

(64t2−64t+16)+(4t2+8t+4)+(4t2−16t+16)=36.(64t^2-64t+16)+(4t^2+8t+4)+(4t^2-16t+16)=36.(64t2−64t+16)+(4t2+8t+4)+(4t2−16t+16)=36. 72t2−72t+36=36.72t^2-72t+36=36.72t2−72t+36=36. 72t2−72t=072t^2-72t=072t2−72t=0 72t(t−1)=0.72t(t-1)=0.72t(t−1)=0.

Hence,

t=0ort=1.t=0 \quad \text{or} \quad t=1.t=0ort=1.

So the two required points are:

  • For t=0t=0t=0,
P=(−3,4,−1)P=(-3,4,-1)P=(−3,4,−1)
  • For t=1t=1t=1,
Q=(5,6,1)Q=(5,6,1)Q=(5,6,1)
  1. Find the centroid of triangle PQRPQRPQR

Here

R=(1,2,3).R=(1,2,3).R=(1,2,3).

Centroid is

(−3+5+13,4+6+23,−1+1+33)=(1,4,1).\left(\frac{-3+5+1}{3},\frac{4+6+2}{3},\frac{-1+1+3}{3}\right) =\left(1,4,1\right).(3−3+5+1​,34+6+2​,3−1+1+3​)=(1,4,1).

Thus,

α=1,β=4,γ=1.\alpha=1,\quad \beta=4,\quad \gamma=1.α=1,β=4,γ=1.
  1. Compute α2+β2+γ2\alpha^2+\beta^2+\gamma^2α2+β2+γ2
α2+β2+γ2=12+42+12=1+16+1=18.\alpha^2+\beta^2+\gamma^2=1^2+4^2+1^2=1+16+1=18.α2+β2+γ2=12+42+12=1+16+1=18.
  1. Match with options
18\boxed{18}18​

So the correct option is A.

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