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MCQMediumJEE 2024Equation of Line in 3D

JEE Mathematics 2024 Question with Solution

Let P and 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} which are at a distance of 66 units from the point R(1,2,3)R(1, 2, 3). If the centroid of the triangle PQR is (α,β,γ)(\alpha, \beta, \gamma), then α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2 is:

  • A

    2626

  • B

    3636

  • C

    1818

  • D

    2424

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The line is

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

and points P and Q on this line are at a distance 66 units from R(1,2,3)R(1,2,3).

Find: If the centroid of triangle PQR is (α,β,γ)(\alpha, \beta, \gamma), find α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2.

Let the common value be tt. Then any point on the line is

x=8t−3,y=2t+4,z=2t−1x = 8t - 3, \quad y = 2t + 4, \quad z = 2t - 1

So a general point on the line is

(8t−3,  2t+4,  2t−1)(8t-3,\;2t+4,\;2t-1)

Since P and Q are each at distance 66 from R(1,2,3)R(1,2,3), we use the distance condition:

(8t−3−1)2+(2t+4−2)2+(2t−1−3)2=6\sqrt{(8t-3-1)^2 + (2t+4-2)^2 + (2t-1-3)^2} = 6

This becomes

(8t−4)2+(2t+2)2+(2t−4)2=6\sqrt{(8t-4)^2 + (2t+2)^2 + (2t-4)^2} = 6

Squaring both sides,

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

Expanding and simplifying,

(64t2−64t+16)+(4t2+8t+4)+(4t2−16t+16)=36(64t^2 - 64t + 16) + (4t^2 + 8t + 4) + (4t^2 - 16t + 16) = 36 72t2−72t+36=3672t^2 - 72t + 36 = 36 72t2−72t=072t^2 - 72t = 0 72t(t−1)=072t(t-1) = 0

Hence,

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

Therefore the two required points are:

  • for t=0t=0, (−3,4,−1)(-3,4,-1)
  • for t=1t=1, (5,6,1)(5,6,1)

So we take

P(−3,4,−1),Q(5,6,1),R(1,2,3)P(-3,4,-1), \quad Q(5,6,1), \quad R(1,2,3)

The centroid of triangle PQR is

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

Thus,

α=1,β=4,γ=1\alpha = 1, \quad \beta = 4, \quad \gamma = 1

Now,

α2+β2+γ2=12+42+12=18\alpha^2 + \beta^2 + \gamma^2 = 1^2 + 4^2 + 1^2 = 18

Therefore, the correct option is C, and the value of α2+β2+γ2\alpha^2 + \beta^2 + \gamma^2 is 1818.

Common mistakes

  • Using the point RR as if it lies on the given line. This is wrong because only P and Q lie on the line, while R(1,2,3)R(1,2,3) is an external point used in the distance condition. First parametrize the line, then impose the distance equation.

  • Taking P and Q to be the same point. The quadratic gives two parameter values, t=0t=0 and t=1t=1, corresponding to two distinct points on the line. These two points must both be used as the line-intersection points with the sphere centered at RR.

  • Computing the centroid of triangle PQR using only P and Q. This is incorrect because the centroid of a triangle is the average of all three vertices P, Q, and R. Include the coordinates of R(1,2,3)R(1,2,3) as well.

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