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JEE Advanced Mathematics Three Dimensional Geometry 2026 JEE Advanced 2026 (Paper 2)

JEE Advanced Mathematics Question (2026) — Solution

Question

Let L be the straight line joining the points P(1, 2, -1) and Q(2, 3, 1). Let S be the foot of the perpendicular drawn from the point R(4, -1, 5) to the line L. Another line passing through R intersects L at a point T such that the point S divides the line segment PT internally in the ratio |PS| : |ST| = 1 : 2, where |PS| and |ST| are the lengths of the line segments PS and ST, respectively. Then which of the following statements is (are) TRUE ?

Options

  1. A. The orthocentre of the triangle PRT is ( 23 5 , -4, 31 5 )
  2. B. The orthocentre of the triangle PRT is (4, 3, 5)
  3. C. The area of the triangle PRT is 6 5
  4. D. The area of the triangle PRT is 18 5

Answer

D. The area of the triangle PRT is 18 5

Step-by-step solution

The direction ratios of line L passing through P(1, 2, -1) and Q(2, 3, 1) are (2-1, 3-2, 1-(-1)) = (1, 1, 2). The equation of line L is x-1 1 = y-2 1 = z+1 2 = . Any point on L can be written as S( +1, +2, 2 -1). Since S is the foot of the perpendicular from R(4, -1, 5) to L, the direction ratios of RS are ( -3, +3, 2 -6). As RS L, the dot product of their direction ratios is zero: 1( -3) + 1( +3) + 2(2 -6) = 0 6 - 12 = 0 = 2 Thus, the coordinates of S are (3, 4, 3). The length of PS is (3-1)^2 + (4-2)^2 + (3-(-1))^2 = 4+4+16 = 2 6 . The length of the altitude RS is (3-4)^2 + (4-(-1))^2 + (3-5)^2 = 1+25+4 = 30 . Since S divides PT internally in the ratio 1:2, we have: S = T + 2P 3 T = 3S - 2P T = 3(3, 4, 3) - 2(1, 2, -1) = (7, 8, 11) The length of the base PT is 3|PS| = 6 6 . The area of PRT is 1 2 |PT| |RS| = 1 2 6 6 30 = 18 5 . The orthocentre H lies on the altitude RS. The line RS passes through S(3, 4, 3) and has direction ratios proportional to R-S = (1, -5, 2). Let the coordinates of H be (3+t, 4-5t, 3+2t). Since H is the orthocentre, PH RT. The direction ratios of RT are (7-4, 8-(-1), 11-5) = (3, 9, 6), which is proportional to (1, 3, 2). The direction ratios of PH are (3+t-1, 4-5t-2, 3+2t-(-1)) = (t+2, 2-5t, 2t+4). Taking the dot product of PH and RT: 1(t+2) + 3(2-5t) + 2(2t+4) = 0 t + 2 + 6 - 15t + 4t + 8 = 0 16 - 10t = 0 t = 8 5 Substituting t = 8 5 into the coordinates of H, we get: H = (3 + 8 5 , 4 - 5 ( 8 5 ), 3 + 2 ( 8 5 ) ) = ( 23 5 , -4, 31 5 ) Answer: The orthocentre of the triangle PRT is ( 23 5 , -4, 31 5 ); The area of the triangle PRT is 18 5

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